0,2

Partial sums of A106510 . Inverse binomial transform of A024495 (without leading zeros). [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 26 2008]

a(n) = A130196(n)-A022003(n) = A080425(n)-A130196(n)+2 = A153727(n)/A130196(n). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Nov 12 2009]

a(n)=(1/9)*{4*(n mod 3)+7*[(n+1) mod 3]+[(n+2) mod 3]}, with n>=0. - Paolo P. Lava (ppl(AT)spl.at), Aug 28 2007

G.f.: -(x+1)^2/(x-1)/(x^2+x+1). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 14 2007

a(n)=4/3+(2/3)*cos(2*pi*(n+2)/3) - Jaume Oliver Lafont (joliverlafont(AT)gmail.com), May 09 2008

a(n)=A101825(n+1). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 13 2008

a(n)=4/3+1/3*(-1/2-(1/2*I)*sqrt(3))^(-2)*(-1/2-(1/2*I)*sqrt(3))^n+1/3*(-1/2+(1/2*I) *sqrt(3))^(-2)*(-1/2+(1/2*I)*sqrt(3))^n+1/3*(-1/2-(1/2*I)*sqrt(3))^n+1/3*(-1/2 +(1/2*I)*sqrt(3))^n+2/3*(-1/2-(1/2*I)*sqrt(3))^(-1)*(-1/2-(1/2*I)*sqrt(3))^n+2/3 *(-1/2+(1/2*I)*sqrt(3))^(-1)*(-1/2+(1/2*I)*sqrt(3))^n, with n>=0 and I=sqrt(-1) [From Paolo P. Lava (ppl(AT)spl.at), Nov 27 2008]

G.f.: (1+x)^2/(1-x^3) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Mar 23 2009]

(PARI) a(n)=1+(n%3==1) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Mar 20 2009]

A167808. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Nov 12 2009]

Sequence in context: A101825 A057079 A087204 this_sequence A061347 A115579 A115573

Adjacent sequences: A131531 A131532 A131533 this_sequence A131535 A131536 A131537

nonn

Paul Curtz (bpcrtz(AT)free.fr), Aug 26 2007