OFFSET
1,1
COMMENTS
Conjecture: (i) a(n) is a positive integer for each n > 0, and a(n) is odd if and only if n is a power of two. Moreover, we have the identity Sum_{k>=0}(39480*k+7321)(-29700)^(-k)*T_k(14,1)*T_k(11,-11)^2 = 6795*sqrt(5)/Pi.
(ii) For any prime p > 5, we have the congruence Sum_{k=0..p-1}(39480*k+7321)(-29700)^(-k)*T_k(14,1)*T_k(11,-11)^2 == p*(1513 + 70*Leg(3/p) + 5738*Leg(-5/p)) (mod p^2), where Leg(a/p) denotes the Legendre symbol.
(iii) Let p > 5 be a prime different from 11, and set S(p) = Sum_{k=0..p-1} T_k(14,1)*T_k(11,-11)^2/(-29700)^k. If Leg(-165/p) = -1, then S(p) == 0 (mod p^2). If Leg(-1/p) = Leg(p/3) = Leg(p/5) = Leg(p/11) = 1 and p = x^2 + 165*y^2 (with x and y integers), then S(p) == 4*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/3) = Leg(p/5) = Leg(p/11) = -1 and 2p = x^2 + 165*y^2, then S(p) == 2*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/5) = -1, Leg(p/3) = Leg(p/11) = 1, and p = 3*x^2 + 55*y^2, then S(p) == 12*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/5) = 1, Leg(p/3) = Leg(p/11) = -1, and 2p = 3*x^2 + 55*y^2, then S(p) == 6*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/11) = 1, Leg(p/3) = Leg(p/5) = -1, and p = 5*x^2 + 33*y^2, then S(p) == 2p-20*x^2 (mod p^2). If Leg(-1/p) = Leg(p/11) = -1, Leg(p/3) = Leg(p/5) = 1, and 2p = 5*x^2 + 33*y^2, then S(p) == 2p-10*x^2 (mod p^2). If Leg(-1/p) = Leg(p/3) = -1, Leg(p/5) = Leg(p/11) = 1, and p = 11*x^2 + 15*y^2, then S(p) == 44*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/3) = 1, Leg(p/5) = Leg(p/11) = -1, and 2p = 11*x^2 + 15*y^2, then S(p) == 22*x^2-2p (mod p^2).
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..40
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) = 7321 since ((-1)^0*(39480*0+7321)*29700^(1-1-0)*T_0(14,1)*T_0(11,-11)^2)/1 = 7321.
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[(39480k+7321)T[14, 1, k]T[11, -11, k]^2*(-1)^k*29700^(n-1-k), {k, 0, n-1}]/n;
Table[a[n], {n, 1, 10}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Nov 05 2019
STATUS
approved