1,2

We solve 0.5*(3*p^2+3*p+2)=0.5*(5*r^2+5*r+2) i.e 3*(2*p+1)^2=5*(2*r+1)^2-2.

The diophantine equation 3*X^2=5*Y^2-2is such that : X is given by A057080 which satisfies the new formula a(n+1)=4*a(n)+(15*a(n)^2+10)^0.5, Y is given by A070997 which satisfies the new formula a(n+1)=4*a(n)+(15*a(n)^2-6)^0.5 while r is given by the sequence 0,3,27,216,1704,... which satisfies a(n+2)=8*a(n+1)-a(n)+3 and a(n+1)=4*a(n)+1.5+0.5*(60*a(n)^2+60*a(n)+9)^0.5, p is given by the sequence 0,4,35,279,2200,... which satisfies a(n+2)=8*a(n+1)-a(n)+3 and a(n+1)=4*a(n)+1.5+0.5*(60*a(n)^2+60*a(n)+25)^0.5

a(n+2)=62*a(n+1)-a(n)-30, a(n+1)=31*a(n)-15+(960*a(n)^2-960*a(n)+225)^0.5. g.f.: f(z)=a(1)*z+a(2)*z^2+...=((z*(1-32*z+z^2))/((1-z)*(1-62*z+z^2))

A005891 INTERSECT A005448. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 24 2007

a(n)=(1/2)+(1/16)*sqrt(15)*[31+8*sqrt(15)]^n-(1/16)*sqrt(15)*[31-8*sqrt(15)]^n+(1/4)*[31+8 *sqrt(15)]^n+(1/4)*[31-8*sqrt(15)]^n, with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Aug 28 2008]

A131751 := proc(n) coeftayl(x*(1-32*x+x^2)/(1-x)/(1-62*x+x^2), x=0, n) ; end: seq(A131751(n), n=1..20) ; - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 24 2007

Cf. A050807 A070997.

Sequence in context: A049081 A069432 A055629 this_sequence A042863 A042860 A072019

Adjacent sequences: A131748 A131749 A131750 this_sequence A131752 A131753 A131754

nonn

Richard Choulet (richardchoulet(AT)yahoo.fr), Sep 20 2007

Corrected and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 24 2007