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%I #40 Nov 01 2024 05:11:03
%S 1,-3,7,-18,47,-123,322,-843,2207,-5778,15127,-39603,103682,-271443,
%T 710647,-1860498,4870847,-12752043,33385282,-87403803,228826127,
%U -599074578,1568397607,-4106118243,10749957122,-28143753123,73681302247,-192900153618,505019158607
%N Alternating row sums of Riordan triangle A110162.
%C If a(0) is put to 2 instead of 1 this becomes a(n) = (-1)^n*A005248(n), n >= 0. These are then the alternating row sums of triangle A127677.
%C Also abs(a(n)) is the number of rounded area of pentagon or pentagram in series arrangement. - _Kival Ngaokrajang_, Mar 27 2013
%H Colin Barker, <a href="/A219233/b219233.txt">Table of n, a(n) for n = 0..1000</a>
%H Richard M. Low and Ardak Kapbasov, <a href="https://www.emis.de/journals/JIS/VOL20/Low/low2.html">Non-Attacking Bishop and King Positions on Regular and Cylindrical Chessboards</a>, Journal of Integer Sequences, Vol. 20 (2017), Article 17.6.1, Table 8.
%H Kival Ngaokrajang, <a href="/A219233/a219233.jpg">Pentagram for n = 1..6</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Pentagram.html">Pentagram</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (-3,-1).
%F a(0) = 1 and a(n) = (-1)^n*(F(2*(n+1)) - F(2*(n-1))) = (-1)^n*L(2*n), n>=1, with F=A000045 (Fibonacci) and L=A000032 (Lucas).
%F O.g.f.: (1-x^2)/(1+3*x+x^2).
%F G.f.: (W(0) -6)/(5*x) -1 , where W(k) = 5*x*k + x + 6 - 6*x*(5*k-9)/W(k+1) ; (continued fraction). - _Sergei N. Gladkovskii_, Aug 19 2013
%F From _Colin Barker_, Oct 14 2015: (Start)
%F a(n) = -3*a(n-1) - a(n-2) for n>2.
%F a(n) = (1/2*(-3-sqrt(5)))^n+(1/2*(-3+sqrt(5)))^n for n>0.
%F (End)
%F E.g.f.: 2*exp(-3*x/2)*cosh(sqrt(5)*x/2) - 1. - _Stefano Spezia_, Dec 26 2021
%o (PARI) Vec((1-x^2)/(1+3*x+x^2) + O(x^40)) \\ _Colin Barker_, Oct 14 2015
%Y Cf. A099837 (row sums of A110162).
%Y Cf. A000032, A000045, A005248, A127677.
%K sign,easy
%O 0,2
%A _Wolfdieter Lang_, Nov 16 2012