A289801 p-INVERT of the tetrahedral numbers (A000292), where p(S) = 1 - S - S^2.

1, 6, 29, 133, 597, 2661, 11856, 52878, 235986, 1053345, 4701627, 20985035, 93662073, 418038721, 1865820223, 8327671681, 37168717729, 165894342774, 740432630793, 3304756826019, 14750048986898, 65833571645931, 293833543748968, 1311460845206801

Offset: 0

Author

Clark Kimberling, Aug 12 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 (9, -31, 62, -74, 57, -28, 8, -1)

Crossrefs

Cf. A000292, A289780.

Programs

Mathematica

z = 60; s = x/(1 - x)^4; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000292 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289801 *)

Formula

G.f.: (1 - 3 x + 6 x^2 - 4 x^3 + x^4)/(1 - 9 x + 31 x^2 - 62 x^3 + 74 x^4 - 57 x^5 + 28 x^6 - 8 x^7 + x^8).

a(n) = 9*a(n-1) - 31*a(n-2) + 62*a(n-3) - 74*a(n-4) + 57*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).