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%I A289787 #19 Aug 19 2017 13:30:56
%S A289787 2,12,62,312,1570,7908,39838,200688,1010978,5092860,25655582,
%T A289787 129241512,651061762,3279762132,16521995710,83230530528,419278719938,
%U A289787 2112141348588,10640036959358,53599815453720,270012240337762,1360202629711812,6852101192007262
%N A289787 p-INVERT of the even positive integers (A005843), where p(S) = 1 - S - S^2.
%C A289787 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 A289787 See A289780 for a guide to related sequences.
%H A289787 Clark Kimberling, <a href="/A289787/b289787.txt">Table of n, a(n) for n = 0..1000</a>
%H A289787 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6, -6, 6, -1)
%F A289787 G.f.: (2 (1 + x^2))/(1 - 6 x + 6 x^2 - 6 x^3 + x^4).
%F A289787 a(n) = 6*a(n-1) - 6*a(n-2) + 6*a(n-3) - a(n-4).
%t A289787 z = 60; s = 2*x/(1 - x)^2; p = 1 - s - s^2;
%t A289787 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005843 *)
%t A289787 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289787 *)
%t A289787 u/2 (* A289788 *)
%Y A289787 Cf. A005843, A289780, A289788.
%K A289787 nonn,easy
%O A289787 0,1
%A A289787 _Clark Kimberling_, Aug 10 2017

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