A290900 p-INVERT of the positive integers, where p(S) = 1 - S^2 - S^3.

0, 1, 5, 17, 51, 149, 439, 1308, 3916, 11728, 35093, 104943, 313773, 938199, 2805439, 8389163, 25086356, 75016104, 224321012, 670787533, 2005857561, 5998122649, 17936209267, 53634716681, 160384099011, 479597177352, 1434141243492, 4288517958652

Offset

0,3

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 (6, -14, 19, -14, 6, -1)

Formula

a(n) = 6*a(n-1) - 14*a(n-2) + 19*a(n-3) - 14*a(n-4) + 6*a(n-5) - a(n-6).

G.f.: x*(1 - x + x^2) / (1 - 6*x + 14*x^2 - 19*x^3 + 14*x^4 - 6*x^5 + x^6). - Colin Barker, Aug 18 2017

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290900 *)

Prog

(PARI) concat(0, Vec(x*(1 - x + x^2) / (1 - 6*x + 14*x^2 - 19*x^3 + 14*x^4 - 6*x^5 + x^6) + O(x^40))) \\ Colin Barker, Aug 18 2017

Crossrefs

Cf. A000027, A290890.

Sequence in context: A039783 A103685 A116521 * A137500 A146814 A034335

Adjacent sequences: A290897 A290898 A290899 * A290901 A290902 A290903

Keyword

nonn,easy

Author

Clark Kimberling, Aug 17 2017

Status

approved