Ramanujan-Style Congruences for primes up to, 15
for the eta-function raised up to the power, 3
By Shalosh B. Ekhad
-----------------------------------------------------
Theorem Number, 1
(5 n + 4)
For each non-neg. integer n, the coefficients of, q , in
infinity
--------'
' | | 1
| | ------
| | i
| | 1 - q
i = 1
is divisible by , 5
Proof: This theorem is not that deep and
a Ramanujan-style proof suffices.
Let :
infinity
--------'
' | | i
E(q) = | | (1 - q )
| |
| |
i = 1
Thanks to Jacobi, we have
infinity /(n + 1) n\
----- |---------|
3 \ n \ 2 /
E(q) = ) (-1) (2 n + 1) q
/
-----
n = 0
by considering all the residue classes modulo, 5
(n + 1) n
of , ---------, such that, 2 n + 1, is not 0 mod, 5
2
we easily see that only the residue classes, [0, 1], show up
so we can write:
3
E(q) = J[0] + J[1]
where J[i] consists of those terms in which the power of q
is congruent to i mod, 5
Now modulo, 5
1 3
----, equals , E(q) , to the power , 3
E(q)
5
divided by, E(q ), to the power, 2
that equals
3
(J[0] + J[1])
5
times a formal power series in powers of, q
if we expand, and extract the powers that are congruent to
4, modulo , 5
none of them show up!, QED!
-----------------------------------------------------
Theorem Number, 2
(7 n + 5)
For each non-neg. integer n, the coefficients of, q , in
infinity
--------'
' | | 1
| | ------
| | i
| | 1 - q
i = 1
is divisible by , 7
Proof: This theorem is not that deep and
a Ramanujan-style proof suffices.
Let :
infinity
--------'
' | | i
E(q) = | | (1 - q )
| |
| |
i = 1
Thanks to Jacobi, we have
infinity /(n + 1) n\
----- |---------|
3 \ n \ 2 /
E(q) = ) (-1) (2 n + 1) q
/
-----
n = 0
by considering all the residue classes modulo, 7
(n + 1) n
of , ---------, such that, 2 n + 1, is not 0 mod, 7
2
we easily see that only the residue classes, [0, 1, 3], show up
so we can write:
3
E(q) = J[0] + J[1] + J[3]
where J[i] consists of those terms in which the power of q
is congruent to i mod, 7
Now modulo, 7
1 3
----, equals , E(q) , to the power , 2
E(q)
7
divided by, E(q ), to the power, 1
that equals
2
(J[0] + J[1] + J[3])
7
times a formal power series in powers of, q
if we expand, and extract the powers that are congruent to
5, modulo , 7
none of them show up!, QED!
-----------------------------------------------------
Theorem Number, 3
(11 n + 6)
For each non-neg. integer n, the coefficients of, q , in
infinity
--------'
' | | 1
| | ------
| | i
| | 1 - q
i = 1
is divisible by , 11
Proof: We will give a Hirschhorn-style proof.
Let :
infinity
--------'
' | | i
E(q) = | | (1 - q ), that equals, thanks to Euler to
| |
| |
i = 1
infinity /(3 n + 1) n\
----- |-----------|
\ n \ 2 /
) (-1) q
/
-----
n = -infinity
by considering all the residue classes modulo, 11
(3 n + 1) n
of , -----------
2
we easily see that only the residue classes, {0, 1, 2, 4, 5, 7}, show up
So we can write
E(q) = E[0] + E[1] + E[2] + E[4] + E[5] + E[7]
where E[i] consists of those terms in which the power of q
is congruent to i mod, 11
Thanks to Jacobi, we have
infinity /(n + 1) n\
----- |---------|
3 \ n \ 2 /
E(q) = ) (-1) (2 n + 1) q
/
-----
n = 0
by considering all the residue classes modulo, 11
(n + 1) n
of , ---------, such that, 2 n + 1, is not 0 mod, 11
2
we easily see that only the residue classes, [0, 1, 3, 6, 10], show up
so we can write:
3
E(q) = J[0] + J[1] + J[3] + J[6] + J[10]
where J[i] consists of those terms in which the power of q
is congruent to i mod, 11
Now
3
E(q) , to the power, 4, equals
12
E(q)
11
that equals E(q) times, E(q ), to the power, 1
Hence, modulo , 11
4
(J[0] + J[1] + J[3] + J[6] + J[10]) =
11
E(q ) (E[0] + E[1] + E[2] + E[4] + E[5] + E[7])
If we consider the powers of q congruent to, 3, 6, 8, 9, 10
we find that, modulo, 11
3 3 2 2
4 J[0] J[3] + 4 J[0] J[1] + 6 J[1] J[6] + 2 J[0] J[1] J[3] J[10]
2 3
+ J[3] J[6] J[10] + 4 J[6] J[10] = 0
3 2 2 3
4 J[0] J[6] + 6 J[0] J[3] + 4 J[1] J[3] + 2 J[0] J[1] J[6] J[10]
2 3
+ J[1] J[3] J[10] + 4 J[6] J[10] = 0
2 2 2 3 3
6 J[1] J[3] + J[0] J[1] J[6] + 4 J[1] J[6] + 4 J[3] J[10]
3
+ 2 J[0] J[3] J[6] J[10] + 4 J[0] J[10] = 0
3 3 2 2 2
4 J[0] J[3] + 4 J[1] J[6] + J[0] J[3] J[6] + 6 J[0] J[10]
3
+ 2 J[1] J[3] J[6] J[10] + 4 J[1] J[10] = 0
3 3 3
4 J[0] J[10] + 4 J[1] J[3] + 2 J[0] J[1] J[3] J[6] + 4 J[3] J[6]
2 2 2
+ J[0] J[1] J[10] + 6 J[6] J[10] = 0
Now modulo, 11
1 3
----, equals , E(q) , to the power , 7
E(q)
11
divided by, E(q ), to the power, 2
that equals
7
(J[0] + J[1] + J[3] + J[6] + J[10])
11
times a formal power series in powers of, q
if we expand, and extract the powers that are congruent to
6, modulo , 11
11
we get a formal power series in , q , times
a polynomial,P, of (total) degree, 7, in , [J[0], J[1], J[3], J[6], J[10]]
where P equals
2 2 2 4 3 3
3 J[0] J[1] J[3] J[6] + J[0] J[1] J[3] J[10] + 8 J[0] J[1] J[3]
3 3 3 3 2 3
+ 8 J[0] J[6] J[10] + 8 J[1] J[3] J[6] + 2 J[0] J[1] J[3] J[6]
2 2 2 2 2 2
+ 3 J[0] J[1] J[6] J[10] + 3 J[0] J[3] J[6] J[10]
2 2 2 4 4
+ 3 J[0] J[1] J[3] J[10] + J[0] J[1] J[3] J[6] + J[0] J[3] J[6] J[10]
3 2 2 3
+ 2 J[1] J[3] J[6] J[10] + 2 J[0] J[3] J[6] J[10]
3 2 2 2 2
+ 2 J[0] J[1] J[6] J[10] + 3 J[1] J[3] J[6] J[10]
4 6 5 2 6
+ J[1] J[3] J[6] J[10] + 7 J[0] J[6] + 10 J[0] J[3] + 7 J[0] J[1]
5 2 2 5 6 6
+ 10 J[1] J[6] + 10 J[1] J[3] + 7 J[3] J[10] + 7 J[3] J[6]
2 5 5 2 6 3 3
+ 10 J[0] J[10] + 10 J[6] J[10] + 7 J[1] J[10] + 8 J[0] J[3] J[10]
3 3 4 3 2
+ 8 J[1] J[6] J[10] + J[0] J[1] J[6] J[10] + 2 J[0] J[1] J[3] J[10]
Now it is easily verified that, modulo, 11, P equals
3 2 3 3
(10 J[1] + 4 J[0] J[3] + 10 J[1] J[3] J[10]) (4 J[0] J[3] + 4 J[0] J[1]
2 2 2
+ 6 J[1] J[6] + 2 J[0] J[1] J[3] J[10] + J[3] J[6] J[10]
3 3 2
+ 4 J[6] J[10] ) + (10 J[0] + 10 J[10] J[0] J[1] + 4 J[10] J[6] ) (
3 2 2 3
4 J[0] J[6] + 6 J[0] J[3] + 4 J[1] J[3] + 2 J[0] J[1] J[6] J[10]
2 3
+ J[1] J[3] J[10] + 4 J[6] J[10]) +
3 2 2 2
(10 J[3] + 10 J[0] J[3] J[6] + 4 J[0] J[10] ) (6 J[1] J[3]
2 3 3
+ J[0] J[1] J[6] + 4 J[1] J[6] + 4 J[3] J[10] + 2 J[0] J[3] J[6] J[10]
3 2 3
+ 4 J[0] J[10] ) + (4 J[1] J[6] + 10 J[3] J[6] J[10] + 10 J[10] ) (
3 3 2 2 2
4 J[0] J[3] + 4 J[1] J[6] + J[0] J[3] J[6] + 6 J[0] J[10]
3
+ 2 J[1] J[3] J[6] J[10] + 4 J[1] J[10] ) +
2 3 3 3
(4 J[1] J[3] + 10 J[0] J[1] J[6] + 10 J[6] ) (4 J[0] J[10] + 4 J[1] J[3]
3 2 2 2
+ 2 J[0] J[1] J[3] J[6] + 4 J[3] J[6] + J[0] J[1] J[10] + 6 J[6] J[10]
)
that is 0 mod , 11, QED!
-----------------------------------------------------
Theorem Number, 4
(5 n + 2)
For each non-neg. integer n, the coefficients of, q , in
infinity
--------'
' | | 1
| | ---------
| | i 2
| | (1 - q )
i = 1
is divisible by , 5
Proof: This theorem is not that deep and
a Ramanujan-style proof suffices.
Let :
infinity
--------'
' | | i
E(q) = | | (1 - q )
| |
| |
i = 1
Thanks to Jacobi, we have
infinity /(n + 1) n\
----- |---------|
3 \ n \ 2 /
E(q) = ) (-1) (2 n + 1) q
/
-----
n = 0
by considering all the residue classes modulo, 5
(n + 1) n
of , ---------, such that, 2 n + 1, is not 0 mod, 5
2
we easily see that only the residue classes, [0, 1], show up
so we can write:
3
E(q) = J[0] + J[1]
where J[i] consists of those terms in which the power of q
is congruent to i mod, 5
Now modulo, 5
1 3
-----, equals , E(q) , to the power , 1
2
E(q)
5
divided by, E(q ), to the power, 1
that equals
J[0] + J[1]
5
times a formal power series in powers of, q
if we expand, and extract the powers that are congruent to
2, modulo , 5
none of them show up!, QED!
-----------------------------------------------------
Theorem Number, 5
(5 n + 3)
For each non-neg. integer n, the coefficients of, q , in
infinity
--------'
' | | 1
| | ---------
| | i 2
| | (1 - q )
i = 1
is divisible by , 5
Proof: This theorem is not that deep and
a Ramanujan-style proof suffices.
Let :
infinity
--------'
' | | i
E(q) = | | (1 - q )
| |
| |
i = 1
Thanks to Jacobi, we have
infinity /(n + 1) n\
----- |---------|
3 \ n \ 2 /
E(q) = ) (-1) (2 n + 1) q
/
-----
n = 0
by considering all the residue classes modulo, 5
(n + 1) n
of , ---------, such that, 2 n + 1, is not 0 mod, 5
2
we easily see that only the residue classes, [0, 1], show up
so we can write:
3
E(q) = J[0] + J[1]
where J[i] consists of those terms in which the power of q
is congruent to i mod, 5
Now modulo, 5
1 3
-----, equals , E(q) , to the power , 1
2
E(q)
5
divided by, E(q ), to the power, 1
that equals
J[0] + J[1]
5
times a formal power series in powers of, q
if we expand, and extract the powers that are congruent to
3, modulo , 5
none of them show up!, QED!
-----------------------------------------------------
Theorem Number, 6
(5 n + 4)
For each non-neg. integer n, the coefficients of, q , in
infinity
--------'
' | | 1
| | ---------
| | i 2
| | (1 - q )
i = 1
is divisible by , 5
Proof: This theorem is not that deep and
a Ramanujan-style proof suffices.
Let :
infinity
--------'
' | | i
E(q) = | | (1 - q )
| |
| |
i = 1
Thanks to Jacobi, we have
infinity /(n + 1) n\
----- |---------|
3 \ n \ 2 /
E(q) = ) (-1) (2 n + 1) q
/
-----
n = 0
by considering all the residue classes modulo, 5
(n + 1) n
of , ---------, such that, 2 n + 1, is not 0 mod, 5
2
we easily see that only the residue classes, [0, 1], show up
so we can write:
3
E(q) = J[0] + J[1]
where J[i] consists of those terms in which the power of q
is congruent to i mod, 5
Now modulo, 5
1 3
-----, equals , E(q) , to the power , 1
2
E(q)
5
divided by, E(q ), to the power, 1
that equals
J[0] + J[1]
5
times a formal power series in powers of, q
if we expand, and extract the powers that are congruent to
4, modulo , 5
none of them show up!, QED!
-----------------------------------------------------
Theorem Number, 7
(11 n + 7)
For each non-neg. integer n, the coefficients of, q , in
infinity
--------'
' | | 1
| | ---------
| | i 3
| | (1 - q )
i = 1
is divisible by , 11
Proof: We will give a Hirschhorn-style proof.
Let :
infinity
--------'
' | | i
E(q) = | | (1 - q ), that equals, thanks to Euler to
| |
| |
i = 1
infinity /(3 n + 1) n\
----- |-----------|
\ n \ 2 /
) (-1) q
/
-----
n = -infinity
by considering all the residue classes modulo, 11
(3 n + 1) n
of , -----------
2
we easily see that only the residue classes, {0, 1, 2, 4, 5, 7}, show up
So we can write
E(q) = E[0] + E[1] + E[2] + E[4] + E[5] + E[7]
where E[i] consists of those terms in which the power of q
is congruent to i mod, 11
Thanks to Jacobi, we have
infinity /(n + 1) n\
----- |---------|
3 \ n \ 2 /
E(q) = ) (-1) (2 n + 1) q
/
-----
n = 0
by considering all the residue classes modulo, 11
(n + 1) n
of , ---------, such that, 2 n + 1, is not 0 mod, 11
2
we easily see that only the residue classes, [0, 1, 3, 6, 10], show up
so we can write:
3
E(q) = J[0] + J[1] + J[3] + J[6] + J[10]
where J[i] consists of those terms in which the power of q
is congruent to i mod, 11
Now
3
E(q) , to the power, 4, equals
12
E(q)
11
that equals E(q) times, E(q ), to the power, 1
Hence, modulo , 11
4
(J[0] + J[1] + J[3] + J[6] + J[10]) =
11
E(q ) (E[0] + E[1] + E[2] + E[4] + E[5] + E[7])
If we consider the powers of q congruent to, 3, 6, 8, 9, 10
we find that, modulo, 11
3 3 2 2
4 J[0] J[3] + 4 J[0] J[1] + 2 J[0] J[1] J[3] J[10] + 6 J[1] J[6]
2 3
+ J[3] J[6] J[10] + 4 J[6] J[10] = 0
3 2 2 3 2
4 J[0] J[6] + 6 J[0] J[3] + 4 J[1] J[3] + J[1] J[3] J[10]
3
+ 2 J[0] J[1] J[6] J[10] + 4 J[6] J[10] = 0
2 2 2 3
J[0] J[1] J[6] + 6 J[1] J[3] + 2 J[0] J[3] J[6] J[10] + 4 J[1] J[6]
3 3
+ 4 J[3] J[10] + 4 J[0] J[10] = 0
2 3 3
J[0] J[3] J[6] + 4 J[0] J[3] + 4 J[1] J[6] + 2 J[1] J[3] J[6] J[10]
2 2 3
+ 6 J[0] J[10] + 4 J[1] J[10] = 0
3 3 2
2 J[0] J[1] J[3] J[6] + 4 J[0] J[10] + 4 J[1] J[3] + J[0] J[1] J[10]
3 2 2
+ 4 J[3] J[6] + 6 J[6] J[10] = 0
Now modulo, 11
1 3
-----, equals , E(q) , to the power , 10
3
E(q)
11
divided by, E(q ), to the power, 3
that equals
10
(J[0] + J[1] + J[3] + J[6] + J[10])
11
times a formal power series in powers of, q
if we expand, and extract the powers that are congruent to
7, modulo , 11
11
we get a formal power series in , q , times
a polynomial,P, of (total) degree, 10, in , [J[0], J[1], J[3], J[6], J[10]]
where P equals
8 2 7 5 4
2 J[1] J[3] J[10] + 8 J[0] J[3] J[10] + 6 J[3] J[6] J[10]
3 2 5 2 2 6 4 5
+ J[0] J[6] J[10] + 6 J[0] J[1] J[10] + 6 J[0] J[3] J[10]
7 2 2 5 3 2 6 2
+ 8 J[0] J[6] J[10] + J[1] J[6] J[10] + 6 J[3] J[6] J[10]
3 4 2 6
+ 10 J[0] J[1] J[3] J[6] J[10] + J[0] J[1] J[6] J[10]
2 6 2 2 5
+ J[1] J[3] J[6] J[10] + 3 J[0] J[3] J[6] J[10]
3 5 3 2 4
+ 2 J[1] J[3] J[6] J[10] + 5 J[1] J[3] J[6] J[10]
3 3 3 2 5 2
+ 3 J[0] J[3] J[6] J[10] + 3 J[0] J[1] J[6] J[10]
2 3 2 3 3 3 3
+ 10 J[0] J[3] J[6] J[10] + 3 J[0] J[1] J[6] J[10]
2 3 4 5 3
+ 5 J[0] J[1] J[3] J[10] + 2 J[0] J[3] J[6] J[10]
3 2 2 3 2 6
+ 10 J[1] J[3] J[6] J[10] + J[1] J[3] J[6] J[10]
4 3 2 3 7 4 6
+ 5 J[1] J[3] J[6] J[10] + 10 J[1] J[10] + J[6] J[10]
7 3 3 7 6 4 4 6
+ 10 J[3] J[10] + 10 J[3] J[6] + J[0] J[10] + J[1] J[6]
7 3 4 6 6 4 2 6
+ 10 J[0] J[6] + J[0] J[3] + J[1] J[3] + J[0] J[1] J[3] J[6]
2 4 3 4 2 3
+ 5 J[0] J[3] J[6] J[10] + 5 J[0] J[1] J[3] J[10]
3 5 6 2
+ 2 J[0] J[1] J[3] J[6] + J[0] J[3] J[6] J[10]
5 2 2 3 3 3
+ 3 J[1] J[3] J[6] J[10] + 3 J[1] J[3] J[6] J[10]
3 2 3 2 3 3 3
+ 10 J[0] J[1] J[6] J[10] + 3 J[0] J[1] J[3] J[10]
3 2 4 2 4 3
+ 5 J[0] J[3] J[6] J[10] + 5 J[0] J[1] J[6] J[10]
6 2 5 3
+ J[0] J[3] J[6] J[10] + 2 J[0] J[1] J[6] J[10]
4 3 2 3 3 2 2
+ 5 J[0] J[1] J[6] J[10] + 10 J[0] J[1] J[3] J[10]
3 2 4 2 2 3 3
+ 5 J[0] J[1] J[3] J[6] + 10 J[0] J[1] J[3] J[6]
2 6 6 2
+ J[0] J[1] J[3] J[10] + J[0] J[1] J[3] J[10]
2 5 2 6 2
+ 3 J[0] J[1] J[3] J[6] + J[0] J[1] J[6] J[10]
5 2 2 4 3 2
+ 3 J[0] J[1] J[3] J[10] + 5 J[0] J[1] J[3] J[6]
3 5 3 3 3
+ 2 J[0] J[1] J[3] J[10] + 3 J[0] J[1] J[3] J[6]
6 2 2 2 4
+ J[0] J[1] J[3] J[6] + 4 J[0] J[1] J[3] J[6] J[10]
2 4 2 4 3
+ 4 J[0] J[1] J[3] J[6] J[10] + 10 J[0] J[1] J[3] J[6] J[10]
2 2 2 2 2 3 4
+ J[0] J[1] J[3] J[6] J[10] + 10 J[0] J[1] J[3] J[6] J[10]
2 4 2 4 2 2
+ 4 J[0] J[1] J[3] J[6] J[10] + 4 J[0] J[1] J[3] J[6] J[10]
3 4 2 4 2
+ 10 J[0] J[1] J[3] J[6] J[10] + 4 J[0] J[1] J[3] J[6] J[10]
4 3 4 5 8
+ 10 J[0] J[1] J[3] J[6] J[10] + 6 J[0] J[1] J[10] + 2 J[0] J[3] J[6]
5 4 4 5 2 6 2
+ 6 J[1] J[6] J[10] + 6 J[0] J[6] J[10] + 6 J[1] J[3] J[10]
5 4 8 5 3 2
+ 6 J[1] J[3] J[6] + 2 J[3] J[6] J[10] + J[0] J[3] J[10]
5 4 7 2 5 2 3
+ 6 J[0] J[1] J[6] + 8 J[1] J[6] J[10] + J[1] J[3] J[6]
4 5 2 7 6 2 2
+ 6 J[1] J[3] J[10] + 8 J[1] J[3] J[6] + 6 J[0] J[3] J[6]
5 4 2 6 2 2 3 5
+ 6 J[0] J[3] J[6] + 6 J[0] J[1] J[6] + J[0] J[1] J[3]
8 8 7 2
+ 2 J[0] J[1] J[10] + 2 J[0] J[1] J[6] + 8 J[0] J[1] J[3]
5 4 3 7
+ 6 J[0] J[1] J[3] + 10 J[0] J[1]
Now it is easily verified that, modulo, 11, P equals
2 4 4 5 3 2
(8 J[0] J[1] + 4 J[0] J[1] J[3] + J[0] J[3] + 3 J[6] J[1] J[3]
2 3 2 3 2 3
+ J[6] J[0] J[3] + 8 J[6] J[0] J[1] + 8 J[6] J[0] J[3]
2 4 5 2 2
+ 4 J[1] J[6] + 6 J[10] J[1] + 9 J[10] J[0] J[1] J[3]
3 2
+ 4 J[10] J[1] J[3] J[6] + 4 J[10] J[0] J[1] J[3] J[6]
4 2 3 2 2 2
+ 9 J[10] J[3] J[6] + 7 J[10] J[1] J[3] + 5 J[10] J[0] J[3]
3 3 3 3 3
+ 3 J[10] J[0] J[1] J[6] + 3 J[10] J[6] ) (4 J[0] J[3] + 4 J[0] J[1]
2 2 2
+ 2 J[0] J[1] J[3] J[10] + 6 J[1] J[6] + J[3] J[6] J[10]
3 5 3 3 2 4
+ 4 J[6] J[10] ) + (6 J[0] J[1] + 6 J[1] J[3] + 9 J[0] J[3]
3 3 2 4 2
+ 7 J[6] J[0] J[1] J[3] + 4 J[6] J[0] J[3] + 8 J[0] J[6]
3 2 4 2
+ 2 J[10] J[0] J[1] + 8 J[10] J[1] J[3] + 8 J[10] J[0] J[1] J[3] J[6]
2 2 2 3 4
+ 4 J[10] J[0] J[1] J[6] + J[10] J[3] J[6] + 4 J[10] J[0] J[6]
2 3 2 3 2 2 2
+ 5 J[10] J[0] J[1] + 10 J[10] J[0] J[3] + J[10] J[1] J[6]
3 3 2 2 3
+ 6 J[10] J[0] J[1] J[3]) (4 J[0] J[6] + 6 J[0] J[3] + 4 J[1] J[3]
2 3
+ J[1] J[3] J[10] + 2 J[0] J[1] J[6] J[10] + 4 J[6] J[10]) + (
4 2 2 3 4
8 J[1] J[3] + 7 J[0] J[1] J[3] + 2 J[6] J[0] J[1]
3 5 3 2
+ 5 J[6] J[0] J[1] J[3] + 6 J[6] J[3] + 8 J[0] J[3] J[6]
3 3 3 2 5 2 3
+ 9 J[6] J[1] + 3 J[6] J[0] J[3] + 3 J[10] J[0] + 4 J[10] J[1] J[3]
2 2 3 2 4
+ 9 J[10] J[0] J[1] J[3] J[6] + 7 J[10] J[0] J[1] + 8 J[10] J[3]
2 2 2 2 2 3 2
+ 6 J[10] J[0] J[3] J[6] + 3 J[10] J[0] J[6] + 6 J[10] J[0] J[1]
4 2 2 2
+ 4 J[0] J[3] J[10] ) (J[0] J[1] J[6] + 6 J[1] J[3]
3 3 3
+ 2 J[0] J[3] J[6] J[10] + 4 J[1] J[6] + 4 J[3] J[10] + 4 J[0] J[10] )
3 2 3 3 2 3
+ (J[0] J[1] J[3] + 5 J[0] J[3] + 3 J[6] J[0] J[1]
4 2 2 2 2 3
+ 9 J[6] J[0] J[3] + 3 J[1] J[3] J[6] + 6 J[0] J[1] J[6]
5 3 4
+ 10 J[1] J[6] + 6 J[10] J[0] J[1] J[3] + 4 J[10] J[6] J[1]
2 4 5 2 3
+ 8 J[1] J[10] + 6 J[3] J[10] ) (J[0] J[3] J[6] + 4 J[0] J[3]
3 2 2 3
+ 4 J[1] J[6] + 2 J[1] J[3] J[6] J[10] + 6 J[0] J[10] + 4 J[1] J[10] )
5 4 2 4 5
+ (7 J[1] J[3] + 4 J[6] J[1] J[3] + 8 J[3] J[6] + 6 J[0] J[6] ) (
3 3 2
2 J[0] J[1] J[3] J[6] + 4 J[0] J[10] + 4 J[1] J[3] + J[0] J[1] J[10]
3 2 2
+ 4 J[3] J[6] + 6 J[6] J[10] )
that is 0 mod , 11, QED!
-----------------------------------------------------
We discovered, 7, interesting congruences
We were able to prove all of them!
We found, 5, Ramanujan-style
relatively easy, proofs, and, 2, deeper Hirschhorn-style
proofs
To sum up the following 3-element list of lists
consisting of pairs [a,[r,p]] for which
the statement that for all n>=0, the coeffcient of, q^(n*p+r)
in the power series expansion of
Product(1/(1-q^i)^a,i=1..infinity)
is divisible by p is:
(i) probably true, but we were unable to prove it (ii) true and we found a s\
imple Ramanujan-style proof
(iii) true and we found a more complicated Hirschhorn-style proof
[[], [[1, [4, 5]], [1, [5, 7]], [2, [2, 5]], [2, [3, 5]], [2, [4, 5]]], [[1, [6
, 11]], [3, [7, 11]]]]
This finishes this gripping article, that took, 3.155, seconds
to produce.