OFFSET
0,3
COMMENTS
a(n+1) is the determinant of the n X n Hankel matrix [C(i+j+3)]_{i,j=1..n} where C(n) = A000108(n), the n-th Catalan number. - Michael Somos, Jun 27 2023
REFERENCES
R. P. Stanley, Enumerative Combinatorics, volume 1 (1986), p. 221, Example 4.5.18.
LINKS
T. D. Noe, Table of n, a(n) for n = 0..1000
Myriam de Sainte-Catherine, Couplages et Pfaffiens en Combinatoire, Physique et Informatique, PhD Dissertation, Université Bordeaux I, 1983. (Annotated scanned copy of pages III.42-III.45)
G. Kreweras and H. Niederhausen, Solution of an enumerative problem connected with lattice paths, European J. Combin., 2 (1981), 55-60.
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).
FORMULA
a(n) = binomial(2*n+6, 7)*(2*n+3)*(n+1)*(n+2)/240.
G.f.: x*(1 + 31*x + 187*x^2 + 330*x^3 + 187*x^4 + 31*x^5 + x^6)/(1-x)^11. - Colin Barker, May 07 2012
a(n) = det(A*Transpose(A))/36, where A is the 2 X (n+1) matrix whose (i,j)-th element is j^(2*i-1). - Lechoslaw Ratajczak, Oct 01 2017
a(n) = binomial(2*n+4, 3)*binomial(2*n+6, 7)/160. - G. C. Greubel, Dec 17 2021
a(n) = a(-3-n) for all n in Z. - Michael Somos, Jun 27 2023
a(n) ~ n^10/4725. - Stefano Spezia, Dec 09 2023
EXAMPLE
G.f. = x + 42*x^2 + 594*x^3 + 4719*x^4 + 26026*x^5 + 111384*x^6 + ... - Michael Somos, Jun 27 2023
MATHEMATICA
LinearRecurrence[{11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 1}, {0, 1, 42, 594, 4719, 26026, 111384, 395352, 1215126, 3331251, 8321170}, 30] (* Harvey P. Dale, Apr 15 2017 *)
PROG
(PARI) a(n) = binomial(2*n+6, 7)*(2*n+3)*(n+1)*(n+2)/240; \\ Michel Marcus, Oct 13 2016
(Sage) [product(binomial(2*(n+j+2), 4*j+3) for j in (0..1))/160 for n in (0..30)] # G. C. Greubel, Dec 17 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Philippe Deléham, Mar 13 2004
STATUS
approved