A291182 p-INVERT of the positive integers, where p(S) = 1 - 3*S + 2*S^2.

3, 13, 52, 203, 781, 2976, 11267, 42469, 159596, 598499, 2241165, 8383872, 31340691, 117100285, 437378260, 1633244795, 6097779229, 22763575008, 84971451155, 317161317781, 1183776154124, 4418211213011, 16489770106653, 61542705525504, 229685859522339

Offset: 0

Author

Clark Kimberling, Aug 19 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 (7,-14,7,-1)

Crossrefs

Cf. A000027, A290890.

Programs

Magma

I:=[3,13,52,203]; [n le 4 select I[n] else 7*Self(n-1)-14*Self(n-2)+7*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Aug 20 2017

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 3 s + 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291182 *)
LinearRecurrence[{7, -14, 7, -1}, {3, 13, 52, 203}, 40] (* Vincenzo Librandi, Aug 20 2017 *)

Formula

G.f.: (3 - 8 x + 3 x^2)/(1 - 7 x + 14 x^2 - 7 x^3 + x^4).

a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3) - a(n-4).