OFFSET
0,2
COMMENTS
The author introduced this sequence in arXiv:1911.05456 and made the following conjecture.
Conjecture: Let p be an odd prime and let S = Sum_{k=0..p-1}a(k)/(-4)^k. If p == 1 (mod 12) and p = x^2 + 9*y^2 with x and y integers, then S == 4*x^2-2*p (mod p^2). If p == 5 (mod 12) and p = x^2 + y^2 with x == y (mod 3), then S == 4*x*y (mod p^2). If p == 3 (mod 4), then S == 0 (mod p^2).
Note that if p > 3 is a prime, then a(p-1) == Sum_{k=0..p-1} T(k)*T(p-1-k) == Legendre(p/3)*Sum_{k=0..p-1}T(k)^2/(-3)^k == 1 (mod p) by (1.7) and (2.3) of the author's 2014 paper in Sci. China Math.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..931 (terms 0..150 from Zhi-Wei Sun)
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.
Zhi-Wei Sun, Characterizing rational Ramanujan-type series for 1/Pi via congruences, arXiv:1911.05456 [math.NT], 2019.
FORMULA
a(n) ~ (3/2)*12^n/(n*Pi)^(3/2) as n tends to the infinity.
EXAMPLE
a(1) = 2 since Sum_{k=0,1} C(1,k)^2*T(k)*T(1-k) = C(1,0)^2*T(0)*T(1) + C(1,1)^2*T(1)*T(0) = 2*T(0)*T(1) = 2*1*1 = 2.
MATHEMATICA
T[0]=1; T[1]=1; T[n_]:=T[n]=((2n-1)T[n-1]+3*(n-1)*T[n-2])/n;
a[n_]:=a[n]=Sum[Binomial[n, k]^2*T[k]*T[n-k], {k, 0, n}];
Table[a[n], {n, 0, 21}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Nov 13 2019
STATUS
approved