A110512 Expansion of (1 + x)/(1 + x + 2x^2).

1, 0, -2, 2, 2, -6, 2, 10, -14, -6, 34, -22, -46, 90, 2, -182, 178, 186, -542, 170, 914, -1254, -574, 3082, -1934, -4230, 8098, 362, -16558, 15834, 17282, -48950, 14386, 83514, -112286, -54742, 279314, -169830, -388798, 728458

Offset: 0

Author

Paul Barry, Jul 24 2005

Keywords

• easy
• sign

Comments

Row sums of number triangle A110511.

The sequences A110512 and A001607 are conjugated by one of the relations ((-1 + i*sqrt(7))/2)^n = a(n) + A001607(n)*(-1 + i*sqrt(7))/2 or ((-1 - i*sqrt(7))/2)^n = a(n) + A001607(n)*(-1 - i*sqrt(7))/2. These relations are connected with the Gauss sums; for example, if e := exp(i*2Pi/7) then e + e^2 + e^4 = (-1 + i*sqrt(7))/2 and e^3 + e^5 + e^6 = (-1 - i*sqrt(7))/2 -- for details see Witula's book. We also have a(n+1) = -2*A001607(n), which implies the Binet formula for a(n) (from the respective Binet formula for A001607(n) given in A001607), and A001607(n+1) = a(n) - A001607(n). - Roman Witula, Jul 27 2012

Pisano period lengths: 1, 1, 8, 1, 24, 8, 42, 1, 24, 24, 10, 8, 168, 42, 24, 2, 144, 24, 360, 24, ... - R. J. Mathar, Aug 10 2012

References

• R. Witula, On some applications of formulas for unimodular complex numbers, Jacek Skalmierski's Press, Gliwice 2011 (in Polish).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Tejo V. Madhavarapu, The Most Malicious Maître D', arXiv:2407.09000 [math.CO], 2024. See p. 3.
• Index entries for linear recurrences with constant coefficients, signature (-1,-2).

Programs

Mathematica

CoefficientList[Series[(1 + x)/(1 + x + 2*x^2), {x,0,50}], x] (* G. C. Greubel, Aug 29 2017 *)
LinearRecurrence[{-1,-2},{1,0},40] (* Harvey P. Dale, Dec 30 2024 *)

PARI

my(x='x+O('x^50)); Vec((1 + x)/(1 + x + 2*x^2)) \\ G. C. Greubel, Aug 29 2017

Formula

a(n) = Sum_{k=0..n} Sum_{j=0..n} (-1)^(n-j)*C(n, j)*(-2)^(j-k)*C(k, j-k).

a(n) = (-1)^n*A078020(n). - R. J. Mathar, Feb 04 2009

a(n+2) + a(n+1) + 2*a(n) = 0. - Roman Witula, Jul 27 2012

G.f.: 2 - x + 2*x^2 + 3*x/Q(0), where Q(k)= 1 - 1/(4^k - 2*x*16^k/(2*x*4^k + 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013

From Ammar Khatab, Jul 11 2025: (Start)

a(n) = ((-sqrt(2))^(n+3)/sqrt(7)) * sin((n-1) * arctan(sqrt(7))).

x^n = A001607(n) * x + a(n) in Z[x]/(x^2 + x + 2).

a(n) = -2 * A001607(n-1), for n > 0. (End)