A329105 a(n) = (Sum_{k=0..n-1} (1435*k+113)*3240^(n-1-k)*T_k(7,1)*T_k(10,10)^2)/(n*10^(n-1)), where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+b*x+c)^k.

113, 72486, 22959360, 6667719680, 1907342690028, 546566353351560, 157644511058113920, 45818502502241488320, 13418569988503429983660, 3957929725047766692949256, 1175070020071246825359359232, 350933963579387124964687828224, 105365902497675176184788931496400, 31787539718100094004136084118514400

Offset

1,1

Comments

Conjecture 1: (i) a(n) is an integer for each n > 0. Moreover, we have Sum_{k>=0}((1435*k+113)/3240^k)*T_k(7,1)*T_k(10,10)^2 = 1452*sqrt(5)/Pi.

(ii) Let p > 3 be a prime. Then Sum_{k=0..p-1}((1435*k+113)/3240^k)*T_k(7,1)*T_k(10,10)^2 == p/9*(2420*Leg(-5/p) + 105*Leg(5/p) - 1508) (mod p^2), where Leg(a/p) denotes the Legendre symbol. For the sum S(p) = Sum_{k=0..p-1}T_k(7,1)*T_k(10,10)^2/3240^k, if Leg(-15/p) = -1 then S(p) == 0 (mod p^2); if p == 1,4 (mod 15) 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 with x and y integers then S(p) == 12*x^2-2p (mod p^2).

Conjecture 2: (i) For each n > 0, the number (Sum_{k=0..n-1}(1435*k+1322)*50^(n-1-k)*T_k(7,1)*T_k(10,10)^2)*3/(2n*10^(n-1)) is an integer.

(ii) Let p > 5 be a prime. Then Sum_{k=0..p-1}((1435*k+1322)/50^k)*T_k(7,1)*T_k(10,10)^2 == p/3*(3432*Leg(5/p) + 968*Leg(-1/p) - 434) (mod p^2). For the sum T(p) = Sum_{k=0..p-1}T_k(7,1)*T_k(10,10)^2/50^k, if Leg(-15/p) = -1 then T(p) == 0 (mod p^2); if p == 1,4 (mod 15) and p = x^2 + 15*y^2 with x and y integers then T(p) == 4*x^2-2p (mod p^2); if p == 2,8 (mod 15) and p = 3*x^2 + 5*y^2 with x and y integers then T(p) == 2p-12*x^2 (mod p^2).

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..70

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) = 113 since ((1435*0+113)*3240^(1-1-0)*T_0(7,1)*T_0(10,10)^2)/(1*10^(1-1)) = 113.

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[(1435k+113)T[7, 1, k]T[10, 10, k]^2*3240^(n-1-k), {k, 0, n-1}]/(n*10^(n-1));
Table[a[n], {n, 1, 14}]

Crossrefs

Cf. A104454, A328786, A329073, A329107.

Sequence in context: A176825 A187703 A352400 * A304324 A180052 A106748

Adjacent sequences: A329102 A329103 A329104 * A329106 A329107 A329108

Keyword

nonn

Author

Zhi-Wei Sun, Nov 04 2019

Status

approved