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%I #8 Feb 10 2014 01:30:30
%S 0,1,50,2354,110595,5195620,244083556,11466731525,538692298134,
%T 25307071280790,1188893657899015,55852694849972936,
%U 2623887764290829000,123266872226818990089,5790919106896201705210,272049931151894661154810
%N One ninth of the sum of the squares of the first n Fibonacci numbers with index divisible by 4.
%C Natural bilateral extension (brackets mark index 0): ..., -110595, -2354, -50, -1, 0, [0], 1, 50, 2354, 110595, 5195620, ... This is (-A156087)-reversed followed by A156087. That is, A156087(-n) = -A156087(n-1).
%F Let F(n) be the Fibonacci number A000045(n).
%F a(n) = sum_{k=1..n} F(4k)^2.
%F Closed form: a(n) = F(8n+4)/135 - (2n + 1)/45.
%F Recurrence: a(n) - 48 a(n-1) + 48 a(n-2) - a(n-3) = 2.
%F Recurrence: a(n) - 49 a(n-1) + 96 a(n-2) - 49 a(n-3) + a(n-4) = 0.
%F G.f.: A(x) = (x + x^2)/(1 - 49 x + 96 x^2 - 49 x^3 + x^4) = x (1 + x)/((1 - x)^2 (1 - 47 x + x^2)).
%t a[n_Integer] := If[ n >= 0, Sum[ (1/9) Fibonacci[4k]^2, {k, 1, n} ], -Sum[ (1/9) Fibonacci[-4k]^2, {k, 1, -n - 1} ] ]
%t Accumulate[Fibonacci[4Range[0,20]]^2]/9 (* _Harvey P. Dale_, Sep 22 2011 *)
%Y Cf. A156086, A156092, A156093.
%K nonn,easy
%O 0,3
%A _Stuart Clary_, Feb 04 2009
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