A317499 Coefficients in expansion of 1/(1 + 2*x - 3*x^3).

1, -2, 4, -5, 4, 4, -23, 58, -104, 139, -104, -104, 625, -1562, 2812, -3749, 2812, 2812, -16871, 42178, -75920, 101227, -75920, -75920, 455521, -1138802, 2049844, -2733125, 2049844, 2049844, -12299063, 30747658, -55345784, 73794379, -55345784, -55345784

Offset: 0

Author

Zagros Lalo, Jul 31 2018

Keywords

• sign
• easy

Comments

The coefficients in the expansion of 1/(1 + 2*x - 3*x^3) are given by the sequence generated by the row sums in triangle A317503.

Coefficients in expansion of 1/(1 + 2*x - 3*x^3) are given by the sum of numbers along second Layer skew diagonals pointing top-left in triangle A303901 ((3-2*x)^n) and by the sum of numbers along second Layer skew diagonals pointing top-right in triangle A317498 ((-2+3*x)^n), see links.

References

• Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 396, 397.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Zagros Lalo, Second layer skew diagonals in center-justified triangle of coefficients in expansion of (3 - 2x)^n
• Zagros Lalo, Second layer skew diagonals in center-justified triangle of coefficients in expansion of (-2 + 3x)^n
• Index entries for linear recurrences with constant coefficients, signature (-2,0,3).

Crossrefs

Cf. A317502, A317503.

Cf. A303901, A317498.

Programs

GAP

a:=[1,-2,4];; for n in [4..40] do a[n]:=-2*a[n-1]+3*a[n-3]; od; a; # Muniru A Asiru, Aug 01 2018

Maple

seq(coeff(series(1/(1+2*x-3*x^3), x,n+1),x,n),n=0..40); # Muniru A Asiru, Aug 01 2018

Mathematica

CoefficientList[Series[1/(1 + 2 x - 3 x^3), {x, 0, 40}], x]
a[0] = 1; a[n_] := a[n] = If[n < 0, 0, -2 * a[n - 1] + 3 * a[n - 3]]; Table[a[n], {n, 0, 40}] // Flatten
LinearRecurrence[{-2, 0, 3}, {1, -2, 4}, 41]

PARI

Vec(1 / ((1 - x)*(1 + 3*x + 3*x^2)) + O(x^40)) \\ Colin Barker, Aug 02 2018

Formula

a(0)=1, a(n) = -2*a(n-1) + 3*a(n-3) for n = 0,1...; a(n)=0 for n < 0.

a(n) = (2^(-n)*(2^n + (-3-i*sqrt(3))^n*(3-2*i*sqrt(3)) + (-3+i*sqrt(3))^n*(3+2*i*sqrt(3)))) / 7 where i=sqrt(-1). - Colin Barker, Aug 02 2018