A155581 a(n)=If[IntegerQ[((6*n - 4)/( n + 1))*a(n - 1)], ((6*n - 4)/(n + 1))* a(n - 1), If[IntegerQ[((4*n - 2)/(n + 1))*a( n - 1)], ((4*n - 2)/(n + 1))*a(n - 1), n*a(n - 1)]]

1, 1, 2, 7, 28, 84, 384, 1824, 6080, 30400, 304000, 1064000, 12768000, 67488000, 359936000, 1934656000, 30954496000, 526226432000, 2880397312000, 15842185216000, 316843704320000, 1757042360320000, 38654931927040000

Offset

0,3

Comments

Catalan recursion is:

a[0] = 1; a[n_] := a[n] = ((4*n - 2)/(n + 1))*a[n - 1];

The object here is to get a sequence that is Catalan like, but lower ( bifurcates higher).

Links

Table of n, a(n) for n=0..22.

Formula

a(n)=If[IntegerQ[((6*n - 4)/( n + 1))*a(n - 1)], ((6*n - 4)/(n + 1))* a(n - 1),

If[IntegerQ[((4*n - 2)/(n + 1))*a( n - 1)], ((4*n - 2)/(n + 1))*a(n - 1), n*a(n - 1)]]

Mathematica

Clear [a, n]; a[0] = 1;
a[n_] := a[n] = If[IntegerQ[((6*n - 4)/(n + 1))*a[n - 1]], ((3*n - 2)/(n + 1))* a[n - 1],
If[IntegerQ[((6*n - 4)/(n + 1))*a[n - 1]], ((4*n - 2)/(n + 1))*a[n - 1], n*a[n - 1]]];
Table[a[n], {n, 0, 30}]

Crossrefs

A000108

Sequence in context: A099815 A123363 A102961 * A349329 A048504 A092465

Adjacent sequences: A155578 A155579 A155580 * A155582 A155583 A155584

Keyword

nonn,uned

Author

Roger L. Bagula, Jan 24 2009

Status

approved