0,2

In general, for the recurrence a(n)=a(n-1)*a(n-2)/a(n-3), all terms are integers iff a(0) divides a(2) and first three terms are positive integers, since a(2n+k)=a(k)*(a(2)/a(0))^n for all nonnegative integers n and k

a(n) = A140740(n+2,2). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), May 26 2008

See A133626 for an essentially identical sequence. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 08 2008

T. D. Noe, Table of n, a(n) for n=0..400

Index entries for sequences related to linear recurrences with constant coefficients

a(n)=a(n-1)*a(n-2)/a(n-3) with a(0)=1, a(1)=2, a(2)=3.

a(2n)=(3/2)*a(2n-1)=3^n, a(2n+1)=2*a(2n)=2*3^n.

a(1)=1, a(n)=a(n-1)+1 if a(n-1) is odd, or a(n-1)=3/2*a(n-1) if a(n-1) is even. - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 27 2003

a(n)=(1/6)*(5-(-1)^n)*3^floor(n/2); a(2n)=a(2n-1)+a(2n-2)+a(2n-3); a(2n+1)=a(2n)+a(2n-1) - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 27 2003

G.f.: (1+2x)/(1-3x^2) - Paul Barry (pbarry(AT)wit.ie), Aug 25 2003

a(n) = (1 + n mod 2) * 3^floor(n/2). a(n) = A087503(n) - A087503(n-1). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Sep 11 2003

a(n)=sqrt(3)(2+sqrt(3))(sqrt(3))^n/6-sqrt(3)(2-sqrt(3))(-sqrt(3))^n/6 - Paul Barry (pbarry(AT)wit.ie), Sep 16 2003

a(n+1) = a(n) + a(n - n mod 2). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), May 26 2008

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=3*a[n-2]+2 od: seq(a[n]+1, n=0..34); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 20 2008

(PARI) a(n)=(1/6)*(5-(-1)^n)*3^floor(n/2)

Cf. Somos sequences A006720, A006721, A006722, A006723.

a(n) = A094718(5, n).

Cf. A000079, A133464, A140730, A037124.

Sequence in context: A035522 A018311 A018481 this_sequence A133626 A165647 A066313

Adjacent sequences: A038751 A038752 A038753 this_sequence A038755 A038756 A038757

easy,nice,nonn

Henry Bottomley (se16(AT)btinternet.com), May 03 2000