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 A179277 A(x) = C(x) * C(x^2) * C(x^4) * C(x^8) *...; C = Catalan, A000108; A(x)/A(x^2) satisfies C(x) 2

%I

%S 1,1,3,6,19,50,158,492,1635,5466,18794,65332,230414,820052,2945436,

%T 10654808,38795523,142045610,522694866,1931912036,7169014298,

%U 26698782108,99756713732,373839656616,1404795235438,5292114330180

%N A(x) = C(x) * C(x^2) * C(x^4) * C(x^8) *...; C = Catalan, A000108; A(x)/A(x^2) satisfies C(x)

%C Let the sequence = A(x) = (1 + x + 3x^2 + 6x^3 + 19x^4 + ...).

%C Then A(x)/A(x^2) = C(x) = (1 + x + 2x^2 + 5x^3 + 14x^4 + ...).

%F Let M = an infinite lower triangular matrix with A000108 in each column but

%F shifted down twice from the previous column, for k>0. Lim_{n->inf.} M^n =

%F A179277, the left shifted vector considered as a sequence: (1 + x + 3x^2 + ...)

%F a(n) = sum_{l=0..n/2} a(l)*A000108(n-2*l). [From _R. J. Mathar_, Jul 09 2010]

%e The generating triangle = M:

%e 1;

%e 1;

%e 2, 1;

%e 5, 1;

%e 14, 2, 1;

%e 42, 5, 1;

%e 132, 14, 2, 1;

%e 429, 42, 5, 1;

%e 1430, 132, 14, 2, 1;

%e 4862, 429, 42, 5, 1;

%e 16796, 1430, 132, 14, 2, 1;

%e ...

%e Then take powers of this matrix, obtaining a left-shifted vector considered

%e as a sequence = A179277.

%p Contribution from _R. J. Mathar_, Jul 09 2010: (Start)

%p A000108 := proc(n) binomial(2*n,n)/(n+1) ; end proc:

%p A179277 := proc(n) if n <= 1 then 1; else add( procname(l)*A000108(n-2*l),l=0..n/2) ; end if; end proc:

%p seq(A179277(n),n=0..80) ; (End)

%Y A000108

%K nonn

%O 0,3

%A _Gary W. Adamson_, Jul 07 2010

%E More terms from _R. J. Mathar_, Jul 09 2010

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Last modified February 19 14:22 EST 2018. Contains 299333 sequences. (Running on oeis4.)