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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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9
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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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Vincenzo Librandi, Table of n, a(n) for n = 0..200
D. Daly and L. Pudwell, Pattern avoidance in rook monoids, 2013.
T. Denton, Algebraic and Affine Pattern Avoidance, arXiv preprint arXiv:1303.3767 [math.CO], 2013.
Z.-W. Sun, Congruences for Franel numbers, arXiv preprint arXiv:1112.1034 [math.NT], 2011. See (1.22).
Z.-W. Sun, On sums of Apery polynomials and related congruences, J. Number Theory 132(2012), 2673-2699.
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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
a(n) ~ 3^(5/2)/(8*Pi) * 9^n/n^2. - Vaclav Kotesovec, Oct 06 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
a(n) = hypergeom([1/2,-n,-n], [1,2], 4). - Vladimir Reshetnikov, Oct 03 2016
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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, binomial(n-1, k)^2*binomial(2*k, k)/(k+1)) \\ Charles R Greathouse IV, Sep 12 2012
(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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Cf. A086617 (table), A086615 (antidiagonal sums), A003046 (determinants).
Cf. A000108.
Cf. A228456.
Sequence in context: A055724 A301433 A054727 * A224769 A302285 A249636
Adjacent sequences: A086615 A086616 A086617 * A086619 A086620 A086621
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KEYWORD
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nonn
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AUTHOR
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Paul D. Hanna, Jul 24 2003
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EXTENSIONS
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Edited by N. J. A. Sloane, Sep 14 2012. The formula in the new definition was first sent in by Michael Somos, Feb 19 2004
Minor edits Vaclav Kotesovec, Mar 31 2014
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STATUS
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approved
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