A291028 p-INVERT of the positive integers, where p(S) = 1 - 6*S + S^2.

6, 47, 362, 2787, 21456, 165180, 1271644, 9789793, 75367038, 580215573, 4466808294, 34387867640, 264736107506, 2038079457267, 15690220398162, 120791667500967, 929918545909756, 7159007901103540, 55113853093361544, 424295774604244773, 3266454697733704038

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 (10,-19,10,-1).

Crossrefs

Cf. A000027, A290890.

Programs

Magma

I:=[6,47,362,2787]; [n le 4 select I[n] else 10*Self(n-1)-19*Self(n-2)+10*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 - 6 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291028 *)
LinearRecurrence[{10, -19, 10, -1}, {6, 47, 362, 2787}, 40] (* Vincenzo Librandi, Aug 20 2017 *)

Formula

G.f.: (6 - 13 x + 6 x^2)/(1 - 10 x + 19 x^2 - 10 x^3 + x^4).

a(n) = 10*a(n-1) - 19*a(n-2) + 10*a(n-3) - a(n-4).