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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.

%I A329478 #9 Dec 18 2024 21:01:08
%S A329478 4,-67,1640,-37725,565296,11056402,-1580442016,96102180805,
%T A329478 -4456155445400,168095261788962,-4821193706309376,61671590987433918,
%U A329478 4332508360801598880,-462368336475965777100,28320921191994637110240,-1347995180149692947542005,51430890880452230248836840
%N 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.
%C A329478 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.
%C A329478 (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.
%C A329478 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).
%H A329478 Zhi-Wei Sun, <a href="/A329478/b329478.txt">Table of n, a(n) for n = 1..100</a>
%H A329478  Zhi-Wei Sun, <a href="http://arxiv.org/abs/1911.05456">Characterizing rational Ramanujan-type series for 1/Pi via congruences</a>, arXiv:1911.05456 [math.NT], 2019.
%e A329478 a(1) = ((-1)^0*(15*0+8)*beta(0)*t(0))/(2*1) = (1*8*1*1)/2 = 4.
%t A329478 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;
%t A329478 beta[n_]:=beta[n]=Sum[Binomial[n,k]^2*Binomial[n+k,k],{k,0,n}];
%t A329478 a[n_]:=a[n]=Sum[(-1)^k*(15k+8)*beta[k]*T[4,-1,k],{k,0,n-1}]/(2*n);
%t A329478 Table[a[n],{n,1,17}]
%Y A329478 Cf. A002426, A005258.
%K A329478 sign
%O A329478 1,1
%A A329478 _Zhi-Wei Sun_, Nov 13 2019