A290902 p-INVERT of the positive integers, where p(S) = 1 - 3*S.

3, 15, 72, 345, 1653, 7920, 37947, 181815, 871128, 4173825, 19997997, 95816160, 459082803, 2199597855, 10538906472, 50494934505, 241935766053, 1159183895760, 5553983712747, 26610734667975, 127499689627128, 610887713467665, 2926938877711197, 14023806675088320

Offset: 0

Author

Clark Kimberling, Aug 17 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.

Examples

			s = (1,2,3,4,...), p(S) = 1-3*S;
S(x) = x + 2 x^2 + 3 x^3 + ... ;
p(S(x)) = 1 - 3(x + 2 x^2 + 3 x^3 + ...);
1/p(S(x)) = 1 + 3 x + 15 x^2 + 72 x^3 + ... ;
(-p(0) + 1/p(S(x)))/x = 3 + 15 x + 72 x^2 + ... ;
t(s) = (3, 15, 72, ...), with offset 0.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (5, -1)

Crossrefs

Cf. A000027, A004254, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 3 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290902 *)
u/3 (* A004254 shifted *)

Formula

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

a(n) = 5*a(n-1) - a(n-2).

a(n) = 3*A004254(n+1) for n >= 0.