A291410 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^2 - 2 S^3.

1, 3, 9, 26, 69, 186, 511, 1401, 3824, 10438, 28521, 77938, 212928, 581700, 1589231, 4341911, 11862339, 32408429, 88541424, 241899801, 660882666, 1805564823, 4932894579, 13476918898, 36819627664, 100593093135, 274825440378, 750837066710, 2051325016200

Offset: 0

Author

Clark Kimberling, Sep 07 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.

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1, 2, 4, 7, 6, 2)

Crossrefs

Cf. A019590, A291382.

Programs

GAP

a:=[1,3,9,26,69,186];;  for n in [7..10^2] do a[n]:=a[n-1]+2*a[n-2]+4*a[n-3]+7*a[n-4]+6*a[n-5]+2*a[n-6]; od; a; # Muniru A Asiru, Sep 12 2017

Mathematica

z = 60; s = x + x^2; p = 1 - s - s^2 - 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291410 *)
CoefficientList[ Series[-(2x^5 +6x^4 +7x^3 +4x^2 +2x +1)/(2x^6 +6x^5 +7x^4 +4x^3 +2x^2 +x- 1), {x, 0, 28}], x] (* or *)
LinearRecurrence[{1, 2, 4, 7, 6, 2}, {1, 3, 9, 26, 69, 186}, 29] (* Robert G. Wilson v, Sep 25 2017 *)

Formula

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

a(n) = a(n-1) + 2*a(n-2) + 4*a(n-3) + 7*a(n-4) + 6*a(n-5) + 2*a(n-6) for n >= 7.