Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #11 Nov 23 2024 08:29:24
%S 2,5,14,38,102,271,714,1868,4858,12569,32374,83058,212350,541219,
%T 1375570,3487384,8821170,22266413,56098206,141087934,354268502,
%U 888238903,2223968666,5561234916,13889778218,34652529473,86361653126,215021205770,534861620718
%N p-INVERT of the Fibonacci sequence (A000045), where p(S) = (1 - S)^2.
%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).
%H Clark Kimberling, <a href="/A292327/b292327.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-4,-1).
%F G.f.: -(2 + x)*(-1 + 2*x)/(-1 + 2*x + x^2)^2.
%F a(n) = 4*a(n-1) - 2*a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.
%F a(n) = A006645(n+1) +2*A000129(n+1). - _R. J. Mathar_, Jul 08 2022
%t z = 60; s = x/(1 - x - x^2); p = (1 - s)^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292327 *)
%Y Cf. A000045, A292328.
%Y Cf. A006645, A000129.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Sep 15 2017