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A275027
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a(n) = Sum_{k=0..n} C(n,k)^2*C(n-k,k), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).
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5
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1, 1, 5, 19, 85, 401, 1931, 9605, 48469, 248365, 1286605, 6726875, 35441275, 187935775, 1002122525, 5369287019, 28889315669, 156015203845, 845330354321, 4593724615175, 25029614166685, 136704935601785, 748273234994675, 4103928115592365, 22549175326327675, 124105065258631651, 684100888645922051, 3776354280849020005
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OFFSET
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0,3
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COMMENTS
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Conjecture: For any prime p > 5 and positive integer n, the number (a(p*n)-a(n))/(p*n)^3 is always a p-adic integer.
The author has proved that for any prime p > 5 and positive integer n the number (a(p*n)-a(n))/(p^3*n^2) is always a p-adic integer.
As a(n) = Sum_{k=0..n} C(n,k)*C(n,2k)*C(2k,k) and C(2k,k) = 2*C(2k-1,k-1) for k = 1,2,3,..., we see that a(n) is always odd. We guess that a(n) is congruent to one of 0, 1, -1 modulo 5.
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LINKS
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FORMULA
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a(n) = Sum_{k=0..n}C(n,k)*C(n,2k)*C(2k,k).
By the Zeilberger algorithm, we have the recurrence (n+3)^2*(23n+25)*a(n+3) = 25*(n+1)^2*(23n+48)*a(n) + (391n^3+1989n^2+3288n+1750)*a(n+1) + (46n^3+280n^2+ 519n+265)*a(n+2) for all n >= 0.
a(n) = hypergeom([-n, 1/2 - n/2, -n/2], [1, 1], -4). - Peter Luschny, Mar 21 2018
a(n) ~ c * d^n / (Pi*n), where d = 5.729031537980930837932235459792820714... is the real root of the equation -25 - 17*d - 2*d^2 + d^3 = 0 and c = 1.107089291883984657933126801836156175486638498732... is the positive real root of the equation -125 + 1048*c^2 - 2576*c^4 + 1472*c^6 = 0. - Vaclav Kotesovec, Jun 09 2019
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EXAMPLE
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a(2) = 5 since a(2) = Sum_{k=0,1,2}C(2,k)^2*C(2-k,k) = C(2,0)^2*C(2,0) + C(2,1)^2*C(1,1) = 1 + 4 = 5.
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MATHEMATICA
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a[n_]:=a[n]=Sum[Binomial[n, k]^2*Binomial[n-k, k], {k, 0, n/2}]
Table[a[n], {n, 0, 27}]
a[n_] := HypergeometricPFQ[{-n, 1/2 - n/2, -n/2}, {1, 1}, -4];
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PROG
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(PARI) a(n) = sum(k=0, n, binomial(n, k)^2*binomial(n-k, k)); \\ Michel Marcus, Nov 13 2016
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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