A290901 p-INVERT of the positive integers, where p(S) = 1 - S^3 - S^4.

0, 0, 1, 7, 29, 93, 260, 689, 1845, 5150, 14897, 43663, 127451, 368383, 1056682, 3022366, 8651672, 24818978, 71319058, 205070493, 589550733, 1694075057, 4866102091, 13975547842, 40139685023, 115298782211, 331216158188, 951506566087, 2733431466995

Offset

0,4

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, -28, 57, -71, 57, -28, 8, -1)

Formula

a(n) = 8*a(n-1) - 28*a(n-2) + 57*a(n-3) - 71*a(n-4) + 57*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).

G.f.: x^2*(1 - x + x^2) / (1 - 8*x + 28*x^2 - 57*x^3 + 71*x^4 - 57*x^5 + 28*x^6 - 8*x^7 + x^8). - Colin Barker, Aug 18 2017

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290901 *)

Prog

(PARI) concat(vector(2), Vec(x^2*(1 - x + x^2) / (1 - 8*x + 28*x^2 - 57*x^3 + 71*x^4 - 57*x^5 + 28*x^6 - 8*x^7 + x^8) + O(x^40))) \\ Colin Barker, Aug 18 2017

Crossrefs

Cf. A000027, A290890.

Sequence in context: A266939 A055798 A002664 * A294843 A042609 A002941

Adjacent sequences: A290898 A290899 A290900 * A290902 A290903 A290904

Keyword

nonn,easy

Author

Clark Kimberling, Aug 17 2017

Status

approved