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<a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-1).

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 *)

for(n=0, 40, print1(a(n), ", "))

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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); A(x) = A052973(x)/(1+x-x^2).

G.f.: (1-2x^2+x^3)/(1-x-3x^2+x^3).

G.f.: A052973(x)/(1+x-x^2).

(PARI) {a(n) = if(n<=1, 1, ( sum(k=1, 0, n-1, a(k))^2 - sum(k=1, 0, n-1, a(k)^2) )/a(n-1))}

for(n=0, 40, print1(a(n), ", "))

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_Paul D. Hanna (pauldhanna(AT)juno.com), _, Sep 15 2003

nonn,new

nonn

Paul D . Hanna (pauldhanna(AT)juno.com), Sep 15 2003

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

0,3

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); A(x) = A052973(x)/(1+x-x^2).

(PARI) a(n) = ( sum(k=1, n-1, a(k))^2 - sum(k=1, n-1, a(k)^2) )/a(n-1)

Cf. A052973.

nonn

Paul D Hanna (pauldhanna(AT)juno.com), Sep 15 2003

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