A122600 Expansion of 1/(1 + 3*x - 4*x^2 + x^3).

1, -3, 13, -52, 211, -854, 3458, -14001, 56689, -229529, 929344, -3762837, 15235416, -61686940, 249765321, -1011279139, 4094585641, -16578638800, 67125538103, -271785755150, 1100438056662, -4455582728689, 18040286167865, -73043627475013, 295747609825188, -1197457625543481

Offset: 0

Author

Roger L. Bagula and Gary W. Adamson, Sep 20 2006

Keywords

• sign
• easy

Comments

Suggested by the Steinbach heptagon polynomial p^3 - 2*p^2*(1 - p) - p(1 - p)^2 + (1 - p)^3 = (1 - 4 p + 3 p^2 + p^3).

B(n):=|a(n-1)| = a(n-1)*(-1)^(n-1) with B(0):=0 (hence the o.g.f. for B(n) is x/(1 + 3*x - 4*x^2 + x^3)) appears in the following formula for the nonnegative powers of rho*sigma, where rho:=2*cos(Pi/7) and sigma:=sin(3*Pi/7)/sin(Pi/7)= rho^2-1 are the ratios of the smaller and larger diagonal length to the side length in a regular 7-gon (heptagon). See the Steinbach reference where the basis <1,rho,sigma> is used in an extension of the rational field. (rho*sigma)^n = C(n) + B(n)*rho + A(n)*sigma,n>=0, with C(n)= A120757(n) with C(0):=1, and A(n)= A181879(n). For the nonpositive powers see A085810*(-1)^n, A181880(n) and A116423(n)*(-1)^n, respectively. See also a comment under A052547.

Links

• Toufik Mansour and Mark Shattuck, Avoidance of vincular patterns by flattened derangements, arXiv:2504.14713 [math.CO], 2025. See p. 6.
• Peter Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.
• Index entries for linear recurrences with constant coefficients, signature (-3,4,-1).

Crossrefs

Cf. A065941.

Programs

Mathematica

p[x_] := 1 - 4 x + 3x^2 + x^3; q[x_] := ExpandAll[x^3*p[1/x]]; Table[ SeriesCoefficient[ Series[x/q[x], {x, 0, 30}], n], {n, 0, 30}]
CoefficientList[Series[1/(1 + 3*x - 4*x^2 + x^3), {x, 0, 50}], x] (* or *) LinearRecurrence[{-3, 4, -1}, {1, -3, 13}, 40] (* Vladimir Joseph Stephan Orlovsky, Jan 31 2012 *)

Formula

a(n)= -3*a(n-1) + 4*a(n-2) - a(n-3), n>=2, a(-1):=0, a(1)=0, a(1)=-3 (from the o.g.f. given in the name).

a(n) = (-1)^n*Sum_{k=0..n} binomial(n+k+2,3*k+2)*7^k. - Emanuele Munarini, Aug 27 2017

From Kai Wang, Jul 05 2020: (Start)

a(n) = Sum_{i+2j+3k=n} (-1)^(i+k)*3^i*4^j*((i+j+k)!)/(i!*j!*k!).

a(n) = (-1)^n*(6*A215076(n+4) - 21*A215076(n+3) - 13*A215076(n+2))/7. (End)

Extensions

Edited by N. J. A. Sloane, Feb 01 2007