A290920 p-INVERT of the positive integers, where p(S) = (1 - S)^5.

5, 25, 110, 450, 1746, 6505, 23465, 82435, 283270, 955258, 3169520, 10368490, 33497790, 107028120, 338582738, 1061557195, 3301399385, 10191612315, 31250047480, 95226980516, 288523285450, 869559080385, 2607834545025, 7785230674580, 23142279699355

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 (15, -95, 330, -685, 873, -685, 330, -95, 15, -1)

Formula

a(n) = 15*a(n-1) - 95*a(n-2) + 330*a(n-3) - 685*a(n-4) + 873*a(n-5) - 695*a(n-6) + 330*a(n-7) - 95*a(n-8) + 15*a(n-9) - a(n-10).

G.f.: (5 - 50*x + 210*x^2 - 475*x^3 + 621*x^4 - 475*x^5 + 210*x^6 - 50*x^7 + 5*x^8) / (1 - 3*x + x^2)^5. - Colin Barker, Aug 24 2017

Mathematica

z = 60; s = x/(1 - x)^2; p = (1 - s)^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290920 *)

Prog

(PARI) Vec((5 - 50*x + 210*x^2 - 475*x^3 + 621*x^4 - 475*x^5 + 210*x^6 - 50*x^7 + 5*x^8) / (1 - 3*x + x^2)^5 + O(x^30)) \\ Colin Barker, Aug 24 2017

Crossrefs

Cf. A000027, A290890.

Sequence in context: A147161 A282077 A282085 * A267228 A183926 A092440

Adjacent sequences: A290917 A290918 A290919 * A290921 A290922 A290923

Keyword

nonn,easy

Author

Clark Kimberling, Aug 18 2017

Status

approved