1,2

We solve r^2+(r+1)^2=0.5*(3*p^2+3*p+2) equivalent to (4*r+2)^2=3*(2*p+1)^2+1.

The diophantine equation X^2=3*Y^2+1 gives X by A001075 and Y by A013453. The return to r gives the sequence 0,6,90,1260,17556,... which satisfies the formulae a(n+2)=14*a(n+1)-a(n)+6 and a(n+1)=7*a(n)+3+(48*a(n)^2+48*a(n)+9)^0.5 and the return to p the sequence A001921 which verifies this new relation : a(n+1)=7*a(n)+(48*a(n)^2+48*a(n)+16)^0.5. Then we obtain the present sequence

a(n+2)=194*a(n+1)-a(n)-108 a(n+1)=97*a(n)-54+14*(48*a(n)^2-54*a(n)+15)^0.5 G.f.: h(z)=a(1)z+a(2)*z^2+...=((z*(1-110*z+z^2)/((1-z)*(1-194*z+z^2))

a(n)=(9/16)+(7/32)*[97-56*sqrt(3)]^n+(7/32)*[97+56*sqrt(3)]^n-(1/8)*[97-56*sqrt(3)]^n*sqrt(3) +(1/8)*sqrt(3)*[97+56*sqrt(3)]^n, with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Sep 26 2008]

A131750 := proc(n) coeftayl(x*(1-110*x+x^2)/(1-x)/(1-194*x+x^2), x=0, n) ; end: seq(A131750(n), n=1..20) ; - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 24 2007

Cf. A001921 A001075 A001353.

Sequence in context: A011813 A006106 A015338 this_sequence A068749 A045916 A033406

Adjacent sequences: A131747 A131748 A131749 this_sequence A131751 A131752 A131753

nonn

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

More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 24 2007