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A086618
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a(n) = Sum{k=0..n} binomial(n,k)^2*C(k), where C() = A000108() are the Catalan numbers.
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11
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1, 2, 7, 33, 183, 1118, 7281, 49626, 349999, 2535078, 18758265, 141254655, 1079364105, 8350678170, 65298467487, 515349097713, 4100346740511, 32858696386766, 265001681344569
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OFFSET
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0,2
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COMMENTS
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Main diagonal of square table A086617 of coefficients, T(n,k), of x^n*y^k in f(x,y) that satisfies f(x,y) = 1/[(1-x)(1-y)] + xy*f(x,y)^2.
a(n) is the number of permutations of length 2n which are invariant under the reverse-complement map and have no decreasing subsequences of length 4. - Eric S. Egge, Oct 21 2008
In 2012, Zhi-Wei Sun proved that for any odd prime p we have the congruence a(1) + ... + a(p-1) == 0 (mod p^2). - Zhi-Wei Sun, Aug 22 2013
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LINKS
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FORMULA
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Recurrence: (n+3)^2*(4*n+7)*a(n+2) = 2*(20*n^3+117*n^2+220*n+135)*a(n+1) - 9*(n+1)^2*(4*n+11)*a(n). - Vaclav Kotesovec, Sep 11 2012
G.f.: (1-(1-9*x)^(1/3)*hypergeom([1/3,1/3],[1],-27*x*(1-x)^2/(1-9*x)^2))/(6*x). - Mark van Hoeij, May 02 2013
D-finite with recurrence (n+1)^2*a(n) +(-19*n^2+8*n+6)*a(n-1) +9*(11*n^2-30*n+21)*a(n-2) -81*(n-2)^2*a(n-3)=0. - R. J. Mathar, Aug 01 2022
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EXAMPLE
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a(5) = 1118 = 1*1^2 + 1*5^2 + 2*10^2 + 5*10^2 + 14*5^2 + 42*1^2.
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MATHEMATICA
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Flatten[{1, RecurrenceTable[{(n+3)^2*(4*n+7)*a[n+2]==2*(20*n^3+117*n^2+220*n+135)*a[n+1]-9*(n+1)^2*(4*n+11)*a[n], a[1]==2, a[2]==7}, a, {n, 1, 20}]}] (* Vaclav Kotesovec, Sep 11 2012 *)
Table[HypergeometricPFQ[{1/2, -n, -n}, {1, 2}, 4], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)
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PROG
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(PARI) a(n)=sum(k=0, n-1, (4*k+3)*sum(i=0, k, binomial(k, i)^2*binomial(2*i, i)))/3/n^2 \\ Charles R Greathouse IV, Sep 12 2012
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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