

A289806


pINVERT of (1,1,2,2,3,3,...) (A008619), where p(S) = 1  S  S^2.


2



1, 3, 9, 26, 74, 211, 600, 1708, 4860, 13832, 39364, 112029, 318827, 907366, 2582312, 7349121, 20915193, 59523497, 169400608, 482104856, 1372044007, 3904762096, 11112739032, 31626246588, 90006565434, 256153755080, 728999555983, 2074692805003, 5904462080604
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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 A289780 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 (3, 1, 5, 2, 2, 1)


FORMULA

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


MATHEMATICA

z = 60; s = x/((1  x) (1  x^2)); p = 1  s  s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A008619 shifted *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289806 *)


CROSSREFS

Cf. A008619, A289780.
Sequence in context: A291000 A276068 A171277 * A303976 A000243 A076264
Adjacent sequences: A289803 A289804 A289805 * A289807 A289808 A289809


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Aug 12 2017


STATUS

approved



