A290921 p-INVERT of the positive integers, where p(S) = (1 - S)^6.

6, 33, 158, 696, 2886, 11425, 43590, 161355, 582340, 2056818, 7130388, 24319054, 81757104, 271353288, 890327048, 2891047695, 9299683770, 29658374355, 93843661530, 294791108106, 919849034686, 2852495485953, 8794877092878, 26971256457596, 82298545175130

Offset

0,1

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 (18, -141, 630, -1770, 3258, -3989, 3258, -1770, 630, -141, 18, -1)

Formula

a(n) = 18*a(n-1) - 141*a(n-2) + 630*a(n-3) - 1770*a(n-4) + 3258*a(n-5) - 3989*a(n-6) + 3258*a(n-7) - 1770*a(n-8) + 630*a(n-9) - 141*a(n-10) + 18*a(n-11) - a(n-12).

G.f.: (2 - x)*(1 - 2*x)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(3 - 15*x + 25*x^2 - 15*x^3 + 3*x^4) / (1 - 3*x + x^2)^6. - Colin Barker, Aug 24 2017

Mathematica

z = 60; s = x/(1 - x)^2; p = (1 - s)^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290921 *)

Prog

(PARI) Vec((2 - x)*(1 - 2*x)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(3 - 15*x + 25*x^2 - 15*x^3 + 3*x^4) / (1 - 3*x + x^2)^6 + O(x^30)) \\ Colin Barker, Aug 24 2017

Crossrefs

Cf. A000027, A290890.

Sequence in context: A022730 A266944 A301272 * A240880 A099432 A072260

Adjacent sequences: A290918 A290919 A290920 * A290922 A290923 A290924

Keyword

nonn,easy

Author

Clark Kimberling, Aug 18 2017

Status

approved