A336982 a(n) = (Sum_{k=0..n-1}(540*k + 137)*3136^(n-1-k)*C(2*k, k)*T_k(2, 81)*T_k(14, 81))/ (2*n*C(2*n, n)), where T_k(b, c) denotes the coefficient of x^k in the expansion of (x^2 + b*x + c)^k.

19481, 15834677, 11228057204, 8565432196217, 6307725016636484, 4757142559658418068, 3551514651027481311824, 2677076362952455673170913, 2013177974581354357341976964, 1521087748999864267161031319444, 1149516234275305699460970109062608

Offset

2,1

Comments

Conjecture 1: a(n) is an integer for each n > 1. Moreover, a(n) is odd if and only if n = 2^k + 1 for some nonnegative integer k.

Conjecture 2: Denote (540*k+137)/3136^k*C(2k,k)*T_k(2,81)*T_k(14,81) by t(k).

(i) We have Sum_{k>=0}t(k) = 98*(10+7*Sqrt(5))/(3*Pi).

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

Sum_{k=0..p-1}t(k) == p/3*(270*(-1/p) - 104*(-2/p) + 245*(-5/p)) (mod p^2), where (a/p) denotes the Legendre symbol.

(iii) For any prime p == 1,-1,9,-9 (mod 40) and positive integer n, the number (T(p*n)-p*(-1/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(2,81)*T_k(14,81).

(1) If (-30/p) = -1, then S(p) == 0 (mod p^2).

(2) If (2/p) = (p/3) = (p/5) = 1 and p = x^2 + 30*y^2 with x and y integers, then S(p) == (-1/p)*(4x^2-2p) (mod p^2).

(3) If (p/3) = 1, (2/p) = (p/5) = -1, and p = 3*x^2 + 10*y^2 with x and y integers, then S(p) == (-1/p)*(2p-12x^2) (mod p^2).

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

(5) If (p/5) = 1, (2/p) = (p/3) = -1, and p = 5*x^2 + 6*y^2 with x and y integers, then S(p) == (-1/p)*(20x^2-2p) (mod p^2).

See also A336981 for similar conjectures.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 2..60

Zhi-Wei Sun, New series for powers of Pi and related congruences, Electron. Res. Arch. 28(2020), no. 3, 1273-1342.

Example

a(2) = 19481 since (Sum_{k=0,1}(540*k+137)*3136^(1-k)*C(2k,k)*T_k(2,81)*T_k(14,81))/(2*2*C(4,2)) = (137*3136 + (540 + 137)*C(2,1)*T_1(2,81)*T_1(14,81))/(4*6) = (137*3136 + 677*2*2*14)/24 = 19481.

Maple

T := (k, b, c) -> coeff((x^2 + b*x + c)^k, x, k);
a := n -> add((540*k + 137)*3136^(n-1-k)*binomial(2*k, k)*T(k, 2, 81)*T(k, 14, 81), k = 0..n-1) / (2*n*binomial(2*n, n)):
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[(540k+137)*3136^(n-1-k)*Binomial[2k, k]*T[2, 81, k]*T[14, 81, k], {k, 0, n-1}]/(2n*Binomial[2n, n]);
Table[a[n], {n, 2, 12}]

Crossrefs

Cf. A000796, A000984, A002426, A336981.

Sequence in context: A254486 A254493 A253858 * A345574 A345575 A345831

Adjacent sequences: A336979 A336980 A336981 * A336983 A336984 A336985

Keyword

nonn

Author

Zhi-Wei Sun, Aug 09 2020

Status

approved