OFFSET
1,2
COMMENTS
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*sqrt(60*a(n)^2+60*a(n)+25).
REFERENCES
Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing (2012), page 420.
LINKS
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for linear recurrences with constant coefficients, signature (63,-63,1).
FORMULA
a(n+2) = 62*a(n+1) - a(n) - 30, a(n+1) = 31*a(n) - 15 + sqrt(960*a(n)^2 - 960*a(n)+225).
G.f.: f(z) = a(1)*z+a(2)*z^2+... = z*(1-32*z+z^2)/((1-z)*(1-62*z+z^2)).
E.g.f.: exp(x)/2 - 1 + exp(31*x)*(4*cosh(8*sqrt(15)*x) - sqrt(15)*sinh(8*sqrt(15)*x))/8. - Stefano Spezia, Nov 04 2025
MAPLE
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, Oct 24 2007
MATHEMATICA
LinearRecurrence[{63, -63, 1}, {1, 31, 1891}, 20] (* Harvey P. Dale, Oct 01 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Richard Choulet, Sep 20 2007
EXTENSIONS
Corrected and extended by R. J. Mathar, Oct 24 2007
STATUS
approved