This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A110512 #43 Jul 22 2025 15:57:36
%S A110512 1,0,-2,2,2,-6,2,10,-14,-6,34,-22,-46,90,2,-182,178,186,-542,170,914,
%T A110512 -1254,-574,3082,-1934,-4230,8098,362,-16558,15834,17282,-48950,14386,
%U A110512 83514,-112286,-54742,279314,-169830,-388798,728458
%N A110512 Expansion of (1 + x)/(1 + x + 2x^2).
%C A110512 Row sums of number triangle A110511.
%C A110512 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
%C A110512 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
%D A110512 R. Witula, On some applications of formulas for unimodular complex numbers, Jacek Skalmierski's Press, Gliwice 2011 (in Polish).
%H A110512 G. C. Greubel, <a href="/A110512/b110512.txt">Table of n, a(n) for n = 0..1000</a>
%H A110512 Tejo V. Madhavarapu, <a href="https://arxiv.org/abs/2407.09000">The Most Malicious Maître D'</a>, arXiv:2407.09000 [math.CO], 2024. See p. 3.
%H A110512 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (-1,-2).
%F A110512 a(n) = Sum_{k=0..n} Sum_{j=0..n} (-1)^(n-j)*C(n, j)*(-2)^(j-k)*C(k, j-k).
%F A110512 a(n) = (-1)^n*A078020(n). - _R. J. Mathar_, Feb 04 2009
%F A110512 a(n+2) + a(n+1) + 2*a(n) = 0. - _Roman Witula_, Jul 27 2012
%F A110512 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
%F A110512 From _Ammar Khatab_, Jul 11 2025: (Start)
%F A110512 a(n) = ((-sqrt(2))^(n+3)/sqrt(7)) * sin((n-1) * arctan(sqrt(7))).
%F A110512 x^n = A001607(n) * x + a(n) in Z[x]/(x^2 + x + 2).
%F A110512 a(n) = -2 * A001607(n-1), for n > 0. (End)
%t A110512 CoefficientList[Series[(1 + x)/(1 + x + 2*x^2), {x,0,50}], x] (* _G. C. Greubel_, Aug 29 2017 *)
%t A110512 LinearRecurrence[{-1,-2},{1,0},40] (* _Harvey P. Dale_, Dec 30 2024 *)
%o A110512 (PARI) my(x='x+O('x^50)); Vec((1 + x)/(1 + x + 2*x^2)) \\ _G. C. Greubel_, Aug 29 2017
%K A110512 easy,sign
%O A110512 0,3
%A A110512 _Paul Barry_, Jul 24 2005