Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured
in the On-Line Encyclopedia Of Integer Sequences
By Shalosh B. Ekhad and Mingjia Yang
In this article we will prove, 33, theorems regarding linear recurrences of c\
ertain algebraic formal power series given in terms of
radicals
-----------------------------------------
Regarding OEIS sequence, A000957, we have the following theorem
Theorem Number, 1,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
1 - (1 - 4 x)
----------------
1/2
3 - (1 - 4 x)
In Maple format:
(1-(1-4*x)^(1/2))/(3-(1-4*x)^(1/2))
Then the sequence satisfies the linear recurrence equation
/ 7 n\
-n a(n) + |-6 + ---| a(n - 1) + (-3 + 2 n) a(n - 2) = 0
\ 2 /
subject to the initial conditions
a(1) = 1, a(2) = 0
and in Maple format
-n*a(n)+(-6+7/2*n)*a(n-1)+(-3+2*n)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2
4 x (2 + x) - 4 (2 + x) (1 + 2 x) P + 4 (2 + x) P = 0
in Maple format:
4*x*(2+x)-4*(2+x)*(1+2*x)*P+4*(2+x)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
2
2 n (n - 1) a(n) - 9 (n - 1) (n - 2) a(n - 1) + (51 - 30 n + 3 n ) a(n - 2)
+ 2 (n - 3) (-5 + 2 n) a(n - 3) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A001006, we have the following theorem
Theorem Number, 2,
: Let a(n) be the sequence whose ordinary generating function is ,
2 1/2
1 - x - (1 - 2 x - 3 x )
---------------------------
2
2 x
In Maple format:
1/2*(1-x-(1-2*x-3*x^2)^(1/2))/x^2
Then the sequence satisfies the linear recurrence equation
(-2 - n) a(n) + (1 + 2 n) a(n - 1) + (-3 + 3 n) a(n - 2) = 0
subject to the initial conditions
a(1) = 1, a(2) = 2
and in Maple format
(-2-n)*a(n)+(1+2*n)*a(n-1)+(-3+3*n)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 4 2
4 x + 4 x (-1 + x) P + 4 x P = 0
and in Maple format
4*x^2+4*x^2*(-1+x)*P+4*x^4*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A003645, we have the following theorem
Theorem Number, 3,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
1 - 4 x - (1 - 8 x)
----------------------
2
8 x
In Maple format:
1/8*(1-4*x-(1-8*x)^(1/2))/x^2
Then the sequence satisfies the linear recurrence equation
(-2 - n) a(n) + (4 + 8 n) a(n - 1) = 0
subject to the initial conditions
a(1) = 4
and in Maple format
(-2-n)*a(n)+(4+8*n)*a(n-1) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 4 2
16 x + 16 x (-1 + 4 x) P + 64 x P = 0
and in Maple format
16*x^2+16*x^2*(-1+4*x)*P+64*x^4*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A004148, we have the following theorem
Theorem Number, 4,
: Let a(n) be the sequence whose ordinary generating function is ,
2 2 3 4 1/2
1 - x + x - (1 - 2 x - x - 2 x + x )
------------------------------------------
2
2 x
In Maple format:
1/2*(1-x+x^2-(1-2*x-x^2-2*x^3+x^4)^(1/2))/x^2
Then the sequence satisfies the linear recurrence equation
(n + 2) a(n) + (-1 - 2 n) a(n - 1) + (1 - n) a(n - 2) + (5 - 2 n) a(n - 3)
+ (n - 4) a(n - 4) = 0
subject to the initial conditions
a(1) = 1, a(2) = 1, a(3) = 2, a(4) = 4
and in Maple format
(n+2)*a(n)+(-1-2*n)*a(n-1)+(1-n)*a(n-2)+(5-2*n)*a(n-3)+(n-4)*a(n-4) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2 4 2
4 x - 4 x (1 - x + x ) P + 4 x P = 0
in Maple format:
4*x^2-4*x^2*(1-x+x^2)*P+4*x^4*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
-n (n + 2) a(n) + n (2 n + 1) a(n - 1) + n (2 n - 5) a(n - 2)
+ (-12 + 6 n) a(n - 3) - (2 n - 3) (n - 4) a(n - 4)
- (n - 4) (2 n - 9) a(n - 5) + (n - 6) (n - 4) a(n - 6) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A014137, we have the following theorem
Theorem Number, 5,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
1 - (1 - 4 x)
----------------
2 x (1 - x)
In Maple format:
1/2*(1-(1-4*x)^(1/2))/x/(1-x)
Then the sequence satisfies the linear recurrence equation
(n + 1) a(n) + (-5 n + 1) a(n - 1) + (4 n - 2) a(n - 2) = 0
subject to the initial conditions
a(1) = 2, a(2) = 4
and in Maple format
(n+1)*a(n)+(-5*n+1)*a(n-1)+(4*n-2)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2
4 x + 4 x (-1 + x) P + 4 x (-1 + x) P = 0
and in Maple format
4*x+4*x*(-1+x)*P+4*x^2*(-1+x)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A014138, we have the following theorem
Theorem Number, 6,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
1 - 2 x - (1 - 4 x)
----------------------
2 x (1 - x)
In Maple format:
1/2*(1-2*x-(1-4*x)^(1/2))/x/(1-x)
Then the sequence satisfies the linear recurrence equation
(n + 1) a(n) + (-5 n + 1) a(n - 1) + (4 n - 2) a(n - 2) = 0
subject to the initial conditions
a(1) = 1, a(2) = 3
and in Maple format
(n+1)*a(n)+(-5*n+1)*a(n-1)+(4*n-2)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2 2
4 x - 4 x (2 x - 1) (-1 + x) P + 4 x (-1 + x) P = 0
and in Maple format
4*x^2-4*x*(2*x-1)*(-1+x)*P+4*x^2*(-1+x)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A023431, we have the following theorem
Theorem Number, 7,
: Let a(n) be the sequence whose ordinary generating function is ,
2 3 1/2
1 - x - (1 - 2 x + x - 4 x )
--------------------------------
3
2 x
In Maple format:
1/2*(1-x-(1-2*x+x^2-4*x^3)^(1/2))/x^3
Then the sequence satisfies the linear recurrence equation
(n + 3) a(n) + (-3 - 2 n) a(n - 1) + n a(n - 2) + (6 - 4 n) a(n - 3) = 0
subject to the initial conditions
a(1) = 1, a(2) = 1, a(3) = 2
and in Maple format
(n+3)*a(n)+(-3-2*n)*a(n-1)+n*a(n-2)+(6-4*n)*a(n-3) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
3 3 6 2
4 x + 4 x (-1 + x) P + 4 x P = 0
in Maple format:
4*x^3+4*x^3*(-1+x)*P+4*x^6*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
2
3 n (n + 3) a(n) + (4 - 7 n - 9 n) a(n - 1) + (5 n + 2) (n - 1) a(n - 2)
2
+ (-2 - 13 n + 21 n) a(n - 3) + 2 (n - 2) (2 n - 5) a(n - 4) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A030238, we have the following theorem
Theorem Number, 8,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
1 - (1 - 4 x)
--------------------------
1/2
x (2 - x + x (1 - 4 x) )
In Maple format:
(1-(1-4*x)^(1/2))/x/(2-x+x*(1-4*x)^(1/2))
Then the sequence satisfies the linear recurrence equation
(n + 1) a(n) + (1 - 5 n) a(n - 1) + (-2 + 4 n) a(n - 2) + (n + 1) a(n - 3)
+ (2 - 4 n) a(n - 4) = 0
subject to the initial conditions
a(1) = 1, a(2) = 3, a(3) = 7, a(4) = 20
and in Maple format
(n+1)*a(n)+(1-5*n)*a(n-1)+(-2+4*n)*a(n-2)+(n+1)*a(n-3)+(2-4*n)*a(n-4) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
3 2 3 2 3 2 2
4 x (1 - x + x ) + 4 x (2 x - 1) (1 - x + x ) P + 4 x (1 - x + x ) P = 0
in Maple format:
4*x*(1-x+x^3)+4*x*(2*x^2-1)*(1-x+x^3)*P+4*x^2*(1-x+x^3)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
-n (n + 1) a(n) + n (7 n - 5) a(n - 1) - 4 (5 n - 6) (n - 2) a(n - 2)
2 2
+ (288 - 215 n + 37 n ) a(n - 3) + (-240 + 150 n - 18 n ) a(n - 4)
- 2 (7 n - 8) (n - 3) a(n - 5) + 12 (2 n - 5) (n - 4) a(n - 6) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A032357, we have the following theorem
Theorem Number, 9,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
1 - (1 - 4 x)
----------------
2 x (1 + x)
In Maple format:
1/2*(1-(1-4*x)^(1/2))/x/(1+x)
Then the sequence satisfies the linear recurrence equation
(-n - 1) a(n) + (3 n - 3) a(n - 1) + (4 n - 2) a(n - 2) = 0
subject to the initial conditions
a(1) = 0, a(2) = 2
and in Maple format
(-n-1)*a(n)+(3*n-3)*a(n-1)+(4*n-2)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2
4 x - 4 x (1 + x) P + 4 x (1 + x) P = 0
and in Maple format
4*x-4*x*(1+x)*P+4*x^2*(1+x)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A049027, we have the following theorem
Theorem Number, 10,
: Let a(n) be the sequence whose ordinary generating function is ,
2
----------------
1
3 - ------------
1/2
(1 - 4 x)
In Maple format:
2/(3-1/(1-4*x)^(1/2))
Then the sequence satisfies the linear recurrence equation
/ 17 n\
n a(n) + |6 - ----| a(n - 1) + (-27 + 18 n) a(n - 2) = 0
\ 2 /
subject to the initial conditions
a(1) = 1, a(2) = 4
and in Maple format
n*a(n)+(6-17/2*n)*a(n-1)+(-27+18*n)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2
4 (-2 + 9 x) (4 x - 1) - 12 (-2 + 9 x) (4 x - 1) P + 4 (-2 + 9 x) P = 0
in Maple format:
4*(-2+9*x)*(4*x-1)-12*(-2+9*x)*(4*x-1)*P+4*(-2+9*x)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
-2 (n - 1) n a(n) + (n - 1) (25 n - 36) a(n - 1)
2
+ (-402 + 410 n - 104 n ) a(n - 2) + 72 (n - 3) (-5 + 2 n) a(n - 3) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A051960, we have the following theorem
Theorem Number, 11,
: Let a(n) be the sequence whose ordinary generating function is ,
1 + 2 x 1
---------------- - ---
1/2 2 x
2 x (1 - 4 x)
In Maple format:
1/2*(1+2*x)/x/(1-4*x)^(1/2)-1/2/x
Then the sequence satisfies the linear recurrence equation
1/4 (n + 1) (3 n - 1) a(n) - 1/2 (3 n + 2) (2 n - 1) a(n - 1) = 0
subject to the initial conditions
a(1) = 5
and in Maple format
1/4*(n+1)*(3*n-1)*a(n)-1/2*(3*n+2)*(2*n-1)*a(n-1) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2 2
4 x (x + 2) (4 x - 1) + 4 (4 x - 1) x P + 4 (4 x - 1) x P = 0
in Maple format:
4*x*(x+2)*(4*x-1)+4*(4*x-1)^2*x*P+4*(4*x-1)^2*x^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
2
n (n + 1) a(n) - 5 n a(n - 1) + 2 (n - 1) (2 n - 3) a(n - 2) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A064062, we have the following theorem
Theorem Number, 12,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
(1 - 8 x)
3/2 - ------------
2
------------------
1 + x
In Maple format:
(3/2-1/2*(1-8*x)^(1/2))/(1+x)
Then the sequence satisfies the linear recurrence equation
-n a(n) + (7 n - 12) a(n - 1) + (8 n - 12) a(n - 2) = 0
subject to the initial conditions
a(1) = 1, a(2) = 3
and in Maple format
-n*a(n)+(7*n-12)*a(n-1)+(8*n-12)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2
8 + 8 x + (-12 x - 12) P + 4 (1 + x) P = 0
and in Maple format
8+8*x+(-12*x-12)*P+4*(1+x)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A064063, we have the following theorem
Theorem Number, 13,
: Let a(n) be the sequence whose ordinary generating function is ,
6
-----------------
1/2
5 + (1 - 12 x)
In Maple format:
6/(5+(1-12*x)^(1/2))
Then the sequence satisfies the linear recurrence equation
-2 n a(n) + (23 n - 36) a(n - 1) + (12 n - 18) a(n - 2) = 0
subject to the initial conditions
a(1) = 1, a(2) = 4
and in Maple format
-2*n*a(n)+(23*n-36)*a(n-1)+(12*n-18)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2
24 + 12 x + (-40 - 20 x) P + 4 (x + 2) P = 0
and in Maple format
24+12*x+(-40-20*x)*P+4*(x+2)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A064310, we have the following theorem
Theorem Number, 14,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
(4 x + 1)
3/4 + ------------
4
------------------
1 - x/2
In Maple format:
(3/4+1/4*(4*x+1)^(1/2))/(1-1/2*x)
Then the sequence satisfies the linear recurrence equation
-2 n a(n) + (-7 n + 12) a(n - 1) + (4 n - 6) a(n - 2) = 0
subject to the initial conditions
a(1) = 1, a(2) = 0
and in Maple format
-2*n*a(n)+(-7*n+12)*a(n-1)+(4*n-6)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2
8 - 4 x + (12 x - 24) P + 4 (-2 + x) P = 0
and in Maple format
8-4*x+(12*x-24)*P+4*(-2+x)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A065096, we have the following theorem
Theorem Number, 15,
: Let a(n) be the sequence whose ordinary generating function is ,
2 1/2 2
(1 - 3 x - (1 - 6 x + x ) )
------------------------------
3
16 x
In Maple format:
1/16*(1-3*x-(1-6*x+x^2)^(1/2))^2/x^3
Then the sequence satisfies the linear recurrence equation
(n + 3) (n - 1) a(n) - 3 n (2 n + 1) a(n - 1) + (n - 1) n a(n - 2) = 0
subject to the initial conditions
a(1) = 1, a(2) = 6
and in Maple format
(n+3)*(n-1)*a(n)-3*n*(2*n+1)*a(n-1)+(n-1)*n*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
4 3 6 2
16 x - 16 x (5 x - 1) (-1 + x) P + 64 x P = 0
and in Maple format
16*x^4-16*x^3*(5*x-1)*(-1+x)*P+64*x^6*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A068875, we have the following theorem
Theorem Number, 16,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
1 - x - (1 - 4 x)
--------------------
x
In Maple format:
(1-x-(1-4*x)^(1/2))/x
Then the sequence satisfies the linear recurrence equation
(n + 1) a(n) + (2 - 4 n) a(n - 1) = 0
subject to the initial conditions
a(1) = 2
and in Maple format
(n+1)*a(n)+(2-4*n)*a(n-1) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2
x (x + 2) + 2 x (-1 + x) P + P x = 0
in Maple format:
x*(x+2)+2*x*(-1+x)*P+P^2*x^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
-n (n + 1) a(n) + 2 n (n + 1) a(n - 1) + 4 (n - 2) (2 n - 3) a(n - 2) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A072547, we have the following theorem
Theorem Number, 17,
: Let a(n) be the sequence whose ordinary generating function is ,
/ 1/2\
| (1 - 4 x) |
x |1/2 + ------------|
\ 2 /
---------------------------------
/ 1/2\
1/2 | (1 - 4 x) |
(1 - 4 x) |3/2 - ------------|
\ 2 /
In Maple format:
x*(1/2+1/2*(1-4*x)^(1/2))/(1-4*x)^(1/2)/(3/2-1/2*(1-4*x)^(1/2))
Then the sequence satisfies the linear recurrence equation
2
1/3 (n - 1) (3 n - 8) a(n) + (- 46/3 + 89/6 n - 7/2 n ) a(n - 1)
- 1/3 (3 n - 5) (-5 + 2 n) a(n - 2) = 0
subject to the initial conditions
a(1) = 1, a(2) = 0
and in Maple format
1/3*(n-1)*(3*n-8)*a(n)+(-46/3+89/6*n-7/2*n^2)*a(n-1)-1/3*(3*n-5)*(-5+2*n)*a(n-2
) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
3 2 2 2 2
x (x + 2) (4 x - 1) - 2 x (x + 2) (4 x - 1) P + (x + 2) (4 x - 1) P = 0
in Maple format:
x^3*(x+2)*(4*x-1)-2*x*(x+2)*(4*x-1)^2*P+(x+2)^2*(4*x-1)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
-2 (n - 2) (n - 1) a(n) + (3 n + 5) (n - 2) a(n - 1)
2
+ (164 - 106 n + 18 n ) a(n - 2) + 4 (n - 3) (-7 + 2 n) a(n - 3) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A076035, we have the following theorem
Theorem Number, 18,
: Let a(n) be the sequence whose ordinary generating function is ,
1
-------------------
1/2
-1 + 2 (1 - 4 x)
In Maple format:
1/(-1+2*(1-4*x)^(1/2))
Then the sequence satisfies the linear recurrence equation
3 n a(n) + (-28 n + 18) a(n - 1) + (64 n - 96) a(n - 2) = 0
subject to the initial conditions
a(1) = 4, a(2) = 20
and in Maple format
3*n*a(n)+(-28*n+18)*a(n-1)+(64*n-96)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2
-3 + 16 x + (32 x - 6) P + (-3 + 16 x) P = 0
and in Maple format
-3+16*x+(32*x-6)*P+(-3+16*x)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A090826, we have the following theorem
Theorem Number, 19,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
1 - (1 - 4 x)
----------------
2
2 - 2 x - 2 x
In Maple format:
(1-(1-4*x)^(1/2))/(2-2*x-2*x^2)
Then the sequence satisfies the linear recurrence equation
n a(n) + (-5 n + 6) a(n - 1) + (3 n - 6) a(n - 2) + (4 n - 6) a(n - 3) = 0
subject to the initial conditions
a(1) = 1, a(2) = 2, a(3) = 5
and in Maple format
n*a(n)+(-5*n+6)*a(n-1)+(3*n-6)*a(n-2)+(4*n-6)*a(n-3) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2 2
4 x + (4 x - 4 + 4 x ) P + 4 (x - 1 + x ) P = 0
and in Maple format
4*x+(4*x-4+4*x^2)*P+4*(x-1+x^2)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A102898, we have the following theorem
Theorem Number, 20,
: Let a(n) be the sequence whose ordinary generating function is ,
2 x
-------------------------
2 1/2
3 (1 - 4 x ) + 2 x - 3
In Maple format:
2*x/(3*(1-4*x^2)^(1/2)+2*x-3)
Then the sequence satisfies the linear recurrence equation
/ 4 n \ /40 n \
1/3 n a(n) - 10/9 n a(n - 1) + |- --- + 4| a(n - 2) + |---- - 40/3| a(n - 3) =
\ 3 / \ 9 /
0
subject to the initial conditions
a(1) = 3, a(2) = 9, a(3) = 30
and in Maple format
1/3*n*a(n)-10/9*n*a(n-1)+(-4/3*n+4)*a(n-2)+(40/9*n-40/3)*a(n-3) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2
4 x (10 x - 3) - 4 (10 x - 3) (2 x - 3) P + 4 (10 x - 3) P = 0
in Maple format:
4*x*(10*x-3)-4*(10*x-3)*(2*x-3)*P+4*(10*x-3)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
-3 (n - 1) n a(n) + 2 (n - 1) (14 n - 27) a(n - 1)
- 48 (n - 1) (n - 3) a(n - 2) - 8 (14 n - 41) (n - 3) a(n - 3)
+ 240 (n - 4) (n - 3) a(n - 4) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A111053, we have the following theorem
Theorem Number, 21,
: Let a(n) be the sequence whose ordinary generating function is ,
2
2 x (1 - 4 x - x )
-------------------------------------------------
2 1/2
(1 - x) (1 - 4 x - 2 x ) + (1 - 5 x) (1 - 4 x)
In Maple format:
2*x*(1-4*x-x^2)/((1-x)*(1-4*x-2*x^2)+(1-5*x)*(1-4*x)^(1/2))
Then the sequence satisfies the linear recurrence equation
2 2
n (n + 5) a(n) + (60 - 39 n - 9 n ) a(n - 1) + (-300 + 80 n + 22 n ) a(n - 2)
2
+ (120 - 33 n - 9 n ) a(n - 3) + 2 (n + 6) (-5 + 2 n) a(n - 4) = 0
subject to the initial conditions
a(1) = 1, a(2) = 2, a(3) = 6, a(4) = 22
and in Maple format
n*(n+5)*a(n)+(60-39*n-9*n^2)*a(n-1)+(-300+80*n+22*n^2)*a(n-2)+(120-33*n-9*n^2)*
a(n-3)+2*(n+6)*(-5+2*n)*a(n-4) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 3 2
4 x (x + 4 x - 1) (x - 2 x + 5 x - 1)
2 3 2
+ 4 (-1 + x) (2 x + 4 x - 1) (x - 2 x + 5 x - 1) P
3 2 2 2
+ 4 (x - 2 x + 5 x - 1) P = 0
in Maple format:
4*x*(x^2+4*x-1)*(x^3-2*x^2+5*x-1)+4*(-1+x)*(2*x^2+4*x-1)*(x^3-2*x^2+5*x-1)*P+4*
(x^3-2*x^2+5*x-1)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
n (n - 1) a(n) - (n - 1) (17 n - 30) a(n - 1) + 3 (n - 2) (37 n - 95) a(n - 2)
2 2
+ (-348 n + 2238 n - 3600) a(n - 3) + (570 n - 4872 n + 10530) a(n - 4)
2
+ (-675 n + 7641 n - 21840) a(n - 5)
2 2
+ (37560 + 818 n - 11078 n) a(n - 6) + (4330 n - 14940 - 310 n ) a(n - 7)
+ 60 (n - 8) (-13 + 2 n) a(n - 8) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A118376, we have the following theorem
Theorem Number, 22,
: Let a(n) be the sequence whose ordinary generating function is ,
2 1/2
-1 + (1 - 8 x + 8 x )
------------------------
-4 + 4 x
In Maple format:
(-1+(1-8*x+8*x^2)^(1/2))/(-4+4*x)
Then the sequence satisfies the linear recurrence equation
n a(n) + (12 - 9 n) a(n - 1) + (-36 + 16 n) a(n - 2) + (-8 n + 24) a(n - 3) = 0
subject to the initial conditions
a(1) = 1, a(2) = 2, a(3) = 6
and in Maple format
n*a(n)+(12-9*n)*a(n-1)+(-36+16*n)*a(n-2)+(-8*n+24)*a(n-3) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2
-8 x (-1 + x) + (8 x - 8) P + 16 (-1 + x) P = 0
in Maple format:
-8*x*(-1+x)+(8*x-8)*P+16*(-1+x)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
2
(n - 1) n a(n) - (n - 1) (11 n - 18) a(n - 1) + (162 - 148 n + 34 n ) a(n - 2)
- 8 (5 n - 14) (n - 3) a(n - 3) + 16 (n - 4) (n - 3) a(n - 4) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A120010, we have the following theorem
Theorem Number, 23,
: Let a(n) be the sequence whose ordinary generating function is ,
/ 2 \1/2
| 4 (x - x )|
|1 - ----------|
| 2|
\ 1 - x + x /
1/2 - -------------------
2
In Maple format:
1/2-1/2*(1-4*(x-x^2)/(1-x+x^2))^(1/2)
Then the sequence satisfies the linear recurrence equation
n a(n) + (8 - 6 n) a(n - 1) + (-26 + 11 n) a(n - 2) + (30 - 10 n) a(n - 3)
+ (5 n - 20) a(n - 4) = 0
subject to the initial conditions
a(1) = 1, a(2) = 1, a(3) = 1, a(4) = 2
and in Maple format
n*a(n)+(8-6*n)*a(n-1)+(-26+11*n)*a(n-2)+(30-10*n)*a(n-3)+(5*n-20)*a(n-4) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2
-4 x (-1 + x) + (-4 - 4 x + 4 x) P + (4 + 4 x - 4 x) P = 0
in Maple format:
-4*x*(-1+x)+(-4-4*x^2+4*x)*P+(4+4*x^2-4*x)*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
-n (n - 1) a(n) + 2 (n - 1) (4 n - 7) a(n - 1) - (n - 2) (23 n - 55) a(n - 2)
+ 4 (8 n - 21) (n - 3) a(n - 3) - 5 (5 n - 13) (n - 4) a(n - 4)
+ 10 (n - 5) (n - 3) a(n - 5) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A129775, we have the following theorem
Theorem Number, 24,
: Let a(n) be the sequence whose ordinary generating function is ,
2
2 x
1 + ------------------------------
2 1/2
-1 + 4 x - 2 x + (1 - 4 x)
In Maple format:
1+2*x^2/(-1+4*x-2*x^2+(1-4*x)^(1/2))
Then the sequence satisfies the linear recurrence equation
(-1 + n) a(n) + (15 - 9 n) a(-1 + n) + (-54 + 24 n) a(n - 2)
+ (41 - 17 n) a(n - 3) + (-10 + 4 n) a(n - 4) = 0
subject to the initial conditions
a(1) = 1, a(2) = 2, a(3) = 6, a(4) = 21
and in Maple format
(-1+n)*a(n)+(15-9*n)*a(-1+n)+(-54+24*n)*a(n-2)+(41-17*n)*a(n-3)+(-10+4*n)*a(n-4
) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 3
4 x (4 x - 1) (-1 + 5 x - 4 x + x )
2 2 3
+ 4 x (x - 2) (4 x - 1) (-1 + 5 x - 4 x + x ) P
2 2 3 2 2
+ 4 x (-1 + 5 x - 4 x + x ) P = 0
in Maple format:
4*x^2*(4*x-1)*(-1+5*x-4*x^2+x^3)+4*x^2*(x-2)*(4*x-1)*(-1+5*x-4*x^2+x^3)*P+4*x^2
*(-1+5*x-4*x^2+x^3)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
n (-1 + n) a(n) - 3 (-1 + n) (5 n - 8) a(-1 + n)
2 2
+ (-355 n + 372 + 85 n ) a(n - 2) + (1240 n - 1764 - 220 n ) a(n - 3)
2 2
+ (-1564 n + 2544 + 238 n ) a(n - 4) + (53 n + 756 - 47 n ) a(n - 5)
2
+ (-1704 + 538 n - 40 n ) a(n - 6) + 8 (2 n - 11) (n - 6) a(n - 7) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A144700, we have the following theorem
Theorem Number, 25,
: Let a(n) be the sequence whose ordinary generating function is ,
/ 1/2\4
| (1 - 4 x) |
|1/2 - ------------|
\ 2 /
---------------------
4 4
(1 - x) x
In Maple format:
(1/2-1/2*(1-4*x)^(1/2))^4/(1-x)^4/x^4
Then the sequence satisfies the linear recurrence equation
2
-n (n + 4) a(n) + 2 (n + 3) (3 n + 2) a(n - 1) + (-36 - 44 n - 9 n ) a(n - 2)
+ 2 (2 n + 7) (n + 3) a(n - 3) = 0
subject to the initial conditions
a(1) = 8, a(2) = 40, a(3) = 164
and in Maple format
-n*(n+4)*a(n)+2*(n+3)*(3*n+2)*a(n-1)+(-36-44*n-9*n^2)*a(n-2)+2*(2*n+7)*(n+3)*a(
n-3) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
4 4 2 4 8 8 2
4 x - 4 x (2 x - 4 x + 1) (-1 + x) P + 4 x (-1 + x) P = 0
and in Maple format
4*x^4-4*x^4*(2*x^2-4*x+1)*(-1+x)^4*P+4*x^8*(-1+x)^8*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A167479, we have the following theorem
Theorem Number, 26,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
1 - (1 - 4 x)
----------------
2 x (1 + 2 x)
In Maple format:
1/2*(1-(1-4*x)^(1/2))/x/(1+2*x)
Then the sequence satisfies the linear recurrence equation
(-n - 1) a(n) + (2 n - 4) a(n - 1) + (8 n - 4) a(n - 2) = 0
subject to the initial conditions
a(1) = -1, a(2) = 4
and in Maple format
(-n-1)*a(n)+(2*n-4)*a(n-1)+(8*n-4)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2
4 x - 4 x (1 + 2 x) P + 4 x (1 + 2 x) P = 0
and in Maple format
4*x-4*x*(1+2*x)*P+4*x^2*(1+2*x)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A185010, we have the following theorem
Theorem Number, 27,
: Let a(n) be the sequence whose ordinary generating function is ,
2 1/2 1/2
(2 - 8 x - 2 (1 - 8 x - 48 x ) )
-------------------------------------
8 x
In Maple format:
1/8*(2-8*x-2*(1-8*x-48*x^2)^(1/2))^(1/2)/x
Then the sequence satisfies the linear recurrence equation
-n (n + 1) a(n) + 4 n (2 n - 1) a(n - 1) + 12 (2 n - 1) (2 n - 3) a(n - 2) = 0
subject to the initial conditions
a(1) = 2, a(2) = 14
and in Maple format
-n*(n+1)*a(n)+4*n*(2*n-1)*a(n-1)+12*(2*n-1)*(2*n-3)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2 4 4
64 x + 64 (4 x - 1) x P + 1024 P x = 0
and in Maple format
64*x^2+64*(4*x-1)*x^2*P^2+1024*P^4*x^4 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A200312, we have the following theorem
Theorem Number, 28,
: Let a(n) be the sequence whose ordinary generating function is ,
2 1/2 1/2
(1 - 2 x - (1 - 4 x - 48 x ) )
-----------------------------------
26 x
In Maple format:
1/26*(1-2*x-(1-4*x-48*x^2)^(1/2))^(1/2)/x
Then the sequence satisfies the linear recurrence equation
-n (n + 1) a(n) + 2 n (2 n - 1) a(n - 1) + 12 (2 n - 1) (2 n - 3) a(n - 2) = 0
subject to the initial conditions
1/2 1/2
26 4 26
a(1) = -----, a(2) = -------
26 13
and in Maple format
-n*(n+1)*a(n)+2*n*(2*n-1)*a(n-1)+12*(2*n-1)*(2*n-3)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2 4 4
52 x + 1352 x (-1 + 2 x) P + 456976 P x = 0
and in Maple format
52*x^2+1352*x^2*(-1+2*x)*P^2+456976*P^4*x^4 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A200375, we have the following theorem
Theorem Number, 29,
: Let a(n) be the sequence whose ordinary generating function is ,
2 1/2 1/2
(2 - 4 x - 2 (1 - 4 x - 32 x ) )
-------------------------------------
6 x
In Maple format:
1/6*(2-4*x-2*(1-4*x-32*x^2)^(1/2))^(1/2)/x
Then the sequence satisfies the linear recurrence equation
-n (n + 1) a(n) + 2 n (2 n - 1) a(n - 1) + 8 (2 n - 1) (2 n - 3) a(n - 2) = 0
subject to the initial conditions
a(1) = 1, a(2) = 6
and in Maple format
-n*(n+1)*a(n)+2*n*(2*n-1)*a(n-1)+8*(2*n-1)*(2*n-3)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2 4 4
36 x + 36 x (-1 + 2 x) P + 324 P x = 0
and in Maple format
36*x^2+36*x^2*(-1+2*x)*P^2+324*P^4*x^4 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A210474, we have the following theorem
Theorem Number, 30,
: Let a(n) be the sequence whose ordinary generating function is ,
2 1/2
(1 - x) (1 + 3 x - (9 x - 10 x + 1) )
----------------------------------------
8 x
In Maple format:
1/8*(1-x)*(1+3*x-(9*x^2-10*x+1)^(1/2))/x
Then the sequence satisfies the linear recurrence equation
(n + 1) a(n) + (6 - 10 n) a(n - 1) + (-27 + 9 n) a(n - 2) = 0
subject to the initial conditions
a(1) = 0, a(2) = 4
and in Maple format
(n+1)*a(n)+(6-10*n)*a(n-1)+(-27+9*n)*a(n-2) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2
16 x (-1 + x) + 16 x (3 x + 1) (-1 + x) P + 64 P x = 0
in Maple format:
16*x*(-1+x)^2+16*x*(3*x+1)*(-1+x)*P+64*P^2*x^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
n (n + 1) a(n) - 14 (n - 1) n a(n - 1) + 2 (13 n - 14) (2 n - 5) a(n - 2)
- 6 (-25 + 11 n) (n - 4) a(n - 3) + 27 (n - 5) (n - 4) a(n - 4) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A219314, we have the following theorem
Theorem Number, 31,
: Let a(n) be the sequence whose ordinary generating function is ,
2 1/2 2
(1 + 2 x) (1 - 2 x - 3 x ) - 1 + x + 2 x
--------------------------------------------
2
2 (1 + x) (1 - 2 x - 4 x )
In Maple format:
1/2*((1+2*x)*(1-2*x-3*x^2)^(1/2)-1+x+2*x^2)/(1+x)/(1-2*x-4*x^2)
Then the sequence satisfies the linear recurrence equation
2
1/5 n (5 n - 7) a(n) + (- 24/5 + 48/5 n - 4 n ) a(n - 1)
2 2
+ (6 + 11/5 n - 3 n ) a(n - 2) + (132/5 - 238/5 n + 14 n ) a(n - 3)
+ 12/5 (5 n - 2) (n - 3) a(n - 4) = 0
subject to the initial conditions
a(1) = 1, a(2) = 0, a(3) = 3, a(4) = 3
and in Maple format
1/5*n*(5*n-7)*a(n)+(-24/5+48/5*n-4*n^2)*a(n-1)+(6+11/5*n-3*n^2)*a(n-2)+(132/5-\
238/5*n+14*n^2)*a(n-3)+12/5*(5*n-2)*(n-3)*a(n-4) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
2 2 2
4 x (1 + x) (4 x + 2 x - 1) + 4 (-1 + 2 x) (4 x + 2 x - 1) (1 + x) P
2 2 2 2
+ 4 (1 + x) (4 x + 2 x - 1) P = 0
in Maple format:
4*x*(1+x)*(4*x^2+2*x-1)+4*(-1+2*x)*(4*x^2+2*x-1)*(1+x)^2*P+4*(1+x)^2*(4*x^2+2*x
-1)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
2
-n (n - 1) a(n) + 8 (n - 1) (n - 2) a(n - 1) + (-11 n + 65 n - 78) a(n - 2)
2 2
+ (-34 n + 146 n - 148) a(n - 3) + (38 n - 262 n + 416) a(n - 4)
+ 4 (19 n - 59) (n - 4) a(n - 5) + 24 (n - 4) (n - 5) a(n - 6) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
-----------------------------------------
Regarding OEIS sequence, A228404, we have the following theorem
Theorem Number, 32,
: Let a(n) be the sequence whose ordinary generating function is ,
/ 1/2\2 / 1/2\4
| (1 - 4 x) | | (1 - 4 x) |
2 |1/2 - ------------| |1/2 - ------------|
\ 2 / \ 2 /
1 - x + ----------------------- + ---------------------
x 3
x
In Maple format:
1-x+2*(1/2-1/2*(1-4*x)^(1/2))^2/x+(1/2-1/2*(1-4*x)^(1/2))^4/x^3
Then the sequence satisfies the linear recurrence equation
2
-3 n (n + 3) a(n) + (-12 + 18 n + 18 n ) a(n - 1)
2 2
+ (44 + 78 n - 38 n ) a(n - 2) + (76 n - 372 n + 224) a(n - 3)
2
+ (-1216 - 92 n + 708 n) a(n - 4) + 24 (2 n - 9) (n - 6) a(n - 5) = 0
subject to the initial conditions
a(1) = 2, a(2) = 8
and in Maple format
-3*n*(n+3)*a(n)+(-12+18*n+18*n^2)*a(n-1)+(44+78*n-38*n^2)*a(n-2)+(76*n^2-372*n+
224)*a(n-3)+(-1216-92*n^2+708*n)*a(n-4)+24*(2*n-9)*(n-6)*a(n-5) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
3 5 3 4 2
4 x (1 - 2 x + x - 3 x + 2 x + 4 x )
3 3 2 4 6 2
+ 4 x (-1 + 4 x + 2 x - 4 x + 2 x ) P + 4 x P = 0
and in Maple format
4*x^3*(1-2*x+x^5-3*x^3+2*x^4+4*x^2)+4*x^3*(-1+4*x+2*x^3-4*x^2+2*x^4)*P+4*x^6*P^
2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which is equivalent to the stated recu\
rrence. QED
-----------------------------------------
Regarding OEIS sequence, A263986, we have the following theorem
Theorem Number, 33,
: Let a(n) be the sequence whose ordinary generating function is ,
1/2
1 - (1 - 4 x) x
---------------- - ----------
2 x 2
1 - x - x
In Maple format:
1/2*(1-(1-4*x)^(1/2))/x-x/(1-x-x^2)
Then the sequence satisfies the linear recurrence equation
/120 389 276 2 3\
1/11 (11 n - 20) (n - 3) (n + 1) a(n) + |--- - --- n + --- n - 5 n | a(n - 1)
\11 11 11 /
/ 240 375 192 2 3\
+ |- --- + --- n - --- n + 3 n | a(n - 2)
\ 11 11 11 /
+ 2/11 (11 n - 9) (2 n - 5) (n - 2) a(n - 3) = 0
subject to the initial conditions
a(1) = 0, a(2) = 1, a(3) = 3
and in Maple format
1/11*(11*n-20)*(n-3)*(n+1)*a(n)+(120/11-389/11*n+276/11*n^2-5*n^3)*a(n-1)+(-240
/11+375/11*n-192/11*n^2+3*n^3)*a(n-2)+2/11*(11*n-9)*(2*n-5)*(n-2)*a(n-3) = 0
Proof: First, by clearing radicals using procedure radtoalg, we found P(x)=F\
satisfies the algebraic equation
3 4 2 2
4 x (1 - 3 x + 4 x + x ) - 4 x (3 x + x - 1) (-1 + x + x ) P
2 2 2 2
+ 4 x (-1 + x + x ) P = 0
in Maple format:
4*x*(1-3*x+4*x^3+x^4)-4*x*(3*x^2+x-1)*(-1+x+x^2)*P+4*x^2*(-1+x+x^2)^2*P^2 = 0
Then, by Comtet's algorithm we found a linear differential operator with pol\
ynomial coefficients
annihilating the generating function, which implies the following linear re\
currence equation with polynomial coefficients
2
n (n + 1) a(n) - 9 n (n - 1) a(n - 1) + (72 - 97 n + 29 n ) a(n - 2)
2 2
+ (-336 + 232 n - 38 n ) a(n - 3) + (288 - 103 n + 5 n ) a(n - 4)
2 2
+ (27 n - 300 + 9 n ) a(n - 5) + (372 + 9 n - 117 n) a(n - 6)
+ 6 (n - 6) (2 n - 13) a(n - 7) = 0
However, this recurrence is not minimal. By using the Euclidean division alg\
orithm we proved that the stated recurrence
is also satisfied by our sequence, hence it is rigorously proved, since they\
have the same initial conditions (that we verified) QED.
This ends this automatically generated article that took, 3.180,
seconds to generate.