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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).
6
1, 20, 362, 6504, 114686, 1992536, 34231540, 583027920, 9862508790, 165918037560, 2778642667020, 46358257249200, 770951008563372, 12785838603285104, 211540243555702376, 3492587812271418784, 57557091526140668070, 946970607665938615032, 15557339429900195819164, 255246113991506558429936
OFFSET
0,2
LINKS
Pedro J. Miana, 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.
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)
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 *)
PROG
(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
(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
(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
(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
CROSSREFS
Cf. A000108, A039598, A024492 (h(0)), A000894 (h(1)), A228329 (h(2)), A000515 (h(3)), this sequence (h(4)), A228331 (h(5)), A228332 (h(6)), A228333 (h(7)).
Sequence in context: A060918 A115100 A049683 * A014901 A290581 A000564
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Aug 26 2013
STATUS
approved