A245088 a(n) = (Sum_{k=0..n-1} (8*k^2+12*k+5)*A244973(k))/n^2.
5, -5, -9, 115, -397, 85, 6625, -36181, 63915, 377365, -3357175, 10579245, 12408269, -295386005, 1383403247, -1317964405, -23051599205, 159765455077, -392146366775, -1358686166755, 16622987639325, -64283150090725, -10522422920465, 1532928077704325, -8583492124492507
Offset: 1
Keywords
Examples
a(2) = -5 since (5*A244973(0)+(8+12+5)*A244973(1))/2^2 = (5*1+25*(-1))/4 = -5.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..200
- V. J. W. Guo, S.-M. Guo and H. Pan, Proof of a conjecture involving Sun polynomials, arXiv preprint arXiv:1511.04005 [math.NT], 2015.
- Z.-W. Sun, Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k, preprint, arXiv:1407.0967 [math.NT], 2014-2015.
- Z.-W. Sun, Congruences involving g_n(x)=sum_{k=0..n} binomial(n,k)^2*binomial(2k,k)*x^k, Ramanujan J., in press. Doi: 10.1007/s11139-015-9727-3.
Crossrefs
Cf. A244973.
Programs
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Mathematica
s[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k](-1)^k,{k,0,n}] S[n_]:=Sum[(8k^2+12k+5)*s[k],{k,0,n-1}]/n^2 Table[S[n],{n,1,25}]
Formula
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
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
Lim sup n->infinity |a(n)|^(1/n) = 5. - Vaclav Kotesovec, Jul 13 2014
Comments