OFFSET
0,2
COMMENTS
Number of standard tableaux of shape (2n+1,1^5). - Emeric Deutsch, May 30 2004
The inverse binomial transform is the (quasi finite) 1, 20, 85, 146, 112, 32, 0,0,0..., the 6th row in A056242. - R. J. Mathar, Mar 14 2026
a(n) is equal to the self convolution of 2*n+1 consecutive terms of the triangular numbers starting with A000217(1): a(0) = 1*1 = 1; a(1) = 1*6 + 3*3 + 6*1 = 21 and so forth. - Klaus Purath, May 04 2026
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..5000 (terms 0..200 from Vincenzo Librandi)
Milan Janjic, Two Enumerative Functions.
J. M. Landsberg and L. Manivel, The sextonions and E7 1/2, Adv. Math., Vol. 201, No. 1 (2006), pp. 143-179. [Th. 7.2(i), case a=1]
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
FORMULA
a(n) = A000389(2*n+5).
G.f.: (1+15*x+15*x^2+x^3)/(1-x)^6 = (1+x)*(x^2+14*x+1)/(1-x)^6.
E.g.f.: (30 + 600*x + 1275*x^2 + 730*x^3 + 140*x^4 + 8*x^5)*exp(x)/30. - G. C. Greubel, Nov 23 2017
Sum_{n>=0} (-1)^n/a(n) = 5*(10/3 - Pi). - Matthieu Pluntz, Oct 08 2019
Sum_{n>=0} 1/a(n) = 40*log(2) - 80/3. - Amiram Eldar, Jan 03 2022
From Peter Bala, Sep 03 2023: (Start)
a(n) = Sum_{0 <= i <= j <= n} (j+1)*(2*i+1)^2.
a(n) = (n+2)*(2*n+5)/(n*(2*n-1))*a(n-1) with a(0) := 1. (End)
a(n) = A053126(n+4)*(2*n+1)/5 . - R. J. Mathar, Mar 14 2026
MATHEMATICA
Table[Binomial[2*n + 5, 5], {n, 0, 50}] (* G. C. Greubel, Nov 23 2017 *)
PROG
(Magma) [Binomial(2*n+5, 5): n in [0..30]]; // Vincenzo Librandi, Oct 07 2011
(PARI) a(n)=n*(8*n^4+60*n^3+170*n^2+225*n+137)/30+1 \\ Charles R Greathouse IV, Apr 18 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Eric Lane
STATUS
approved