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%I #15 Aug 18 2017 19:09:31
%S 5,35,240,1645,11275,77280,529685,3630515,24883920,170556925,
%T 1169014555,8012544960,54918800165,376419056195,2580014593200,
%U 17683683096205,121205767080235,830756686465440,5694091038177845,39027880580779475,267501073027278480
%N p-INVERT of the positive integers, where p(S) = 1 - 5*S.
%C 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).
%C See A290890 for a guide to related sequences.
%H Clark Kimberling, <a href="/A290903/b290903.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7, -1)
%F G.f.: 5/(1 - 7 x + x^2).
%F a(n) = 7*a(n-1) - a(n-2).
%F a(n) = 5*A004187(n+1) for n >= 0.
%e (See the example at A290902.)
%t z = 60; s = x/(1 - x)^2; p = 1 - 5 s;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290903 *)
%t u/5 (* A004187 shifted *)
%Y Cf. A000027, A004187, A290890.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Aug 17 2017
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