%I #19 Aug 19 2017 13:30:56

%S 2,12,62,312,1570,7908,39838,200688,1010978,5092860,25655582,

%T 129241512,651061762,3279762132,16521995710,83230530528,419278719938,

%U 2112141348588,10640036959358,53599815453720,270012240337762,1360202629711812,6852101192007262

%N p-INVERT of the even positive integers (A005843), where p(S) = 1 - S - 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).

%C See A289780 for a guide to related sequences.

%H Clark Kimberling, <a href="/A289787/b289787.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 (6, -6, 6, -1)

%F G.f.: (2 (1 + x^2))/(1 - 6 x + 6 x^2 - 6 x^3 + x^4).

%F a(n) = 6*a(n-1) - 6*a(n-2) + 6*a(n-3) - a(n-4).

%t z = 60; s = 2*x/(1 - x)^2; p = 1 - s - s^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005843 *)

%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289787 *)

%t u/2 (* A289788 *)

%Y Cf. A005843, A289780, A289788.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 10 2017