A336981 a(n) = (Sum_{k=0..n-1} (4290*k + 367)*3136^(n-1-k)*C(2*k, k)*T_k(14, 1)*T_k(17, 16)) / (n*C(2*n-1, n-1)), where T_k(b, c) denotes the coefficient of x^k in the expansion of (x^2 + b*x + c)^k.

367, 561274, 465761738, 347992898596, 253672374192058, 184472558346073676, 134741252587315803972, 99021561483595207492616, 73215620625604449084882202, 54432892306811842643034599356, 40662211372552333974451185020716, 30499994580401713594837984852435832

Offset: 1

Author

Zhi-Wei Sun, Aug 09 2020

Keywords

• nonn

Comments

Conjecture 1: a(n) is an integer for each n > 0. Moreover, a(n) is even for every n > 1.

Conjecture 2: Denote (4290k+367)/3136^k*C(2k,k)*T_k(14,1)*T_k(17,16) by t(k).

(i) We have Sum_{k>=0}t(k) = 5390/Pi.

(ii) For any odd prime p different from 7, we have

Sum_{k=0..p-1}t(k) == p/2*(1430*(-1/p) + 30*(3/p) - 375) (mod p^2), where (a/p) denotes the Legendre symbol.

(iii) For any prime p == 1 (mod 12) and positive integer n, the number (T(p*n)-p*T(n))/((p*n)^2*C(2k,k)) is a p-adic integer, where T(m) denotes the Sum_{k=0..m-1} t(k).

Conjecture 3. Let p > 7 be a prime and let S(p) denote the sum Sum_{k=0..p-1}C(2k,k)*T_k(14,1)*T_k(17,16).

(1) If (-15/p) = -1, then S(p) == 0 (mod p^2).

(2) If p == 1,4 (mod 15) and p = x^2 + 15*y^2 with x and y integers, then S(p) == (-1/p)*(4x^2-2p) (mod p^2).

(3) If p == 2,8 (mod 15) and p = 3x^2 + 5y^2 with x and y integers, then S(p) == (-1/p)*(2p-12x^2) (mod p^2).

See also A336982 for similar conjectures.

Examples

			a(1) = 367 since C(0,0) = T_0(14,1) = T_0(17,16) = 1.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..60
• Zhi-Wei Sun, New series for powers of Pi and related congruences, Electron. Res. Arch. 28(2020), no. 3, 1273-1342.

Crossrefs

Cf. A000796, A000984, A002426, A336982.

Programs

Maple

T := (k, b, c) -> coeff((x^2 + b*x + c)^k, x, k):
a := n -> add((4290*k + 367)*3136^(n - 1 - k)*binomial(2*k, k)*T(k, 14, 1)*T(k, 17, 16), k = 0..n-1) / (n*binomial(2*n-1, n-1)):
seq(a(n), n=1..14); # Peter Luschny, Aug 10 2020

Mathematica

T[b_,c_,0] = 1; T[b_,c_,1] = b;
T[b_,c_,n_] := T[b,c,n] = (b(2n-1)T[b,c,n-1] - (b^2-4c)(n-1)T[b,c,n-2])/n;
a[n_] := a[n] = Sum[(4290k+367)*3136^(n-1-k)*Binomial[2k,k]*T[14,1,k]*T[17,16,k],{k,0,n-1}]/(n*Binomial[2n-1,n-1]);
Table[a[n], {n,1,10}]