A291389 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 5 S)^2.

10, 85, 650, 4700, 32750, 222375, 1481250, 9721875, 63062500, 405175000, 2582687500, 16353078125, 102955156250, 644991640625, 4023367968750, 25002220312500, 154848222656250, 956155732421875, 5888138769531250, 36171585068359375, 221714776953125000

Offset: 0

Author

Clark Kimberling, Sep 04 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 A291382 for a guide to related sequences.

All terms = 0 mod 5. - Muniru A Asiru, Sep 07 2017

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (10, -15, -50, -25)

Crossrefs

Cf. A019590, A291382, A291390.

Programs

GAP

a:=5*[2,17,130,940];; for n in [5..10^2] do a[n]:=10*a[n-1]-15*a[n-2]-50*a[n-3]-25*a[n-4]; od; a;  # Muniru A Asiru, Sep 07 2017

Mathematica

z = 60; s = x + x^2; p = (1 - 5 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291389 *)
u / 5  (* A291390 *)

Formula

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

a(n) = 10*a(n-1) - 15*a(n-2) - 50*a(n-3) - 25*a(n-4) for n >= 5.