A290906 p-INVERT of the positive integers, where p(S) = 1 - 3*S^2.

0, 3, 12, 39, 132, 456, 1572, 5409, 18612, 64053, 220440, 758640, 2610840, 8985147, 30922188, 106418031, 366235308, 1260390744, 4337606988, 14927778921, 51373622388, 176801189997, 608457401520, 2093992746720, 7206429919920, 24800769855603, 85351303248012

Offset: 0

Author

Clark Kimberling, Aug 17 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

• Index entries for linear recurrences with constant coefficients, signature (4, -3, 4, -1)

Crossrefs

Cf. A000027, A290890, A290907.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 3 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290906 *)
u/3 (* A290907 *)

Formula

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

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

a(n) = 3*A290907(n) for n >= 0.