A131750 Numbers that are both centered triangular and centered square.

1, 85, 16381, 3177721, 616461385, 119590330861, 23199907725541, 4500662508423985, 873105326726527441, 169377932722437899461, 32858445842826225967885, 6374369115575565399870121

Offset: 1

Author

Richard Choulet, Sep 20 2007

Keywords

• nonn

Comments

We solve r^2+(r+1)^2=0.5*(3*p^2+3*p+2), which is 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 formulas 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 satisfies this new relation: a(n+1)=7*a(n)+sqrt(48*a(n)^2+48*a(n)+16). Then we obtain the present sequence.

Links

• Robert Israel, Table of n, a(n) for n = 1..394
• Index entries for linear recurrences with constant coefficients, signature (195,-195,1).

Crossrefs

Cf. A001921, A001075, A001353.

Intersection of A001844 and A005448.

Programs

Magma

[n le 2 select 1 else Floor(97*Self(n-2) - 54 + 14*Sqrt(48*Self(n-2)^2-54*Self(n-2)+15)): n in [2..30]]; // Vincenzo Librandi, Aug 26 2015

Maple

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, Oct 24 2007

Mathematica

LinearRecurrence[{195,-195,1},{1,85,16381},20] (* Harvey P. Dale, Apr 26 2018 *)

Formula

a(n+2) = 195*a(n+1)-195*a(n)+a(n-1).

a(n+1) = 97*a(n) - 54 + 14*sqrt(48*a(n)^2-54*a(n)+15).

G.f.: x*(1-110*x+x^2)/((1-x)*(1-194*x+x^2)).

Extensions

More terms from R. J. Mathar, Oct 24 2007

Recurrences corrected by Robert Israel, Aug 26 2015

Name corrected by Daniel Poveda Parrilla, Sep 19 2016