OFFSET
1,1
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.
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.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Nov 04 2019
STATUS
approved