A290922 p-INVERT of the positive integers, where p(S) = 1 - S - 2*S^2.

1, 5, 20, 75, 279, 1040, 3881, 14485, 54060, 201755, 752959, 2810080, 10487361, 39139365, 146070100, 545141035, 2034494039, 7592835120, 28336846441, 105754550645, 394681356140, 1472970873915, 5497202139519, 20515837684160, 76566148597121, 285748756704325

Offset: 0

Author

Clark Kimberling, Aug 18 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 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 (5, -6, 5, -1)

Crossrefs

Cf. A000027, A290890.

Programs

Magma

I:=[1,5,20,75]; [n le 4 select I[n] else 5*Self(n-1)- 6*Self(n-2)+5*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Aug 19 2017

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290922 *)
LinearRecurrence[{5, -6, 5, -1}, {1, 5, 20, 75}, 30] (* Vincenzo Librandi, Aug 19 2017 *)

Formula

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

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