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A289805
p-INVERT of A103889, where p(S) = 1 - S - S^2.
2
2, 9, 36, 153, 624, 2584, 10632, 43865, 180774, 745347, 3072528, 12666854, 52218790, 215273737, 887468000, 3658604277, 15082652352, 62178493132, 256331858332, 1056732372729, 4356396740786, 17959318086575, 74037587378784, 305221185520298, 1258279413185322
OFFSET
0,1
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 p-INVERT 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 p-INVERT is a generalization of the INVERT transform (e.g., A033453).
See A289780 for a guide to related sequences.
FORMULA
G.f.: (-2 - x + 4 x^2 - 7 x^3 + 4 x^4 - 2 x^5)/(-1 + 4 x + 2 x^2 - 8 x^3 + 8 x^4 - 2 x^5 + x^6).
a(n) = 4*a(n-1) + 2*a(n-2) - 8*a(n-3) + 8*a(n-4) - 2*a(n-5) + a(n-6).
MATHEMATICA
z = 60; s = x*(x^2 - x + 2)/((x - 1)^2*(1 + x)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A103889 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289805 *)
LinearRecurrence[{4, 2, -8, 8, -2, 1}, {2, 9, 36, 153, 624, 2584}, 30] (* Harvey P. Dale, Jun 30 2021 *)
CROSSREFS
Sequence in context: A037730 A029874 A052834 * A150967 A344108 A121769
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 12 2017
STATUS
approved