A228304 a(n) = Sum_{k=0..n} C(n,k)^4*(-1)^k.

1, 0, -14, 0, 786, 0, -61340, 0, 5562130, 0, -549676764, 0, 57440496036, 0, -6242164112184, 0, 698300344311570, 0, -79881547652046140, 0, 9301427008157320036, 0, -1098786921802152516024, 0, 131361675994216221116836, 0, -15863471168011822803270200, 0, 1932252897656224864335299400, 0, -237114404923760858875375113840

Offset

0,3

Comments

As (-1)^n*a(n) = a(n), we have a(n) = 0 for n = 1,3,5,... For any odd prime p, the author could show that a(p-1) == 1 + 4*(2^{p-1}-1) + 6*(2^{p-1}-1)^2 (mod p^3).

Conjecture: Let p be any odd prime, and let A(p) be the p X p determinant with (i,j)-entry equal to a(i+j) for all i,j = 0,...,p-1. Then A(p) == (-1)^{(p-1)/2} (mod p). Similarly, if c(n) = sum_{k=0}^n (-1)^k*C(n,k)^2*C(2k,k)*C(2(n-k),n-k) and C(p) is the p X p determinant with (i,j)-entry equal to c(i+j) for all i,j = 0,...,p-1, then we have C(p) == 1 (mod p).

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..100

Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.

Formula

Conjecture: n^3*(n-1)*(12*n^2-63*n+83)*a(n) +(n-2)*(12*n^2-87*n+158)*(n-1)^3*a(n-1) +4*(408*n^6-3774*n^5+13760*n^4-25203*n^3+24465*n^2-11970*n+2340)*a(n-2) +4*(408*n^6-6222*n^5+38750*n^4-126143*n^3+226494*n^2-212867*n+81920)*a(n-3) +16*(n-2)*(12*n^2-15*n+5)*(n-3)^3*a(n-4) +16*(n-3)*(12*n^2-39*n+32)*(n-4)^3*a(n-5)=0. - R. J. Mathar, Aug 21 2013

a(2n) = A050983(n) * (-1)^n. - Vaclav Kotesovec, Feb 01 2014

Mathematica

a[n_]:=Sum[Binomial[n, k]^4*(-1)^k, {k, 0, n}]
Table[a[n], {n, 0, 30}]
Table[HypergeometricPFQ[{-n, -n, -n, -n}, {1, 1, 1}, -1], {n, 0, 20}] (* Vaclav Kotesovec, Feb 01 2014 *)

Crossrefs

Cf. A050983, A228289.

Sequence in context: A161384 A337104 A198806 * A002393 A185284 A262705

Adjacent sequences: A228301 A228302 A228303 * A228305 A228306 A228307

Keyword

sign

Author

Zhi-Wei Sun, Aug 20 2013

Status

approved