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A329478 a(n) = (Sum_{k=0..n-1}(-1)^k*(15*k+8)*beta(k)*t(k))/(2*n), where beta(k) = A005258(k), and t(k) is the coefficient of x^k in the expansion of (x^2+4*x-1)^k. 1
4, -67, 1640, -37725, 565296, 11056402, -1580442016, 96102180805, -4456155445400, 168095261788962, -4821193706309376, 61671590987433918, 4332508360801598880, -462368336475965777100, 28320921191994637110240, -1347995180149692947542005, 51430890880452230248836840 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Conjecture 1: (i) a(n) is an integer for each n > 0. Moreover, a(n) is odd if and only if n is a positive power of two.

(ii) For any prime p, we have a(p) == (27*Leg(p/3) + 5*Leg(p/5))/8 (mod p), where Leg refers to the Legendre symbol.

Conjecture 2: Let p > 5 be a prime and let S(p) = Sum_{k=0..p-1}(-1)^k*beta(k)*t(k). If p == 1,4 (mod 5) and p = x^2 + 15*y^2 (with x and y integers), then S(p) == 4*x^2-2p (mod p^2). If p == 2,8 (mod 15) and p = 3*x^2 + 5*y^2, then S(p) == 12*x^2-2p (mod p^2). If Leg(-15/p) = -1, then S(p) == 0 (mod p^2).

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..100

Zhi-Wei Sun, Characterizing rational Ramanujan-type series for 1/Pi via congruences, arXiv:1911.05456 [math.NT], 2019.

EXAMPLE

a(1) = 4 since (-1)^0*(15*0+8)*beta(0)*t(0))/(2*1) = 8/2 = 4.

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;

beta[n_]:=beta[n]=Sum[Binomial[n, k]^2*Binomial[n+k, k], {k, 0, n}];

a[n_]:=a[n]=Sum[(-1)^k*(15k+8)*beta[k]*T[4, -1, k], {k, 0, n-1}]/(2*n);

Table[a[n], {n, 1, 17}]

CROSSREFS

Cf. A002426, A005258.

Sequence in context: A225940 A003360 A225772 * A241001 A308382 A083931

Adjacent sequences:  A329475 A329476 A329477 * A329479 A329480 A329481

KEYWORD

sign

AUTHOR

Zhi-Wei Sun, Nov 13 2019

STATUS

approved

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Last modified February 27 03:02 EST 2020. Contains 332299 sequences. (Running on oeis4.)