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A179277
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A(x) = C(x) * C(x^2) * C(x^4) * C(x^8) *...; C = Catalan, A000108; A(x)/A(x^2) satisfies C(x)
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2
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1, 1, 3, 6, 19, 50, 158, 492, 1635, 5466, 18794, 65332, 230414, 820052, 2945436, 10654808, 38795523, 142045610, 522694866, 1931912036, 7169014298, 26698782108, 99756713732, 373839656616, 1404795235438, 5292114330180
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OFFSET
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0,3
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COMMENTS
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Let the sequence = A(x) = (1 + x + 3x^2 + 6x^3 + 19x^4 + ...).
Then A(x)/A(x^2) = C(x) = (1 + x + 2x^2 + 5x^3 + 14x^4 + ...).
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LINKS
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Table of n, a(n) for n=0..25.
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FORMULA
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Let M = an infinite lower triangular matrix with A000108 in each column but
shifted down twice from the previous column, for k>0. Lim_{n->inf.} M^n =
A179277, the left shifted vector considered as a sequence: (1 + x + 3x^2 + ...)
a(n) = sum_{l=0..n/2} a(l)*A000108(n-2*l). [From R. J. Mathar, Jul 09 2010]
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EXAMPLE
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The generating triangle = M:
1;
1;
2, 1;
5, 1;
14, 2, 1;
42, 5, 1;
132, 14, 2, 1;
429, 42, 5, 1;
1430, 132, 14, 2, 1;
4862, 429, 42, 5, 1;
16796, 1430, 132, 14, 2, 1;
...
Then take powers of this matrix, obtaining a left-shifted vector considered
as a sequence = A179277.
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MAPLE
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Contribution from R. J. Mathar, Jul 09 2010: (Start)
A000108 := proc(n) binomial(2*n, n)/(n+1) ; end proc:
A179277 := proc(n) if n <= 1 then 1; else add( procname(l)*A000108(n-2*l), l=0..n/2) ; end if; end proc:
seq(A179277(n), n=0..80) ; (End)
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CROSSREFS
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A000108
Sequence in context: A148567 A148568 A003162 * A129417 A132335 A148569
Adjacent sequences: A179274 A179275 A179276 * A179278 A179279 A179280
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KEYWORD
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nonn
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AUTHOR
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Gary W. Adamson, Jul 07 2010
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EXTENSIONS
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More terms from R. J. Mathar, Jul 09 2010
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STATUS
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approved
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