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 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. 4
 1, 1, 2, 5, 10, 23, 48, 107, 228, 501, 1078, 2353, 5086, 11067, 23972, 52087, 112936, 245225, 531946, 1154685, 2505298, 5437407, 11798616, 25605539, 55563980, 120581981, 261668382, 567850345, 1232273510, 2674156163, 5803126348 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 LINKS Index entries for linear recurrences with constant coefficients, signature (1,3,-1). FORMULA a(n) = a(n-1) + 3a(n-2) - a(n-3) for n>3. G.f.: (1-2x^2+x^3)/(1-x-3x^2+x^3). G.f.: A052973(x)/(1+x-x^2). MATHEMATICA 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 *) PROG (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))} for(n=0, 40, print1(a(n), ", ")) CROSSREFS Cf. A052973. Sequence in context: A291249 A260744 A317535 * A116953 A099516 A293741 Adjacent sequences:  A087637 A087638 A087639 * A087641 A087642 A087643 KEYWORD nonn AUTHOR Paul D. Hanna, Sep 15 2003 STATUS approved

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Last modified June 15 17:41 EDT 2021. Contains 345049 sequences. (Running on oeis4.)