A329107 a(n) = (Sum_{k=0..n-1}(840*k+197)(-1)^k*2430^(n-1-k)*T_k(8,1)*T_k(5,-5)^2)/(n*5^(n-1)), where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+b*x+c)^k.

197, 27131, 9162090, 3337679905, 1300603606702, 526423563257310, 219304133423593380, 93259079677243221345, 40287972095635400291790, 17621949843841860346761946, 7785698346885200295051911308, 3468528609790576968835453926954, 1556035297261133424003013368953900

Offset: 1

Author

Zhi-Wei Sun, Nov 04 2019

Keywords

• nonn

Comments

Conjecture: (i) a(n) is a positive integer for each n > 0. Moreover, we have Sum_{k>=0} ((840*k+197)/(-2430)^k)*T_k(8,1)*T_k(5,-5)^2 = 189*sqrt(15)/(2*Pi).

(ii) If p > 3 is a prime, then Sum_{k=0..p-1}((840*k+197)/(-2430)^k)*T_k(8,1)*T_k(5,-5)^2 == p*(52 + 5*Leg(15/p) + 140*(-15/p)) (mod p^2), where Leg(a/p) denotes the Legendre symbol.

(iii) Let p > 7 be a prime and set S(p) = Sum_{k=0..p-1}T_k(8,1)*T_k(5,-5)^2/(-2430)^k. If Leg(-105,p) = -1, then S(p) == 0 (mod p^2). If Leg(-1/p) = Leg(p/3) = Leg(p/5) = Leg(p/7) = 1 and p = x^2 + 105*y^2 (with x and y integers), then S(p) == 4*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/7) = 1, Leg(p/3) = Leg(p/5) = -1 and 2p = x^2 + 105*y^2, then S(p) == 2*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/3) = Leg(p/5) = Leg(p/7) = -1 and p = 3*x^2 + 35*y^2, then S(p) == 12*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/7) = -1, Leg(p/3) = Leg(p/5) = 1, and 2p = 3*x^2 + 35*y^2, then S(p) == 6*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/5) = 1, Leg(p/3) = Leg(p/7) = -1, and p = 5*x^2 + 21*y^2, then S(p) == 2p-20*x^2 (mod p^2). If Leg(-1/p) = Leg(p/3) = 1, Leg(p/5) = Leg(p/7) = -1, and 2p = 5*x^2 + 21*y^2, then S(p) == 2p-10*x^2 (mod p^2). If Leg(-1/p) = Leg(p/5) = -1, Leg(p/3) = Leg(p/7) = 1, and p = 7*x^2 + 15*y^2, then S(p) == 28*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/3) = -1, Leg(p/5) = Leg(p/7) = 1, and 2p = 7*x^2 + 15*y^2, then S(p) == 14*x^2-2p (mod p^2).

One can easily check the identity in part (i) as the series converges very fast. Note also that the imaginary quadratic field Q(sqrt(-105)) has class number 8.

Examples

			a(1) = 197 since (840*0+197)*T_0(8,1)*T_0(5,-5)^2*(-1)^0*2430^(1-1-0)/(1*5^(1-1)) = 197.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..70
• Zhi-Wei Sun, List of conjectural series for powers of Pi and other constants, arXiv:1102.5649 [math.CA], 2011-2014.
• Zhi-Wei Sun, Congruences involving generalized central trinomial coefficients, Sci. China Math. 57(2014), no.7, 1375-1400.
• Zhi-Wei Sun, On sums related to central binomial and trinomial coefficients, 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 arXiv:1101.0600 [math.NT], 2011-2014.

Crossrefs

Cf. A328786, A329073, A329105.

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;
a[n_]:=a[n]=Sum[(840k+197)T[8,1,k]T[5,-5,k]^2*(-1)^k*2430^(n-1-k),{k,0,n-1}]/(n*5^(n-1));
Table[a[n],{n,1,13}]