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

%I #11 Jan 22 2020 20:10:33

%S 0,1,6,43,291,1992,13595,92845,633966,4329023,29560367,201850896,

%T 1378323999,9411785201,64267689006,438847231427,2996636337771,

%U 20462312853336,139725412120339,954104794142789,6515035056168654

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

%C 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.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (7,1,-16,8).

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

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

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

%F a(n) = Sum_{k=0..n} A109466(n,k)*A015565(k). - _Philippe Deléham_, Oct 30 2008

%t LinearRecurrence[{7,1,-16,8},{0,1,6,43},30] (* _Harvey P. Dale_, Jul 19 2016 *)

%Y Cf. A001045.

%K easy,nonn

%O 0,3

%A _Paul Barry_, Nov 17 2004