A328786 - OEIS

%I #17 Nov 07 2019 22:15:05

%S 7321,69076403,1423525024746,31676475535509475,752633551945067097470,

%T 18627509719518121995003486,474204125641606160260805604468,

%U 12323377272130975561953028453412931,325337163371714764552775702345400136950,8696262375383068237957170325229215635055690

%N a(n) = (Sum_{k=0..n-1}(-1)^k*(39480*k+7321)*29700^(n-1-k)*T_k(14,1)*T_k(11,-11)^2)/n, where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+b*x+c)^k.

%C 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.

%C (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.

%C (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).

%C In 2011 the author formulated many conjectural series for 1/Pi involving T_k(b,c). The conjectural series for 1/Pi in A329073, A329105, A329107 and the present one are of a new type which might be called type VIII to distinguish from the 7 known types (types I-VII).

%H Zhi-Wei Sun, <a href="/A328786/b328786.txt">Table of n, a(n) for n = 1..40</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1102.5649">List of conjectural series for powers of Pi and other constants</a>, arXiv:1102.5649 [math.CA], 2011-2014.

%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) = 7321 since ((-1)^0*(39480*0+7321)*29700^(1-1-0)*T_0(14,1)*T_0(11,-11)^2)/1 = 7321.

%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[(39480k+7321)T[14,1,k]T[11,-11,k]^2*(-1)^k*29700^(n-1-k),{k,0,n-1}]/n;

%t Table[a[n],{n,1,10}]

%Y Cf. A329073, A329105, A329107.

%K nonn

%O 1,1

%A _Zhi-Wei Sun_, Nov 05 2019