

A291403


pINVERT of (1,1,0,0,0,0,...), where p(S) = 1  2 S^2  S^4.


2



0, 2, 4, 7, 20, 42, 92, 214, 472, 1062, 2396, 5361, 12052, 27074, 60764, 136497, 306520, 688292, 1545768, 3471224, 7795184, 17505588, 39311608, 88280985, 198250312, 445204610, 999783508, 2245185343, 5041947516, 11322557726, 25426742788, 57100105470
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OFFSET

0,2


COMMENTS

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (p(0) + 1/p(S(x)))/x. The pINVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1  S gives the "INVERT" transform of s, so that pINVERT is a generalization of the "INVERT" transform (e.g., A033453).
See A291382 for a guide to related sequences.


LINKS



FORMULA

G.f.: ((x (1 + x)^2 (2 + x^2 + 2 x^3 + x^4))/(1 + 2 x^2 + 4 x^3 + 3 x^4 + 4 x^5 + 6 x^6 + 4 x^7 + x^8)).
a(n) = 2*a(n2) + 4*a(n3) + 3*a(n4) + 4*a(n5) + 6*a(n6) + 4*a(n7) + a(n8) for n >= 9.


MATHEMATICA

z = 60; s = x + x^2; p = 1  2 s^2  s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291403 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



