A291003 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S)(1 - 2*S)(1 - 3*S)(1 - 4*S).

10, 75, 490, 2956, 16944, 93800, 506600, 2687256, 14064904, 72873880, 374671560, 1914880856, 9741440264, 49378177560, 249583291720, 1258711575256, 6336814854024, 31857331730840, 159980377179080, 802678826106456, 4024508089842184, 20167014882109720

Offset

0,1

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 A291000 for a guide to related sequences.

Links

Clark Kimberling, Table of n, a(n) for n = 0..999

Index entries for linear recurrences with constant coefficients, signature (14,-71,154,-120).

Formula

G.f.: (10 - 65*x + 150*x^2 - 119*x^3)/(1 - 14*x + 71*x^2 - 154*x^3 + 120*x^4).

a(n) = 14*a(n-1) - 71*a(n-2) + 154*a(n-3) - 120*a(n-4) for n >= 5.

a(n) = (-2^n + 16*3^(1+n) - 243*4^n + 256*5^n) / 6. - Colin Barker, Aug 23 2017

Mathematica

z = 60; s = x/(1-x); p = (1 - s)(1 - 2 s)(1 - 3 s)(1 - 4 s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291003 *)
LinearRecurrence[{14, -71, 154, -120}, {10, 75, 490, 2956}, 41] (* G. C. Greubel, May 23 2023 *)

Prog

(Magma) [(-2^n +48*3^n -243*4^n +256*5^n)/6: n in [0..40]]; // G. C. Greubel, May 23 2023
(SageMath) [(-2^n +48*3^n -243*4^n +256*5^n)//6 for n in range(41)] # G. C. Greubel, May 23 2023

Crossrefs

Cf. A000012, A289780, A291000.

Sequence in context: A316462 A022734 A027203 * A305784 A053464 A111998

Adjacent sequences: A291000 A291001 A291002 * A291004 A291005 A291006

Keyword

nonn,easy

Author

Clark Kimberling, Aug 22 2017

Status

approved