A291183 p-INVERT of the positive integers, where p(S) = 1 - 4*S + 2*S^2.

4, 22, 116, 608, 3180, 16618, 86812, 453440, 2368292, 12369174, 64601428, 337397536, 1762142540, 9203221994, 48066074172, 251036784256, 1311100720708, 6847542588950, 35762957380148, 186780746599392, 975507894703660, 5094827328491242, 26608975328086364

Offset: 0

Author

Clark Kimberling, Aug 19 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 A290890 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 (8,-16,8,-1)

Crossrefs

Cf. A000027, A290890.

Programs

Magma

I:=[4,22,116,608]; [n le 4 select I[n] else 8*Self(n-1)-16*Self(n-2)+8*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Aug 20 2017

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 4 s + 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291183 *)
LinearRecurrence[{8, -16, 8, -1}, {4, 22, 116, 608}, 40] (* Vincenzo Librandi, Aug 20 2017 *)

Formula

G.f.: (2 (2 - 5 x + 2 x^2))/(1 - 8 x + 16 x^2 - 8 x^3 + x^4).

a(n) = 8*a(n-1) - 16*a(n-2) + 8*a(n-3) - a(n-4).