A291030 p-INVERT of the positive integers, where p(S) = 1 - S - S^2 - S^3 - S^4.

1, 4, 15, 56, 208, 767, 2812, 10278, 37530, 137044, 500571, 1828818, 6682264, 24416877, 89218462, 325997507, 1191160160, 4352355633, 15902968338, 58107491971, 212317732888, 775783501558, 2834620130881, 10357363200392, 37844566834330, 138279520124262

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 (9,-33,68,-85,68,-33,9,-1).

Crossrefs

Cf. A000027, A290890.

Programs

Magma

I:=[1,4,15,56,208,767,2812,10278]; [n le 8 select I[n] else 9*Self(n-1)-33*Self(n-2)+68*Self(n-3)-85*Self(n-4)+68*Self(n-5)-33*Self(n-6)+9*Self(n-7)-Self(n-8): n in [1..40]]; // Vincenzo Librandi, Aug 20 2017

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - s - s^2 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291030 *)
LinearRecurrence[{9, -33, 68, -85, 68, -33, 9, -1}, {1, 4, 15, 56, 208, 767, 2812, 10278}, 40] (* Vincenzo Librandi, Aug 20 2017 *)

Formula

G.f.: (1 - 5 x + 12 x^2 - 15 x^3 + 12 x^4 - 5 x^5 + x^6)/(1 - 9 x + 33 x^2 - 68 x^3 + 85 x^4 - 68 x^5 + 33 x^6 - 9 x^7 + x^8).

a(n) = 9*a(n-1) - 33*a(n-2) + 68*a(n-3) - 85*a(n-4) + 68*a(n-5) - 33*a(n-6) + 9*a(n-7) - a(n-8).