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A329073 a(n) = (1/n)*Sum_{k=0..n-1} (40k+13)*(-1)^k*50^(n-1-k)*T_k(4,1)*T_k(1,-1)^2, where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+b*x+c)^k.

%I A329073 #23 Aug 23 2023 08:43:47
%S A329073 13,219,7858,221525,9253710,375158958,16882409364,736344816813,
%T A329073 32964312771550,1471835619627770,66910145732699964,
%U A329073 3061043035494001682,141458526138008430124,6567714993530314856700,306628434270114823521000,14370411994543866356077725,676259546148988495771751550
%N A329073 a(n) = (1/n)*Sum_{k=0..n-1} (40k+13)*(-1)^k*50^(n-1-k)*T_k(4,1)*T_k(1,-1)^2, where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+b*x+c)^k.
%C A329073 Conjecture 1: (i) a(n) is a positive integer for each n > 0; also, a(n) is odd if and only if n is a power of two. Moreover, we have the identity Sum_{k>=0} ((40k+13)/(-50)^k)*T_k(4,1)*T_k(1,-1)^2 = 55*sqrt(15)/(9*Pi).
%C A329073 (ii) Let p > 5 be a prime. Then Sum_{k=0..p-1} ((40k+13)/(-50)^k)*T_k(4,1)* T_k(1,-1)^2 == (p/3)*(12 + 5*Leg(3/p) + 22*Leg(p/15)) (mod p^2), where Leg(a/p) denotes the Legendre symbol. Also, for the sum S(p) = Sum_{k=0..p-1} T_k(4,1)* T_k(1,-1)^2/(-50)^k, if Leg(-5/p) = -1 then S(p) == 0 (mod p^2); if p == 1,9 (mod 20) and p = x^2 + 5*y^2 with x and y integers then S(p) == 4x^2-2p (mod p^2); if p == 3,7 (mod 20) and 2p = x^2 + 5*y^2 with x and y integers then S(p) == 2x^2-2p (mod p^2).
%C A329073 Conjecture 2: (i) For any n > 0, the number b(n):=(1/n)*Sum_{k=0..n-1} (40k+27)*(-6)^(n-1-k)*T_k(4,1)*T_k(1,-1)^2 is an integer. Moreover, b(n) is odd if and only if n is a power of two.
%C A329073 (ii) Let p > 3 be a prime. Then Sum_{k=0..p-1} ((40k+27)/(-6)^k)*T_k(4,1)* T_k(1,-1)^2 == (p/9)*(55*Leg(-5/p) + 198*Leg(3/p)-10) (mod p^2). Also, for the sum T(p) = Sum_{k=0..p-1} T_k(4,1)*T_k(1,-1)^2/(-6)^k, if Leg(-5/p) = -1 then T(p) == 0 (mod p^2); if p == 1,9 (mod 20) and p = x^2 + 5*y^2 with x and y integers then T(p) == Leg(p/3)*(4x^2-2p) (mod p^2); if p == 3,7 (mod 20) and 2p = x^2 + 5*y^2 with x and y integers then T(p) == Leg(p/3)(2p-2x^2) (mod p^2).
%H A329073 Zhi-Wei Sun, <a href="/A329073/b329073.txt">Table of n, a(n) for n = 1..100</a>
%H A329073 Zhi-Wei Sun, <a href="https://doi.org/10.1007/978-1-4939-1601-6_18">On sums related to central binomial and trinomial coefficients</a>, in: M. B. Nathanson (ed.), Combinatorial and Additive Number Theory: CANT 2011 and 2012, Springer Proc. in Math. & Stat., Vol. 101, Springer, New York, 2014, pp. 257-312. Also available from <a href="https://arxiv.org/abs/1101.0600v25">arXiv:1101.0600 [math.NT]</a>, 2011-2014.
%e A329073 a(1) = 13 since (40*0+13)*(-1)^0*50^(1-1-0)*T_0(4,1)*T_0(1,-1)^2/1 = 13/1 = 13.
%t A329073 T[b_,c_,0]=1;T[b_,c_,1]=b;
%t A329073 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 A329073 a[n_]:=a[n]=Sum[(40k+13)(-1)^k*50^(n-1-k)*T[4,1,k]*T[1,-1,k]^2,{k,0,n-1}]/n;
%t A329073 Table[a[n],{n,1,20}]
%Y A329073 Cf. A081671, A098331.
%K A329073 nonn
%O A329073 1,1
%A A329073 _Zhi-Wei Sun_, Nov 03 2019