A289787 p-INVERT of the even positive integers (A005843), where p(S) = 1 - S - S^2.

2, 12, 62, 312, 1570, 7908, 39838, 200688, 1010978, 5092860, 25655582, 129241512, 651061762, 3279762132, 16521995710, 83230530528, 419278719938, 2112141348588, 10640036959358, 53599815453720, 270012240337762, 1360202629711812, 6852101192007262

Offset: 0

Author

Clark Kimberling, Aug 10 2017

Keywords

• nonn
• easy

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.

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (6, -6, 6, -1)

Crossrefs

Cf. A005843, A289780, A289788.

Programs

Mathematica

z = 60; s = 2*x/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005843 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289787 *)
u/2 (* A289788 *)

Formula

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

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