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A245088
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a(n) = (sum_{k=0}^{n-1}(8*k^2+12*k+5)*A244973(k))/n^2
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1
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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
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OFFSET
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1,1
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COMMENTS
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Conjecture: All the terms are odd integers. Moreover, a(p) == 3 (mod p) for any prime p.
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
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LINKS
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FORMULA
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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
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EXAMPLE
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a(2) = -5 since (5*A244973(0)+(8+12+5)*A244973(1))/2^2 = (5*1+25*(-1))/4 = -5.
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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