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A027826
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Inverse binomial transform of a_0 = 1, a_1, a_2, etc. is a_0, 0, a_1, 0, a_2, 0, etc.
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4
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1, 1, 2, 4, 9, 21, 50, 120, 290, 706, 1732, 4280, 10644, 26612, 66824, 168384, 425481, 1077529, 2733746, 6945812, 17669149, 44994345, 114682042, 292544200, 746831570, 1907983346, 4877966628, 12479883736, 31951158024
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| The self-convolution equals A051163. - Paul D. Hanna (pauldhanna(AT)juno.com), Nov 23 2004
Equals row sums of triangle A152193. [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Nov 28 2008]
Hankel transform is A166446(n+1). [From Paul Barry (pbarry(AT)wit.ie), Oct 13 2009]
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LINKS
| N. J. A. Sloane, Transforms
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FORMULA
| G.f. A(x) satisfies A(x^2)=A(x/(1+x))/(1+x) and A(x)=A(x^2/(1-x)^2)/(1-x).
Contribution from Paul Barry (pbarry(AT)wit.ie), Jul 05 2009: (Start)
G.f.: (1-x)/((1-x)^2-x^2-x^4/((1-x)^2-x^2-x^4/(1-... (continued fraction);
a(n)=sum{k=0..n, C(n,2k)*A001006(k)}. (End)
G.f.: ((1-x)*(1-2x-sqrt((1-2x)^2-4x^4))/(2x^4). [From Paul Barry (pbarry(AT)wit.ie), Oct 13 2009]
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PROG
| (PARI) a(n)=local(A, m); if(n<0, 0, m=1; A=1+O(x); while(m<=n, m*=2; A=subst(A, x, (x/(1-x))^2)/(1-x)); polcoeff(A, n))
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CROSSREFS
| Cf. A051163.
A152193 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Nov 28 2008]
Sequence in context: A018905 A024537 A171842 * A091964 A092423 A199410
Adjacent sequences: A027823 A027824 A027825 * A027827 A027828 A027829
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KEYWORD
| nonn
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AUTHOR
| Allan Wechsler (acw(AT)alum.mit.edu)
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