A329105 - OEIS

%I #21 Aug 23 2023 08:43:44

%S 113,72486,22959360,6667719680,1907342690028,546566353351560,

%T 157644511058113920,45818502502241488320,13418569988503429983660,

%U 3957929725047766692949256,1175070020071246825359359232,350933963579387124964687828224,105365902497675176184788931496400,31787539718100094004136084118514400

%N 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.

%C 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.

%C (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).

%C 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.

%C (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).

%H Zhi-Wei Sun, <a href="/A329105/b329105.txt">Table of n, a(n) for n = 1..70</a>

%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1007/s11425-014-4809-z">Congruences involving generalized central trinomial coefficients</a>, Sci. China Math. 57(2014), no.7, 1375-1400.

%H 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 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.

%t T[b_,c_,0]=1;T[b_,c_,1]=b;

%t 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 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));

%t Table[a[n],{n,1,14}]

%Y Cf. A104454, A328786, A329073, A329107.

%K nonn

%O 1,1

%A _Zhi-Wei Sun_, Nov 04 2019