A099322 An inverse Catalan transform of J(3n)/J(3).

0, 1, 6, 43, 291, 1992, 13595, 92845, 633966, 4329023, 29560367, 201850896, 1378323999, 9411785201, 64267689006, 438847231427, 2996636337771, 20462312853336, 139725412120339, 954104794142789, 6515035056168654

Offset: 0

Author

Paul Barry, Nov 17 2004

Keywords

• easy
• nonn

Comments

The g.f. is obtained from that of A015565 through the mapping g(x)->g(x(1-x)). A015565 may be retrieved through the mapping g(x)->g(xc(x)), where c(x) is the g.f. of A000108.

Links

• Index entries for linear recurrences with constant coefficients, signature (7,1,-16,8).

Crossrefs

Cf. A001045.

Programs

Mathematica

LinearRecurrence[{7,1,-16,8},{0,1,6,43},30] (* Harvey P. Dale, Jul 19 2016 *)

Formula

G.f.: x(1-x)/(1-7x-x^2+16x^3-8x^4);

a(n) = 7a(n-1) + a(n-2) - 16a(n-3) + 8a(n-4);

a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*J(3n-3k)/J(3).

a(n) = Sum_{k=0..n} A109466(n,k)*A015565(k). - Philippe Deléham, Oct 30 2008