A131751 Numbers that are both centered triangular and centered pentagonal.

1, 31, 1891, 117181, 7263301, 450207451, 27905598631, 1729696907641, 107213302675081, 6645495068947351, 411913480972060651, 25531990325198812981, 1582571486681354344141, 98093900183918770523731

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).

Links

Table of n, a(n) for n=1..14.

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)).

A005891 INTERSECT A005448. - R. J. Mathar, Oct 24 2007

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

Cf. A050807 A070997.

Sequence in context: A069432 A338533 A055629 * A042863 A042860 A184488

Adjacent sequences: A131748 A131749 A131750 * A131752 A131753 A131754

Keyword

nonn,easy

Author

Richard Choulet, Sep 20 2007

Extensions

Corrected and extended by R. J. Mathar, Oct 24 2007

Status

approved