%I #17 Nov 14 2015 14:53:05

%S 5,-5,-9,115,-397,85,6625,-36181,63915,377365,-3357175,10579245,

%T 12408269,-295386005,1383403247,-1317964405,-23051599205,159765455077,

%U -392146366775,-1358686166755,16622987639325,-64283150090725,-10522422920465,1532928077704325,-8583492124492507

%N a(n) = (sum_{k=0}^{n-1}(8*k^2+12*k+5)*A244973(k))/n^2

%C Conjecture: All the terms are odd integers. Moreover, a(p) == 3 (mod p) for any prime p.

%C In 2015 Guo, Mao and Pan confirmed the conjecture partially by proving that n*a(n) is always an integer. - _Zhi-Wei Sun_, Nov 14 2015

%H Zhi-Wei Sun, <a href="/A245088/b245088.txt">Table of n, a(n) for n = 1..200</a>

%H V. J. W. Guo, S.-M. Guo and H. Pan, <a href="http://arxiv.org/abs/1511.04005">Proof of a conjecture involving Sun polynomials</a>, arXiv preprint arXiv:1511.04005 [math.NT], 2015.

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1407.0967">Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k</a>, preprint, arXiv:1407.0967 [math.NT], 2014-2015.

%H Z.-W. Sun, <a href="http://dx.doi.org/10.1007/s11139-015-9727-3">Congruences involving g_n(x)=sum_{k=0..n}binom(n,k)^2*binom(2k,k)*x^k</a>, Ramanujan J., in press. Doi: 10.1007/s11139-015-9727-3.

%F Conjecture: +n^2*(23656*n-2233)*(n-1)*a(n) +(n-1)*(23656*n^3+655645*n^2-2273556*n+1915600)*a(n-1) -2*(n-2) *(177048*n^3-3748011*n^2+13870329*n-15407140)*a(n-2) -2*(n-3) *(2020728*n^3-18138829*n^2+63885382*n-73802216)*a(n-3) -5*(n-4) *(782552*n^3-11993599*n^2+55520403*n-80051816)*a(n-4) +125*(66088*n-203507)*(n-5)*(n-4)^2*a(n-5)=0. - _R. J. Mathar_, Jul 12 2014

%F Recurrence (of order 3): (n-1)*n^2*(8*n^2 - 36*n + 41)*(8*n^2 - 20*n + 13)*a(n) = -(n-1)*(8*n^2 - 36*n + 41)*(40*n^4 - 148*n^3 + 161*n^2 - 42*n + 4)*a(n-1) - (n-2)*(8*n^2 - 4*n + 1)*(152*n^4 - 1188*n^3 + 3427*n^2 - 4293*n + 1960)*a(n-2) + 25*(n-3)*(n-2)^2*(8*n^2 - 20*n + 13)*(8*n^2 - 4*n + 1)*a(n-3). - _Vaclav Kotesovec_, Jul 13 2014

%F Lim sup n->infinity |a(n)|^(1/n) = 5. - _Vaclav Kotesovec_, Jul 13 2014

%e a(2) = -5 since (5*A244973(0)+(8+12+5)*A244973(1))/2^2 = (5*1+25*(-1))/4 = -5.

%t s[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k](-1)^k,{k,0,n}]

%t S[n_]:=Sum[(8k^2+12k+5)*s[k],{k,0,n-1}]/n^2

%t Table[S[n],{n,1,25}]

%Y Cf. A244973.

%K sign

%O 1,1

%A _Zhi-Wei Sun_, Jul 11 2014