

A292534


pINVERT of the squares (A000290), where p(S) = 1 + S  S^2.


1



1, 2, 4, 21, 30, 11, 80, 622, 2055, 4584, 10711, 34354, 115480, 341213, 934750, 2640483, 7874188, 23564242, 68738591, 198108496, 575654335, 1688669686, 4951141372, 14443935957, 42064267934, 122731975243, 358682023576, 1047906654118, 3058580566407
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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 A292479 for a guide to related sequences.


LINKS

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5, 12, 22, 16, 7, 1)


FORMULA

G.f.: ((1 + x) (1 + 4 x  2 x^2 + x^3))/(1  5 x + 12 x^2  22 x^3 + 16 x^4  7 x^5 + x^6).
a(n) = 5*a(n1)  12*a(n2) + 22*a(n3)  16*a(n4) + 7*a(n5)  a(n6) for n >= 7.


MATHEMATICA

z = 60; s = x (x + 1)/(1  x)^3; p = 1 + s  s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292534 *)


CROSSREFS

Cf. A000290, A292479.
Sequence in context: A099179 A102049 A058522 * A122736 A296276 A092458
Adjacent sequences: A292531 A292532 A292533 * A292535 A292536 A292537


KEYWORD

easy,sign


AUTHOR

Clark Kimberling, Oct 04 2017


STATUS

approved



