A087640 - OEIS

The OEIS is supported by the many generous donors to the OEIS Foundation.

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.

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

Table of n, a(n) for n=0..30.

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: A365441 A260744 A317535 * A116953 A099516 A293741

Adjacent sequences: A087637 A087638 A087639 * A087641 A087642 A087643

KEYWORD

nonn

AUTHOR

Paul D. Hanna, Sep 15 2003

STATUS

approved