A013453 -id:A013453 - cp's OEIS frontend

A131750 Numbers that are both centered triangular and centered square.

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

Offset: 1

Richard Choulet, Sep 20 2007

nonn

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.

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

Cf. A001921, A001075, A001353.

Intersection of A001844 and A005448.

[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

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

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

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)).

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

cp's OEIS Frontend

A131750 Numbers that are both centered triangular and centered square.

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