A131751 - OEIS

%I #24 Dec 16 2023 18:00:47

%S 1,31,1891,117181,7263301,450207451,27905598631,1729696907641,

%T 107213302675081,6645495068947351,411913480972060651,

%U 25531990325198812981,1582571486681354344141,98093900183918770523731

%N Numbers that are both centered triangular and centered pentagonal.

%C 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.

%C 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).

%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2016volume16/FG2016volume16.pdf#page=423">Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences</a>, Forum Geometricorum, Volume 16 (2016) 419-427.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (63,-63,1).

%F 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).

%F G.f.: f(z)=a(1)*z+a(2)*z^2+...=((z*(1-32*z+z^2))/((1-z)*(1-62*z+z^2)).

%F A005891 INTERSECT A005448. - _R. J. Mathar_, Oct 24 2007

%p 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

%t LinearRecurrence[{63,-63,1},{1,31,1891},20] (* _Harvey P. Dale_, Oct 01 2017 *)

%Y Cf. A050807 A070997.

%K nonn,easy

%O 1,2

%A _Richard Choulet_, Sep 20 2007

%E Corrected and extended by _R. J. Mathar_, Oct 24 2007