A336982 -id:A336982 - cp's OEIS frontend

A336981 a(n) = (Sum_{k=0..n-1} (4290k + 367)3136^(n-1-k)C(2k, k)T_k(14, 1)T_k(17, 16)) / (nC(2n-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

Zhi-Wei Sun, Aug 09 2020

nonn

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.

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

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.

Cf. A000796, A000984, A002426, A336982.

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

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}]

A337247 a(n) = (Sum_{k=0..n-1} (-1)^k * (4k+1) * 160^(n-1-k) * C(2k,k) * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C(2j,j) * (-20)^(k-j)) / (n * C(2n,n)).

Original entry on oeis.org

25, 809, 23020, 730325, 27867142, 1117643720, 42658771456, 1558395721085, 57260792702050, 2179584653311070, 84835851591609400, 3292250198848240760, 126379831667243976400, 4841030410501144484000, 186842197443136622824960, 7269291788529191112814925, 283472902036823148786161530

Offset: 2

Zhi-Wei Sun, Aug 20 2020

nonn

Conjecture 1: a(n) is a positive 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: The infinite series Sum_{k>=0} (4*k+1)/(-160)^k * C(2k,k) * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C(2j,j) * (-20)^(k-j) has the value sqrt(30)/(5*Pi)*(5+c^(1/3))/c^(1/6), where c = 145 + 30*sqrt(6).
Conjecture 3. Let p be an odd prime different from 5, and let S(p) denote the sum Sum_{k=0..p-1} C(2k,k)/(-160)^k * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C(2j,j) * (-20)^(k-j).
(i) If p == 1,3 (mod 8) and p = x^2 + 2*y^2 with x and y integers, then S(p) == (5/p)*(4x^2 - 2p) (mod p^2), where (5/p) is the Legendre symbol.
(ii) If p == 5,7 (mod 8), then S(p) == 0 (mod p^2).

			a(2) = (160 - (4 + 1)*C(2,1)*(-20 + C(3,2)*C(2,1)))/(2*C(4,2)) = 300/12 = 25.

Zhi-Wei Sun, Table of n, a(n) for n = 2..100
Zhi-Wei Sun, Two curious series for 1/Pi, Question 369569 at MathOverflow, August 19-20, 2020.
Zhi-Wei Sun, New series for powers of Pi and related congruences, Electron. Res. Arch. 28(2020), no. 3, 1273-1342.
Zhi-Wei Sun, Some new series for 1/Pi motivated by congruences, arXiv:2009.04379 [math.NT], 2020.

Cf. A000796, A000984, A336981, A336982.

a[n_]:=a[n]=Sum[(4k+1)(-1)^k*160^(n-1-k)*Binomial[2k,k]*Sum[Binomial[k,j]Binomial[k+2j,2j]Binomial[2j,j](-20)^(k-j),{j,0,k}],{k,0,n-1}]/(n*Binomial[2n,n])
Table[a[n],{n,2,18}]

A337332 a(n) = Sum_{k=0..n}C(n,k)C(n+k,k)C(2k,k)C(2n-2k,n-k)(-8)^(n-k).

Original entry on oeis.org

1, -12, 228, -3504, 44580, -298032, 1407504, -275772096, 21324125988, -966349948080, 32198201397648, -831808446595776, 16275197594916624, -210881419152530112, 1110165241205298240, -28746364298042321664, 4877709692143697517348, -323151109677783574203312, 13976671241536620108719376

Offset: 0

Zhi-Wei Sun, Aug 23 2020

sign

(-1)^n*a(n) > 0, and Sum_{k=0..n} C(n,k)*C(n+k,k)*C(2k,k)*C(2n-2k,n-k)*(-1)^(n-k) = Sum_{k=0..n}C(n,k)^4.
Conjecture 1: Sum_{k>=0}(4k+1) a(k)/(-48)^k = sqrt(72+42*sqrt(3))/Pi.
Conjecture 2: For each n > 0, the number (Sum_{k=0..n-1} (-1)^k*(4k+1)*48^(n-1-k)*a(k))/n is a positive integer.
Conjecture 3: For any prime p > 3, the square of (Sum_{k=0..p-1} (4k+1)a(k)/(-48)^k)/p is congruent to 14*(3/p)-(p/3)-12 modulo p, where (a/p) is the Legendre symbol.
Conjecture 4: Let p > 3 be a prime, and let S(p) = Sum_{k=0..p-1} a(k)/(-48)^k. If p == 1 (mod 4) and p = x^2 + 4y^2 with x and y integers, then S(p) == 4x^2-2p (mod p^2). If p == 3 (mod 4), then S(p) == 0 (mod p^2).

			a(1) = C(1,0)*C(1,0)*C(0,0)*C(2,1)*(-8) + C(1,1)*C(2,1)*C(2,1)*C(0,0) = -16 + 4 = -12.

Zhi-Wei Sun, Table of n, a(n) for n = 0..100
Zhi-Wei Sun, An explicit solution to the congruence x^2 == 14*(3/p)-(p/3)-12 (mod p)?, Question 369963 at MathOverflow, August 23, 2020.
Zhi-Wei Sun, New series for powers of Pi and related congruences, Electron. Res. Arch. 28(2020), no. 3, 1273-1342.
Zhi-Wei Sun, Some new series for 1/Pi motivated by congruences, arXiv:2009.04379 [math.NT], 2020.

Cf. A000796, A000984, A005260, A336981, A336982, A337247.

a[n_]:=Sum[Binomial[n,k]Binomial[n+k,k]Binomial[2k,k]Binomial[2(n-k),n-k](-8)^(n-k),{k,0,n}];
Table[a[n],{n,0,18}]

a(n) = (-8)^n*binomial(2*n, n)*hypergeom([1/2, -n, -n, n + 1], [1, 1, 1/2 - n], 1/8). - Peter Luschny, Aug 24 2020

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

A337247 a(n) = (Sum_{k=0..n-1} (-1)^k * (4k+1) * 160^(n-1-k) * C(2k,k) * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C(2j,j) * (-20)^(k-j)) / (n * C(2n,n)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A337332 a(n) = Sum_{k=0..n}C(n,k)C(n+k,k)C(2k,k)C(2n-2k,n-k)(-8)^(n-k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

cp's OEIS Frontend

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

A337247 a(n) = (Sum_{k=0..n-1} (-1)^k * (4k+1) * 160^(n-1-k) * C(2k,k) * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C(2j,j) * (-20)^(k-j)) / (n * C(2n,n)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A337332 a(n) = Sum_{k=0..n}C(n,k)*C(n+k,k)*C(2k,k)*C(2n-2k,n-k)*(-8)^(n-k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

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

A337332 a(n) = Sum_{k=0..n}C(n,k)C(n+k,k)C(2k,k)C(2n-2k,n-k)(-8)^(n-k).