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a(n) = F(n+1)^2, if n>=0 is even (F=A000045) and a(n) = (L(2n+2)+8)/5, if n is odd (L=A000204).
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%I #27 Feb 13 2013 23:58:29

%S 1,3,4,11,25,66,169,443,1156,3027,7921,20738,54289,142131,372100,

%T 974171,2550409,6677058,17480761,45765227,119814916,313679523,

%U 821223649,2149991426,5628750625,14736260451,38580030724,101003831723,264431464441,692290561602

%N a(n) = F(n+1)^2, if n>=0 is even (F=A000045) and a(n) = (L(2n+2)+8)/5, if n is odd (L=A000204).

%C The row sums of Pascal-like triangle with the left side {1,1,1,...} and the right side (a(0), a(1), a(2),...) are F(n+2)^2 (cf. A007598).

%C Triangle begins:

%C n/k.|..0.....1.....2.....3.....4.....5.....6.....7

%C ==================================================

%C .0..|..1

%C .1..|..1.....3

%C .2..|..1.....4.....4

%C .3..|..1.....5.....8....11

%C .4..|..1.....6....13....19....25

%C .5..|..1.....7....19....32....44....66

%C .6..|..1.....8....26....51....76...110...164

%C .7..|..1.....9....34....77...127...186...279....443

%C .8..|

%C This sequence is associated with the identity arctan(1/(a+b)) = arctan(1/a) - arctan(b/(a^2+a*b+1)) (which is due to Euler, see the reference to Beckman), let a = F(n) and b = F(n+1). - _Gary Detlefs_, Apr 18 2012

%D Petr Beckman, The History of Pi, Golem Press, 1977, p. 154

%H Alois P. Heinz, <a href="/A208176/b208176.txt">Table of n, a(n) for n = 0..725</a>

%F G.f.: -(2*x^3-5*x^2+1)/(x^4-3*x^3+3*x-1). - _Alois P. Heinz_, Feb 24 2012

%F a(n) = F(n)^2 + F(n)*F(n+1) + 1. - _Gary Detlefs_, Apr 18 2012

%p a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <1|-3|0|3>>^n.

%p <<1,3,4,11>>)[1,1]:

%p seq (a(n), n=0..30); # _Alois P. Heinz_, Feb 24 2012

%t rows={{1},{1,3}}; Table[(x=Flatten[{1, 2MovingAverage[rows[[n]], 2]}]; z=If[EvenQ[n],Fibonacci[n+1]^2, (8+LucasL[(2n+2)])/5]; rows=Append[rows, Append[x,z]]), {n,2,15}]; A208176 = Map[Last[#] &, rows]

%Y Cf. A000045, A000204, A007598, A207890.

%K nonn

%O 0,2

%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Feb 24 2012