On computing the Mod, 2, of Many Interesting sequences
by Shalosh B. Ekhad
Theorem Number, 1, : Let A(n) be the constant term, in x, of
n
(1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 2, : Let A(n) be the constant term, in x, of
n
(1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 3, : Let A(n) be the constant term, in x, of
n
(2 + x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 4, : Let A(n) be the constant term, in x, of
n
(2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 5, : Let A(n) be the constant term, in x, of
n
(3 + x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 6, : Let A(n) be the constant term, in x, of
n
(3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 7, : Let A(n) be the constant term, in x, of
n
(1/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 8, : Let A(n) be the constant term, in x, of
n
(1/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 9, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 10, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 11, : Let A(n) be the constant term, in x, of
n
(1/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 12, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600,
40116600, 155117520
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 13, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 14, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 15, : Let A(n) be the constant term, in x, of
n
(1/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 16, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261,
212601015, 1024642875, 4950790605
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 17, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 18, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 19, : Let A(n) be the constant term, in x, of
n
(2/x + x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 20, : Let A(n) be the constant term, in x, of
n
(2/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 21, : Let A(n) be the constant term, in x, of
n
(2/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 22, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 23, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041,
148331665, 716698425, 3462260265
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 24, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 25, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 26, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 27, : Let A(n) be the constant term, in x, of
n
(2/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 28, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 29, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491,
2081258529, 14011445763, 94594402353, 640188979299
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 30, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 31, : Let A(n) be the constant term, in x, of
n
(3/x + x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 32, : Let A(n) be the constant term, in x, of
n
(3/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 33, : Let A(n) be the constant term, in x, of
n
(3/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 34, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 35, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 36, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709,
1714206781, 11255897485, 77419522675, 517370395015
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 37, : Let A(n) be the constant term, in x, of
n
(3/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 38, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 39, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 40, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344,
8981891164, 69111416792, 533463087928, 4126851588176
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 0
all the congruences classes mod, 2, show up
Theorem Number, 41, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
Theorem Number, 42, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 2
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 2, equals , 1
The congruence classes mod, 2, in the following set , {0}, never show up!
------------------------------------------
This ends this fascinating book that took, 0.079, to generate.
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On computing the Mod, 4, of Many Interesting sequences
by Shalosh B. Ekhad
Theorem Number, 1, : Let A(n) be the constant term, in x, of
n
(1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up!
Theorem Number, 2, : Let A(n) be the constant term, in x, of
n
(1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up!
Theorem Number, 3, : Let A(n) be the constant term, in x, of
n
(2 + x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 4, : Let A(n) be the constant term, in x, of
n
(2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 5, : Let A(n) be the constant term, in x, of
n
(3 + x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 6, : Let A(n) be the constant term, in x, of
n
(3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 7, : Let A(n) be the constant term, in x, of
n
(1/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {2, 3}, never show up!
Theorem Number, 8, : Let A(n) be the constant term, in x, of
n
(1/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 9, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up!
Theorem Number, 10, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 11, : Let A(n) be the constant term, in x, of
n
(1/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 12, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600,
40116600, 155117520
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 13, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 14, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 15, : Let A(n) be the constant term, in x, of
n
(1/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 16, : Let A(n) be the constant term, in x, of
n
(3 + 1/x + x)
For the record, the first 15 terms of the sequence are:
3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261,
212601015, 1024642875, 4950790605
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 17, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 18, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 19, : Let A(n) be the constant term, in x, of
n
(2/x + x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {2, 3}, never show up!
Theorem Number, 20, : Let A(n) be the constant term, in x, of
n
(2/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {2, 3}, never show up!
Theorem Number, 21, : Let A(n) be the constant term, in x, of
n
(2/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up!
Theorem Number, 22, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up!
Theorem Number, 23, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041,
148331665, 716698425, 3462260265
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up!
Theorem Number, 24, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up!
Theorem Number, 25, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 26, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 27, : Let A(n) be the constant term, in x, of
n
(2/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 28, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 29, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491,
2081258529, 14011445763, 94594402353, 640188979299
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 30, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 31, : Let A(n) be the constant term, in x, of
n
(3/x + x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 32, : Let A(n) be the constant term, in x, of
n
(3/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {2, 3}, never show up!
Theorem Number, 33, : Let A(n) be the constant term, in x, of
n
(1 + 3/x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up!
Theorem Number, 34, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 35, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2, 3}, never show up!
Theorem Number, 36, : Let A(n) be the constant term, in x, of
n
(1 + 3/x + 3 x)
For the record, the first 15 terms of the sequence are:
1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709,
1714206781, 11255897485, 77419522675, 517370395015
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 37, : Let A(n) be the constant term, in x, of
n
(3/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 38, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 39, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 40, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344,
8981891164, 69111416792, 533463087928, 4126851588176
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 4, equals , 0
The congruence classes mod, 4, in the following set , {3}, never show up!
Theorem Number, 41, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
Theorem Number, 42, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 4
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 4, equals , 1
The congruence classes mod, 4, in the following set , {0, 2}, never show up!
------------------------------------------
This ends this fascinating book that took, 0.093, to generate.
-----------------------------------------
-----------------------------------------------------
On computing the Mod, 8, of Many Interesting sequences
by Shalosh B. Ekhad
Theorem Number, 1, : Let A(n) be the constant term, in x, of
n
(1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 5, 6, 7},
never show up!
Theorem Number, 2, : Let A(n) be the constant term, in x, of
n
(1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 14, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 5, 6, 7},
never show up!
Theorem Number, 3, : Let A(n) be the constant term, in x, of
n
(2 + x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 5, 6, 7},
never show up!
Theorem Number, 4, : Let A(n) be the constant term, in x, of
n
(2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 5, 6, 7},
never show up!
Theorem Number, 5, : Let A(n) be the constant term, in x, of
n
(3 + x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 25, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 5, 6, 7},
never show up!
Theorem Number, 6, : Let A(n) be the constant term, in x, of
n
(3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 5, 6, 7},
never show up!
Theorem Number, 7, : Let A(n) be the constant term, in x, of
n
(1/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {2, 3, 5, 6, 7},
never show up!
Theorem Number, 8, : Let A(n) be the constant term, in x, of
n
(1/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {2, 3, 5, 7},
never show up!
Theorem Number, 9, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 32, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 6, 7},
never show up!
Theorem Number, 10, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 50, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 6},
never show up!
Theorem Number, 11, : Let A(n) be the constant term, in x, of
n
(1/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 5, 6, 7},
never show up!
Theorem Number, 12, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600,
40116600, 155117520
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 9, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 5, 7}, never show up!
Theorem Number, 13, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 4, 5, 6, 7},
never show up!
Theorem Number, 14, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 11, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 5, 7}, never show up!
Theorem Number, 15, : Let A(n) be the constant term, in x, of
n
(1/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 25, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 5, 6, 7},
never show up!
Theorem Number, 16, : Let A(n) be the constant term, in x, of
n
(3 + 1/x + x)
For the record, the first 15 terms of the sequence are:
3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261,
212601015, 1024642875, 4950790605
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 49, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 6},
never show up!
Theorem Number, 17, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 54, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 6},
never show up!
Theorem Number, 18, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 50, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 6},
never show up!
Theorem Number, 19, : Let A(n) be the constant term, in x, of
n
(2/x + x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {2, 3, 5, 6, 7},
never show up!
Theorem Number, 20, : Let A(n) be the constant term, in x, of
n
(2/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {2, 3, 5, 6, 7},
never show up!
Theorem Number, 21, : Let A(n) be the constant term, in x, of
n
(2/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 5, 6, 7},
never show up!
Theorem Number, 22, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 32, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 6, 7},
never show up!
Theorem Number, 23, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041,
148331665, 716698425, 3462260265
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 5, 6, 7},
never show up!
Theorem Number, 24, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 32, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 6, 7},
never show up!
Theorem Number, 25, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 4, 5, 6, 7},
never show up!
Theorem Number, 26, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 4, 5, 6, 7},
never show up!
Theorem Number, 27, : Let A(n) be the constant term, in x, of
n
(2/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 5, 6, 7},
never show up!
Theorem Number, 28, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 54, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 6},
never show up!
Theorem Number, 29, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491,
2081258529, 14011445763, 94594402353, 640188979299
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 5, 6, 7},
never show up!
Theorem Number, 30, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 54, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 6},
never show up!
Theorem Number, 31, : Let A(n) be the constant term, in x, of
n
(3/x + x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {2, 3, 5, 7},
never show up!
Theorem Number, 32, : Let A(n) be the constant term, in x, of
n
(3/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {2, 3, 5, 6, 7},
never show up!
Theorem Number, 33, : Let A(n) be the constant term, in x, of
n
(1 + 3/x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 14, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 5, 6, 7},
never show up!
Theorem Number, 34, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 50, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 6},
never show up!
Theorem Number, 35, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 32, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 3, 4, 6, 7},
never show up!
Theorem Number, 36, : Let A(n) be the constant term, in x, of
n
(1 + 3/x + 3 x)
For the record, the first 15 terms of the sequence are:
1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709,
1714206781, 11255897485, 77419522675, 517370395015
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 49, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 6},
never show up!
Theorem Number, 37, : Let A(n) be the constant term, in x, of
n
(3/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 5, 6, 7},
never show up!
Theorem Number, 38, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 11, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 5, 7}, never show up!
Theorem Number, 39, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 4, 5, 6, 7},
never show up!
Theorem Number, 40, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344,
8981891164, 69111416792, 533463087928, 4126851588176
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 9, states .
For example, A(100000), mudolo , 8, equals , 0
The congruence classes mod, 8, in the following set , {3, 5, 7}, never show up!
Theorem Number, 41, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 50, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 6},
never show up!
Theorem Number, 42, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 8
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 54, states .
For example, A(100000), mudolo , 8, equals , 1
The congruence classes mod, 8, in the following set , {0, 2, 4, 6},
never show up!
------------------------------------------
This ends this fascinating book that took, 0.232, to generate.
-----------------------------------------
-----------------------------------------------------
On computing the Mod, 16, of Many Interesting sequences
by Shalosh B. Ekhad
Theorem Number, 1, : Let A(n) be the constant term, in x, of
n
(1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 11, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 2, : Let A(n) be the constant term, in x, of
n
(1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 52, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 3, : Let A(n) be the constant term, in x, of
n
(2 + x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 21, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 4, : Let A(n) be the constant term, in x, of
n
(2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 21, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 5, : Let A(n) be the constant term, in x, of
n
(3 + x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 181, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15}, never show up!
Theorem Number, 6, : Let A(n) be the constant term, in x, of
n
(3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 17, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15}, never show up!
Theorem Number, 7, : Let A(n) be the constant term, in x, of
n
(1/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{2, 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 8, : Let A(n) be the constant term, in x, of
n
(1/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 29, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{2, 3, 5, 7, 9, 10, 11, 13, 14, 15}, never show up!
Theorem Number, 9, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 155, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 14, 15}, never show up!
Theorem Number, 10, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 366, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 6, 8, 10, 12, 14}, never show up!
Theorem Number, 11, : Let A(n) be the constant term, in x, of
n
(1/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 21, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 12, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600,
40116600, 155117520
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 34, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 5, 7, 9, 10, 11, 13, 14, 15}, never show up!
Theorem Number, 13, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 19, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 14, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 41, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 5, 7, 9, 11, 13, 14, 15}, never show up!
Theorem Number, 15, : Let A(n) be the constant term, in x, of
n
(1/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 181, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15}, never show up!
Theorem Number, 16, : Let A(n) be the constant term, in x, of
n
(3 + 1/x + x)
For the record, the first 15 terms of the sequence are:
3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261,
212601015, 1024642875, 4950790605
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 357, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 6, 8, 10, 12, 14}, never show up!
Theorem Number, 17, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 198, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14}, never show up!
Theorem Number, 18, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 366, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 6, 8, 10, 12, 14}, never show up!
Theorem Number, 19, : Let A(n) be the constant term, in x, of
n
(2/x + x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{2, 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 20, : Let A(n) be the constant term, in x, of
n
(2/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15}, never show up!
Theorem Number, 21, : Let A(n) be the constant term, in x, of
n
(2/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 11, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 22, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 155, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 14, 15}, never show up!
Theorem Number, 23, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041,
148331665, 716698425, 3462260265
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 12, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 24, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 155, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 14, 15}, never show up!
Theorem Number, 25, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 19, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 26, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 17, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 27, : Let A(n) be the constant term, in x, of
n
(2/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 17, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15}, never show up!
Theorem Number, 28, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 198, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14}, never show up!
Theorem Number, 29, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491,
2081258529, 14011445763, 94594402353, 640188979299
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 15, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 30, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 198, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 6, 8, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 31, : Let A(n) be the constant term, in x, of
n
(3/x + x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 29, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{2, 3, 5, 7, 9, 10, 11, 13, 14, 15}, never show up!
Theorem Number, 32, : Let A(n) be the constant term, in x, of
n
(3/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15}, never show up!
Theorem Number, 33, : Let A(n) be the constant term, in x, of
n
(1 + 3/x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 52, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 34, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 366, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 6, 8, 10, 12, 14}, never show up!
Theorem Number, 35, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 155, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 14, 15}, never show up!
Theorem Number, 36, : Let A(n) be the constant term, in x, of
n
(1 + 3/x + 3 x)
For the record, the first 15 terms of the sequence are:
1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709,
1714206781, 11255897485, 77419522675, 517370395015
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 357, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 6, 8, 10, 12, 14}, never show up!
Theorem Number, 37, : Let A(n) be the constant term, in x, of
n
(3/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 21, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 38, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 41, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 5, 7, 9, 11, 13, 14, 15}, never show up!
Theorem Number, 39, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 17, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15}, never show up!
Theorem Number, 40, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344,
8981891164, 69111416792, 533463087928, 4126851588176
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 36, states .
For example, A(100000), mudolo , 16, equals , 0
The congruence classes mod, 16, in the following set ,
{3, 5, 7, 9, 10, 11, 13, 14, 15}, never show up!
Theorem Number, 41, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 366, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 6, 8, 10, 12, 14}, never show up!
Theorem Number, 42, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 16
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 198, states .
For example, A(100000), mudolo , 16, equals , 1
The congruence classes mod, 16, in the following set ,
{0, 2, 4, 6, 8, 9, 10, 11, 12, 13, 14, 15}, never show up!
------------------------------------------
This ends this fascinating book that took, 1.920, to generate.
-----------------------------------------
-----------------------------------------------------
On computing the Mod, 3, of Many Interesting sequences
by Shalosh B. Ekhad
Theorem Number, 1, : Let A(n) be the constant term, in x, of
n
(1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0, 2}, never show up!
Theorem Number, 2, : Let A(n) be the constant term, in x, of
n
(1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0, 2}, never show up!
Theorem Number, 3, : Let A(n) be the constant term, in x, of
n
(2 + x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 4, : Let A(n) be the constant term, in x, of
n
(2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 5, : Let A(n) be the constant term, in x, of
n
(3 + x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 6, : Let A(n) be the constant term, in x, of
n
(3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 7, : Let A(n) be the constant term, in x, of
n
(1/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 8, : Let A(n) be the constant term, in x, of
n
(1/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 9, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 2
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 10, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0, 2}, never show up!
Theorem Number, 11, : Let A(n) be the constant term, in x, of
n
(1/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 12, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600,
40116600, 155117520
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 0
all the congruences classes mod, 3, show up
Theorem Number, 13, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 2
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 14, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 15, : Let A(n) be the constant term, in x, of
n
(1/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 16, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261,
212601015, 1024642875, 4950790605
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 0
all the congruences classes mod, 3, show up
Theorem Number, 17, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 18, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 19, : Let A(n) be the constant term, in x, of
n
(2/x + x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 20, : Let A(n) be the constant term, in x, of
n
(2/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 21, : Let A(n) be the constant term, in x, of
n
(2/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0, 2}, never show up!
Theorem Number, 22, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 2
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 23, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041,
148331665, 716698425, 3462260265
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 24, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0, 2}, never show up!
Theorem Number, 25, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 2
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 26, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 27, : Let A(n) be the constant term, in x, of
n
(2/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 28, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 29, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491,
2081258529, 14011445763, 94594402353, 640188979299
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 0
all the congruences classes mod, 3, show up
Theorem Number, 30, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 31, : Let A(n) be the constant term, in x, of
n
(3/x + x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 32, : Let A(n) be the constant term, in x, of
n
(3/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 33, : Let A(n) be the constant term, in x, of
n
(3/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0, 2}, never show up!
Theorem Number, 34, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0, 2}, never show up!
Theorem Number, 35, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0, 2}, never show up!
Theorem Number, 36, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709,
1714206781, 11255897485, 77419522675, 517370395015
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0, 2}, never show up!
Theorem Number, 37, : Let A(n) be the constant term, in x, of
n
(3/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 38, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 39, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 40, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344,
8981891164, 69111416792, 533463087928, 4126851588176
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 3, equals , 1
The congruence classes mod, 3, in the following set , {0}, never show up!
Theorem Number, 41, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
Theorem Number, 42, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 3
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 3, equals , 0
The congruence classes mod, 3, in the following set , {2}, never show up!
------------------------------------------
This ends this fascinating book that took, 0.087, to generate.
-----------------------------------------
-----------------------------------------------------
On computing the Mod, 9, of Many Interesting sequences
by Shalosh B. Ekhad
Theorem Number, 1, : Let A(n) be the constant term, in x, of
n
(1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 9, equals , 1
The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 2, : Let A(n) be the constant term, in x, of
n
(1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 9, equals , 1
The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 3, : Let A(n) be the constant term, in x, of
n
(2 + x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 31, states .
For example, A(100000), mudolo , 9, equals , 7
The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up!
Theorem Number, 4, : Let A(n) be the constant term, in x, of
n
(2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 9, equals , 7
The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up!
Theorem Number, 5, : Let A(n) be the constant term, in x, of
n
(3 + x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 6, : Let A(n) be the constant term, in x, of
n
(3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 7, : Let A(n) be the constant term, in x, of
n
(1/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 14, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 3, 5, 8},
never show up!
Theorem Number, 8, : Let A(n) be the constant term, in x, of
n
(1/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 3, 4, 5, 7, 8},
never show up!
Theorem Number, 9, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 33, states .
For example, A(100000), mudolo , 9, equals , 2
The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up!
Theorem Number, 10, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 13, states .
For example, A(100000), mudolo , 9, equals , 1
The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 8},
never show up!
Theorem Number, 11, : Let A(n) be the constant term, in x, of
n
(1/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 31, states .
For example, A(100000), mudolo , 9, equals , 7
The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up!
Theorem Number, 12, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600,
40116600, 155117520
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 23, states .
For example, A(100000), mudolo , 9, equals , 0
all the congruences classes mod, 9, show up
Theorem Number, 13, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 33, states .
For example, A(100000), mudolo , 9, equals , 5
The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up!
Theorem Number, 14, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 23, states .
For example, A(100000), mudolo , 9, equals , 7
The congruence classes mod, 9, in the following set , {0, 3, 4, 5, 6},
never show up!
Theorem Number, 15, : Let A(n) be the constant term, in x, of
n
(1/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 16, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261,
212601015, 1024642875, 4950790605
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 27, states .
For example, A(100000), mudolo , 9, equals , 0
all the congruences classes mod, 9, show up
Theorem Number, 17, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 18, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 5, 8}, never show up!
Theorem Number, 18, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 4, 5, 7, 8},
never show up!
Theorem Number, 19, : Let A(n) be the constant term, in x, of
n
(2/x + x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 14, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 3, 5, 8},
never show up!
Theorem Number, 20, : Let A(n) be the constant term, in x, of
n
(2/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 21, : Let A(n) be the constant term, in x, of
n
(2/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 8, states .
For example, A(100000), mudolo , 9, equals , 1
The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 22, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 33, states .
For example, A(100000), mudolo , 9, equals , 2
The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up!
Theorem Number, 23, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041,
148331665, 716698425, 3462260265
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 16, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 5, 8}, never show up!
Theorem Number, 24, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 13, states .
For example, A(100000), mudolo , 9, equals , 1
The congruence classes mod, 9, in the following set , {0, 2, 3, 5, 6, 7, 8},
never show up!
Theorem Number, 25, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 33, states .
For example, A(100000), mudolo , 9, equals , 5
The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up!
Theorem Number, 26, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 23, states .
For example, A(100000), mudolo , 9, equals , 7
The congruence classes mod, 9, in the following set , {0, 3, 4, 5, 6},
never show up!
Theorem Number, 27, : Let A(n) be the constant term, in x, of
n
(2/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 28, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 18, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 5, 8}, never show up!
Theorem Number, 29, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491,
2081258529, 14011445763, 94594402353, 640188979299
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 27, states .
For example, A(100000), mudolo , 9, equals , 0
all the congruences classes mod, 9, show up
Theorem Number, 30, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 31, : Let A(n) be the constant term, in x, of
n
(3/x + x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 3, 4, 5, 7, 8},
never show up!
Theorem Number, 32, : Let A(n) be the constant term, in x, of
n
(3/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 33, : Let A(n) be the constant term, in x, of
n
(3/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 9, equals , 1
The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 34, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 13, states .
For example, A(100000), mudolo , 9, equals , 1
The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 8},
never show up!
Theorem Number, 35, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 13, states .
For example, A(100000), mudolo , 9, equals , 1
The congruence classes mod, 9, in the following set , {0, 2, 3, 5, 6, 7, 8},
never show up!
Theorem Number, 36, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709,
1714206781, 11255897485, 77419522675, 517370395015
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 9, equals , 1
The congruence classes mod, 9, in the following set , {0, 2, 3, 4, 5, 6, 7, 8},
never show up!
Theorem Number, 37, : Let A(n) be the constant term, in x, of
n
(3/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 9, equals , 7
The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up!
Theorem Number, 38, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 23, states .
For example, A(100000), mudolo , 9, equals , 7
The congruence classes mod, 9, in the following set , {0, 3, 4, 5, 6},
never show up!
Theorem Number, 39, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 23, states .
For example, A(100000), mudolo , 9, equals , 7
The congruence classes mod, 9, in the following set , {0, 3, 4, 5, 6},
never show up!
Theorem Number, 40, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344,
8981891164, 69111416792, 533463087928, 4126851588176
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 9, equals , 7
The congruence classes mod, 9, in the following set , {0, 3, 6}, never show up!
Theorem Number, 41, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 4, 5, 7, 8},
never show up!
Theorem Number, 42, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 9
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 3, states .
For example, A(100000), mudolo , 9, equals , 0
The congruence classes mod, 9, in the following set , {2, 4, 5, 6, 7, 8},
never show up!
------------------------------------------
This ends this fascinating book that took, 0.198, to generate.
-----------------------------------------
-----------------------------------------------------
On computing the Mod, 27, of Many Interesting sequences
by Shalosh B. Ekhad
Theorem Number, 1, : Let A(n) be the constant term, in x, of
n
(1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 62, states .
For example, A(100000), mudolo , 27, equals , 1
The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 2, : Let A(n) be the constant term, in x, of
n
(1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 27, equals , 1
The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 3, : Let A(n) be the constant term, in x, of
n
(2 + x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 691, states .
For example, A(100000), mudolo , 27, equals , 25
The congruence classes mod, 27, in the following set ,
{0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up!
Theorem Number, 4, : Let A(n) be the constant term, in x, of
n
(2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 22, states .
For example, A(100000), mudolo , 27, equals , 25
The congruence classes mod, 27, in the following set ,
{0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up!
Theorem Number, 5, : Let A(n) be the constant term, in x, of
n
(3 + x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 6, : Let A(n) be the constant term, in x, of
n
(3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 7, : Let A(n) be the constant term, in x, of
n
(1/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 173, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set ,
{2, 3, 5, 8, 11, 12, 14, 17, 20, 21, 23, 26}, never show up!
Theorem Number, 8, : Let A(n) be the constant term, in x, of
n
(1/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 3, 4, 5, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 9, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 714, states .
For example, A(100000), mudolo , 27, equals , 20
The congruence classes mod, 27, in the following set ,
{0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up!
Theorem Number, 10, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 170, states .
For example, A(100000), mudolo , 27, equals , 1
The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 8, 9,
11, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26}, never show up!
Theorem Number, 11, : Let A(n) be the constant term, in x, of
n
(1/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 691, states .
For example, A(100000), mudolo , 27, equals , 25
The congruence classes mod, 27, in the following set ,
{0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up!
Theorem Number, 12, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600,
40116600, 155117520
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 280, states .
For example, A(100000), mudolo , 27, equals , 0
all the congruences classes mod, 27, show up
Theorem Number, 13, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 714, states .
For example, A(100000), mudolo , 27, equals , 14
The congruence classes mod, 27, in the following set ,
{0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up!
Theorem Number, 14, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 322, states .
For example, A(100000), mudolo , 27, equals , 25
The congruence classes mod, 27, in the following set ,
{0, 3, 4, 5, 6, 7, 9, 11, 12, 13, 14, 15, 16, 18, 20, 21, 22, 23, 24},
never show up!
Theorem Number, 15, : Let A(n) be the constant term, in x, of
n
(1/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 16, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261,
212601015, 1024642875, 4950790605
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 351, states .
For example, A(100000), mudolo , 27, equals , 0
all the congruences classes mod, 27, show up
Theorem Number, 17, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 240, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set ,
{2, 5, 8, 11, 14, 17, 20, 23, 26}, never show up!
Theorem Number, 18, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up!
Theorem Number, 19, : Let A(n) be the constant term, in x, of
n
(2/x + x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 173, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set ,
{2, 3, 5, 8, 11, 12, 14, 17, 20, 21, 23, 26}, never show up!
Theorem Number, 20, : Let A(n) be the constant term, in x, of
n
(2/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 21, : Let A(n) be the constant term, in x, of
n
(2/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 62, states .
For example, A(100000), mudolo , 27, equals , 1
The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 22, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 714, states .
For example, A(100000), mudolo , 27, equals , 20
The congruence classes mod, 27, in the following set ,
{0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up!
Theorem Number, 23, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041,
148331665, 716698425, 3462260265
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 180, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set ,
{2, 5, 8, 11, 14, 17, 20, 23, 26}, never show up!
Theorem Number, 24, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 164, states .
For example, A(100000), mudolo , 27, equals , 1
The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 11, 12, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26}, never show up!
Theorem Number, 25, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 714, states .
For example, A(100000), mudolo , 27, equals , 14
The congruence classes mod, 27, in the following set ,
{0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up!
Theorem Number, 26, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 322, states .
For example, A(100000), mudolo , 27, equals , 25
The congruence classes mod, 27, in the following set ,
{0, 3, 4, 5, 6, 8, 9, 10, 12, 13, 14, 15, 17, 18, 19, 21, 22, 23, 24},
never show up!
Theorem Number, 27, : Let A(n) be the constant term, in x, of
n
(2/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 28, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 240, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set ,
{2, 5, 8, 11, 14, 17, 20, 23, 26}, never show up!
Theorem Number, 29, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491,
2081258529, 14011445763, 94594402353, 640188979299
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 354, states .
For example, A(100000), mudolo , 27, equals , 0
all the congruences classes mod, 27, show up
Theorem Number, 30, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26}, never show up!
Theorem Number, 31, : Let A(n) be the constant term, in x, of
n
(3/x + x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 3, 4, 5, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 32, : Let A(n) be the constant term, in x, of
n
(3/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 33, : Let A(n) be the constant term, in x, of
n
(3/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 5, states .
For example, A(100000), mudolo , 27, equals , 1
The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 34, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 170, states .
For example, A(100000), mudolo , 27, equals , 1
The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 8, 9,
11, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26}, never show up!
Theorem Number, 35, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 164, states .
For example, A(100000), mudolo , 27, equals , 1
The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 11, 12, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26}, never show up!
Theorem Number, 36, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709,
1714206781, 11255897485, 77419522675, 517370395015
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 6, states .
For example, A(100000), mudolo , 27, equals , 1
The congruence classes mod, 27, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26},
never show up!
Theorem Number, 37, : Let A(n) be the constant term, in x, of
n
(3/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 22, states .
For example, A(100000), mudolo , 27, equals , 25
The congruence classes mod, 27, in the following set ,
{0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up!
Theorem Number, 38, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 322, states .
For example, A(100000), mudolo , 27, equals , 25
The congruence classes mod, 27, in the following set ,
{0, 3, 4, 5, 6, 7, 9, 11, 12, 13, 14, 15, 16, 18, 20, 21, 22, 23, 24},
never show up!
Theorem Number, 39, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 322, states .
For example, A(100000), mudolo , 27, equals , 25
The congruence classes mod, 27, in the following set ,
{0, 3, 4, 5, 6, 8, 9, 10, 12, 13, 14, 15, 17, 18, 19, 21, 22, 23, 24},
never show up!
Theorem Number, 40, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344,
8981891164, 69111416792, 533463087928, 4126851588176
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 22, states .
For example, A(100000), mudolo , 27, equals , 25
The congruence classes mod, 27, in the following set ,
{0, 3, 6, 9, 12, 15, 18, 21, 24}, never show up!
Theorem Number, 41, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, never show up!
Theorem Number, 42, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 27
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 7, states .
For example, A(100000), mudolo , 27, equals , 0
The congruence classes mod, 27, in the following set , {2, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26}, never show up!
------------------------------------------
This ends this fascinating book that took, 8.294, to generate.
-----------------------------------------
-----------------------------------------------------
On computing the Mod, 5, of Many Interesting sequences
by Shalosh B. Ekhad
Theorem Number, 1, : Let A(n) be the constant term, in x, of
n
(1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0, 2, 3, 4},
never show up!
Theorem Number, 2, : Let A(n) be the constant term, in x, of
n
(1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0, 2, 3, 4},
never show up!
Theorem Number, 3, : Let A(n) be the constant term, in x, of
n
(2 + x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 4, : Let A(n) be the constant term, in x, of
n
(2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 5, : Let A(n) be the constant term, in x, of
n
(3 + x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 6, : Let A(n) be the constant term, in x, of
n
(3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 7, : Let A(n) be the constant term, in x, of
n
(1/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 5, equals , 0
The congruence classes mod, 5, in the following set , {2, 3}, never show up!
Theorem Number, 8, : Let A(n) be the constant term, in x, of
n
(1/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 5, equals , 0
The congruence classes mod, 5, in the following set , {2, 3}, never show up!
Theorem Number, 9, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 0
all the congruences classes mod, 5, show up
Theorem Number, 10, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 2
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 11, : Let A(n) be the constant term, in x, of
n
(1/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 12, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600,
40116600, 155117520
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 4
all the congruences classes mod, 5, show up
Theorem Number, 13, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 2
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 14, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 0
all the congruences classes mod, 5, show up
Theorem Number, 15, : Let A(n) be the constant term, in x, of
n
(1/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 16, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261,
212601015, 1024642875, 4950790605
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 4
all the congruences classes mod, 5, show up
Theorem Number, 17, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 2
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 18, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 0
all the congruences classes mod, 5, show up
Theorem Number, 19, : Let A(n) be the constant term, in x, of
n
(2/x + x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 5, equals , 0
The congruence classes mod, 5, in the following set , {2, 3}, never show up!
Theorem Number, 20, : Let A(n) be the constant term, in x, of
n
(2/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 0
all the congruences classes mod, 5, show up
Theorem Number, 21, : Let A(n) be the constant term, in x, of
n
(2/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0, 2, 3, 4},
never show up!
Theorem Number, 22, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 0
all the congruences classes mod, 5, show up
Theorem Number, 23, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041,
148331665, 716698425, 3462260265
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 5, equals , 4
The congruence classes mod, 5, in the following set , {2, 3}, never show up!
Theorem Number, 24, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 3
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 25, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 2
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 26, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 4
all the congruences classes mod, 5, show up
Theorem Number, 27, : Let A(n) be the constant term, in x, of
n
(2/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 28, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 2
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 29, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491,
2081258529, 14011445763, 94594402353, 640188979299
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 3
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 30, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 4
all the congruences classes mod, 5, show up
Theorem Number, 31, : Let A(n) be the constant term, in x, of
n
(3/x + x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 5, equals , 0
The congruence classes mod, 5, in the following set , {2, 3}, never show up!
Theorem Number, 32, : Let A(n) be the constant term, in x, of
n
(3/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 0
all the congruences classes mod, 5, show up
Theorem Number, 33, : Let A(n) be the constant term, in x, of
n
(3/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 1, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0, 2, 3, 4},
never show up!
Theorem Number, 34, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 2
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 35, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 3
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 36, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709,
1714206781, 11255897485, 77419522675, 517370395015
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 2, states .
For example, A(100000), mudolo , 5, equals , 4
The congruence classes mod, 5, in the following set , {2, 3}, never show up!
Theorem Number, 37, : Let A(n) be the constant term, in x, of
n
(3/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 1
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 38, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 0
all the congruences classes mod, 5, show up
Theorem Number, 39, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 4
all the congruences classes mod, 5, show up
Theorem Number, 40, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344,
8981891164, 69111416792, 533463087928, 4126851588176
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 3
The congruence classes mod, 5, in the following set , {0}, never show up!
Theorem Number, 41, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 0
all the congruences classes mod, 5, show up
Theorem Number, 42, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 5
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 4, states .
For example, A(100000), mudolo , 5, equals , 4
all the congruences classes mod, 5, show up
------------------------------------------
This ends this fascinating book that took, 0.105, to generate.
-----------------------------------------
-----------------------------------------------------
On computing the Mod, 25, of Many Interesting sequences
by Shalosh B. Ekhad
Theorem Number, 1, : Let A(n) be the constant term, in x, of
n
(1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 18, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24},
never show up!
Theorem Number, 2, : Let A(n) be the constant term, in x, of
n
(1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 14, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24},
never show up!
Theorem Number, 3, : Let A(n) be the constant term, in x, of
n
(2 + x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 4, : Let A(n) be the constant term, in x, of
n
(2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 5, : Let A(n) be the constant term, in x, of
n
(3 + x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 261, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 6, : Let A(n) be the constant term, in x, of
n
(3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 7, : Let A(n) be the constant term, in x, of
n
(1/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 67, states .
For example, A(100000), mudolo , 25, equals , 15
The congruence classes mod, 25, in the following set ,
{2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up!
Theorem Number, 8, : Let A(n) be the constant term, in x, of
n
(1/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 67, states .
For example, A(100000), mudolo , 25, equals , 15
The congruence classes mod, 25, in the following set ,
{2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up!
Theorem Number, 9, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 224, states .
For example, A(100000), mudolo , 25, equals , 0
all the congruences classes mod, 25, show up
Theorem Number, 10, : Let A(n) be the constant term, in x, of
n
(1/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 17
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 11, : Let A(n) be the constant term, in x, of
n
(1/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 12, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600,
40116600, 155117520
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 113, states .
For example, A(100000), mudolo , 25, equals , 9
all the congruences classes mod, 25, show up
Theorem Number, 13, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 17
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 14, : Let A(n) be the constant term, in x, of
n
(1/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 221, states .
For example, A(100000), mudolo , 25, equals , 0
all the congruences classes mod, 25, show up
Theorem Number, 15, : Let A(n) be the constant term, in x, of
n
(1/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 261, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 16, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 11, 45, 195, 873, 3989, 18483, 86515, 408105, 1936881, 9238023, 44241261,
212601015, 1024642875, 4950790605
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 117, states .
For example, A(100000), mudolo , 25, equals , 9
all the congruences classes mod, 25, show up
Theorem Number, 17, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 17
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 18, : Let A(n) be the constant term, in x, of
n
(1/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 221, states .
For example, A(100000), mudolo , 25, equals , 0
all the congruences classes mod, 25, show up
Theorem Number, 19, : Let A(n) be the constant term, in x, of
n
(2/x + x)
For the record, the first 15 terms of the sequence are:
0, 4, 0, 24, 0, 160, 0, 1120, 0, 8064, 0, 59136, 0, 439296, 0
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 67, states .
For example, A(100000), mudolo , 25, equals , 15
The congruence classes mod, 25, in the following set ,
{2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up!
Theorem Number, 20, : Let A(n) be the constant term, in x, of
n
(2/x + 3 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 117, states .
For example, A(100000), mudolo , 25, equals , 15
all the congruences classes mod, 25, show up
Theorem Number, 21, : Let A(n) be the constant term, in x, of
n
(2/x + 1)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 18, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24},
never show up!
Theorem Number, 22, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 5, 13, 49, 161, 581, 2045, 7393, 26689, 97285, 355565, 1305745, 4808545,
17760965, 65753693
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 224, states .
For example, A(100000), mudolo , 25, equals , 0
all the congruences classes mod, 25, show up
Theorem Number, 23, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 9, 25, 145, 561, 2841, 12489, 60705, 281185, 1353769, 6418809, 30917041,
148331665, 716698425, 3462260265
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 67, states .
For example, A(100000), mudolo , 25, equals , 9
The congruence classes mod, 25, in the following set ,
{2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up!
Theorem Number, 24, : Let A(n) be the constant term, in x, of
n
(2/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 13
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 25, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904,
110690816, 515483648, 2406449152
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 17
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 26, : Let A(n) be the constant term, in x, of
n
(2/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 117, states .
For example, A(100000), mudolo , 25, equals , 9
all the congruences classes mod, 25, show up
Theorem Number, 27, : Let A(n) be the constant term, in x, of
n
(2/x + 3)
For the record, the first 15 terms of the sequence are:
3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323,
4782969, 14348907
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 28, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719,
251595969, 1409933619, 7923848253, 44642381823
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 17
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 29, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 17, 99, 609, 3843, 24689, 160611, 1054657, 6975747, 46406097, 310171491,
2081258529, 14011445763, 94594402353, 640188979299
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 261, states .
For example, A(100000), mudolo , 25, equals , 13
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 30, : Let A(n) be the constant term, in x, of
n
(2/x + 3 + 3 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 117, states .
For example, A(100000), mudolo , 25, equals , 9
all the congruences classes mod, 25, show up
Theorem Number, 31, : Let A(n) be the constant term, in x, of
n
(3/x + x)
For the record, the first 15 terms of the sequence are:
0, 6, 0, 54, 0, 540, 0, 5670, 0, 61236, 0, 673596, 0, 7505784, 0
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 67, states .
For example, A(100000), mudolo , 25, equals , 15
The congruence classes mod, 25, in the following set ,
{2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up!
Theorem Number, 32, : Let A(n) be the constant term, in x, of
n
(3/x + 2 x)
For the record, the first 15 terms of the sequence are:
0, 12, 0, 216, 0, 4320, 0, 90720, 0, 1959552, 0, 43110144, 0, 960740352, 0
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 117, states .
For example, A(100000), mudolo , 25, equals , 15
all the congruences classes mod, 25, show up
Theorem Number, 33, : Let A(n) be the constant term, in x, of
n
(1 + 3/x)
For the record, the first 15 terms of the sequence are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 14, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24},
never show up!
Theorem Number, 34, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + x)
For the record, the first 15 terms of the sequence are:
1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909,
34533253, 148803487, 642228139
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 17
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 35, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 2 x)
For the record, the first 15 terms of the sequence are:
1, 13, 37, 289, 1201, 7741, 38053, 227137, 1207009, 6995053, 38591653,
221446369, 1245188881, 7130897437, 40516456357
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 13
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 36, : Let A(n) be the constant term, in x, of
n
(3/x + 1 + 3 x)
For the record, the first 15 terms of the sequence are:
1, 19, 55, 595, 2611, 22141, 119449, 902035, 5420035, 38712169, 246360709,
1714206781, 11255897485, 77419522675, 517370395015
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 67, states .
For example, A(100000), mudolo , 25, equals , 9
The congruence classes mod, 25, in the following set ,
{2, 3, 7, 8, 12, 13, 17, 18, 22, 23}, never show up!
Theorem Number, 37, : Let A(n) be the constant term, in x, of
n
(3/x + 2)
For the record, the first 15 terms of the sequence are:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 1
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 38, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + x)
For the record, the first 15 terms of the sequence are:
2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220,
530479640, 2795917960, 14771797424
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 221, states .
For example, A(100000), mudolo , 25, equals , 0
all the congruences classes mod, 25, show up
Theorem Number, 39, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 2 x)
For the record, the first 15 terms of the sequence are:
2, 16, 80, 520, 3152, 20224, 129152, 838240, 5462720, 35846656, 236191232,
1562588416, 10370408960, 69019648000, 460456939520
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 117, states .
For example, A(100000), mudolo , 25, equals , 9
all the congruences classes mod, 25, show up
Theorem Number, 40, : Let A(n) be the constant term, in x, of
n
(3/x + 2 + 3 x)
For the record, the first 15 terms of the sequence are:
2, 22, 116, 934, 6332, 48124, 352424, 2669062, 20107628, 153277972, 1170192344,
8981891164, 69111416792, 533463087928, 4126851588176
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 265, states .
For example, A(100000), mudolo , 25, equals , 13
The congruence classes mod, 25, in the following set , {0, 5, 10, 15, 20},
never show up!
Theorem Number, 41, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + x)
For the record, the first 15 terms of the sequence are:
3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155,
827418645, 5143397535, 32063180535, 200367960201
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 221, states .
For example, A(100000), mudolo , 25, equals , 0
all the congruences classes mod, 25, show up
Theorem Number, 42, : Let A(n) be the constant term, in x, of
n
(3/x + 3 + 2 x)
For the record, the first 15 terms of the sequence are:
3, 21, 135, 945, 6723, 48789, 358263, 2655585, 19825155, 148853781, 1122869223,
8503237521, 64604559555, 492221474325, 3759348384855
We are interested in A(n) modulo , 25
Then there is a Congruence Tree scheme to compute it in linear-time (in bit-\
size, i.e. log-time in n)
with , 117, states .
For example, A(100000), mudolo , 25, equals , 9
all the congruences classes mod, 25, show up
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This ends this fascinating book that took, 8.567, to generate.
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The whole thing took, 19.611, seconds.