cp's OEIS Frontend

A087640 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.

%I A087640 #19 Jun 12 2025 23:30:16
%S A087640 1,1,2,5,10,23,48,107,228,501,1078,2353,5086,11067,23972,52087,112936,
%T A087640 245225,531946,1154685,2505298,5437407,11798616,25605539,55563980,
%U A087640 120581981,261668382,567850345,1232273510,2674156163,5803126348
%N A087640 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.
%C A087640 For n > 0, a(n) counts the number of walks of length n-1 starting at vertex A in the undirected graph with edge set {(A, B), (A, C), (B, C), (C, D)}. - _Noah Niederklein_, Jun 07 2025
%H A087640 Sean A. Irvine, <a href="https://oeis.org/wiki/User:Sean_A._Irvine/Walks_on_Graphs#4_vertices">Walks on Graphs</a>
%H A087640 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-1).
%F A087640 a(n) = a(n-1) + 3a(n-2) - a(n-3) for n>3.
%F A087640 G.f.: (1-2x^2+x^3)/(1-x-3x^2+x^3).
%F A087640 G.f.: A052973(x)/(1+x-x^2).
%t A087640 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 A087640 (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 A087640 for(n=0,40,print1(a(n),", "))
%Y A087640 Cf. A052973.
%K A087640 nonn
%O A087640 0,3
%A A087640 _Paul D. Hanna_, Sep 15 2003