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A245329
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a(n) = sum_{k=0..n}C(n,k)^3*(-8)^k with C(n,k) = n!/(k!(n-k)!).
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1
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1, -7, 1, 1001, -15359, 30233, 3126529, -61392247, 259448833, 11970181433, -287815672319, 1854020654633, 48800262650881, -1443188813338279, 12410505050039041, 198977188596692681, -7472188661349285887, 80331498114096555641
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OFFSET
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0,2
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COMMENTS
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Conjecture: (i) For any prime p > 3, we have sum_{k=0}^{p-1}(-1)^k*a(k) == (p/3) - p^2/12*B_{p-2}(1/3) (mod p^3) and sum_{k=0}^{p-1}(-1)^k*a(k)*H(k,2) == B_{p-2}(1/3) (mod p), where B_m(x) denotes the Bernoulli polynomial of degree m and H(k,2) stands for sum_{j=1..k} 1/j^2. Also, sum_{k=1}^{p-1}(-1)^k*a(k)/k == - 3*(2^p-2)/p (mod p) for any prime p.
(ii) For any positive integer n, the arithmetic mean (sum_{k=0}^{n-1}(6*k+5)(-1)^k*a(k))/n is an odd integer. Moreover, if p is a prime then sum_{k=0}^{p-1}(6*k+5)(-1)^k*a(k) == 3*p (mod p^2).
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LINKS
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FORMULA
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Recurrence (obtained via the Zeilberger algorithm):
343*(3n+7)*(n+1)^2*a(n) + (3n+5)*(363n^2+1331n+1113)*a(n+1) + 7*(9n^3+57n^2+116n+74)*a(n+2) + (3n+4)*(n+3)^2*a(n+3) = 0.
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EXAMPLE
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a(2) = 1 since sum_{k=0}^2 C(2,k)^3*(-8)^k = 1 + 2^3*(-8) + (-8)^2 = 1.
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MATHEMATICA
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a[n_]:=Sum[Binomial[n, k]^3*(-8)^k, {k, 0, n}]
Table[a[n], {n, 0, 17}]
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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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