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%I #7 Dec 06 2015 11:56:15
%S 1,1,2,5,10,23,48,107,228,501,1078,2353,5086,11067,23972,52087,112936,
%T 245225,531946,1154685,2505298,5437407,11798616,25605539,55563980,
%U 120581981,261668382,567850345,1232273510,2674156163,5803126348
%N To obtain a(n+1), take the square of the n-th partial sum, minus the sum of the first n squared terms, then divide this difference by a(n); for all n>1, starting with a(0)=1, a(1)=1.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-1).
%F a(n) = a(n-1) + 3a(n-2) - a(n-3) for n>3.
%F G.f.: (1-2x^2+x^3)/(1-x-3x^2+x^3).
%F G.f.: A052973(x)/(1+x-x^2).
%t CoefficientList[Series[(1-2x^2+x^3)/(1-x-3x^2+x^3),{x,0,40}],x] (* or *) LinearRecurrence[{1,3,-1},{1,1,2,5},40] (* _Harvey P. Dale_, Dec 06 2015 *)
%o (PARI) {a(n) = if(n<=1,1,( sum(k=0, n-1, a(k))^2 - sum(k=0, n-1, a(k)^2) )/a(n-1))}
%o for(n=0,40,print1(a(n),", "))
%Y Cf. A052973.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Sep 15 2003
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