A131750 - OEIS

%I #46 May 26 2026 00:52:16

%S 1,85,16381,3177721,616461385,119590330861,23199907725541,

%T 4500662508423985,873105326726527441,169377932722437899461,

%U 32858445842826225967885,6374369115575565399870121

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

%C 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.

%C 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.

%H Robert Israel, <a href="/A131750/b131750.txt">Table of n, a(n) for n = 1..394</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (195,-195,1).

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

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

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

%p 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

%t LinearRecurrence[{195,-195,1},{1,85,16381},20] (* _Harvey P. Dale_, Apr 26 2018 *)

%o (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

%o (PARI) a(n)=([0,1,0; 0,0,1; 1,-195,195]^(n-1)*[1;85;16381])[1,1] \\ _Charles R Greathouse IV_, May 26 2026

%Y Cf. A001921, A001075, A001353, A013453.

%Y Intersection of A001844 and A005448.

%K nonn,easy

%O 1,2

%A _Richard Choulet_, Sep 20 2007

%E More terms from _R. J. Mathar_, Oct 24 2007

%E Recurrences corrected by _Robert Israel_, Aug 26 2015

%E Name corrected by _Daniel Poveda Parrilla_, Sep 19 2016