A289797 p-INVERT of the triangular numbers (A000217), where p(S) = 1 - S - S^2.

1, 5, 21, 84, 330, 1291, 5052, 19784, 77500, 303608, 1189372, 4659245, 18252027, 71500068, 280092848, 1097230105, 4298267549, 16837948391, 65960645632, 258392925744, 1012223324455, 3965263584006, 15533444957104, 60850409347588, 238374187312038

Offset: 0

Author

Clark Kimberling, Aug 12 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 A289780 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 (7, -17, 23, -16, 6, -1)

Crossrefs

Cf. A000217, A289780.

Programs

Mathematica

z = 60; s = x/(1 - x)^3; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000217 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289797 *)
LinearRecurrence[{7,-17,23,-16,6,-1},{1,5,21,84,330,1291},30] (* Harvey P. Dale, Jul 10 2020 *)

Formula

G.f.: (1 - 2 x + 3 x^2 - x^3)/(1 - 7 x + 17 x^2 - 23 x^3 + 16 x^4 - 6 x^5 + x^6).

a(n) = 7*a(n-1) - 17*a(n-2) + 23*a(n-3) - 16*a(n-4) + 6*a(n-5) - a(n-6).