cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

4, -67, 1640, -37725, 565296, 11056402, -1580442016, 96102180805, -4456155445400, 168095261788962, -4821193706309376, 61671590987433918, 4332508360801598880, -462368336475965777100, 28320921191994637110240, -1347995180149692947542005, 51430890880452230248836840
Offset: 1

Views

Author

Zhi-Wei Sun, Nov 13 2019

Keywords

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).

Examples

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

Crossrefs

Programs

  • 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}]