A228330 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(4).
1, 20, 362, 6504, 114686, 1992536, 34231540, 583027920, 9862508790, 165918037560, 2778642667020, 46358257249200, 770951008563372, 12785838603285104, 211540243555702376, 3492587812271418784, 57557091526140668070, 946970607665938615032, 15557339429900195819164, 255246113991506558429936
Offset: 0
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 0..825
- Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega4(n) = a(n-1).
- Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv:1305.2017 [math.CO], 2013.
- Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.
Crossrefs
Programs
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GAP
List([0..20], n-> Binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1))); # G. C. Greubel, Mar 02 2019
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Magma
[Binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)): n in [0..20]]; // G. C. Greubel, Mar 02 2019
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Mathematica
Table[4*Sum[(k+1)^6*(Binomial[2n+1, n-k]/(n+k+2))^2, {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *) -
PARI
vector(20, n, n--; binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1))) \\ G. C. Greubel, Mar 02 2019
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Sage
[binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)) for n in (0..20)] # G. C. Greubel, Mar 02 2019
Formula
Conjecture: n*(2*n+1)*(3467*n-4029)*a(n) + 8*(-36721*n^3 + 109040*n^2 - 137926*n + 69822)*a(n-1) + 4*(4*n-9)*(45706*n-7907)*(4*n-7)*a(n-2) = 0. - R. J. Mathar, Sep 08 2013
Recurrence: n*(2*n+1)*(15*n^3 - 30*n^2 + 16*n - 2)*a(n) = 2*(4*n-5)*(4*n-3)*(15*n^3 + 15*n^2 + n - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
From Vaclav Kotesovec, Dec 08 2013: (Start)
a(n) = binomial(4*n,2*n) * (15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)).
a(n) = 4*Sum_{k=0..n} (k+1)^6*(binomial(2*n+1, n-k)/(n+k+2))^2. (End)