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A131750 Numbers that are both centered triangular and centered square.

%I A131750 #44 Dec 16 2023 17:58:36
%S A131750 1,85,16381,3177721,616461385,119590330861,23199907725541,
%T A131750 4500662508423985,873105326726527441,169377932722437899461,
%U A131750 32858445842826225967885,6374369115575565399870121
%N A131750 Numbers that are both centered triangular and centered square.
%C A131750 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 A131750 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 A131750 Robert Israel, <a href="/A131750/b131750.txt">Table of n, a(n) for n = 1..394</a>
%H A131750 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (195,-195,1).
%F A131750 a(n+2) = 195*a(n+1)-195*a(n)+a(n-1).
%F A131750 a(n+1) = 97*a(n) - 54 + 14*sqrt(48*a(n)^2-54*a(n)+15).
%F A131750 G.f.: x*(1-110*x+x^2)/((1-x)*(1-194*x+x^2)).
%p A131750 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 A131750 LinearRecurrence[{195,-195,1},{1,85,16381},20] (* _Harvey P. Dale_, Apr 26 2018 *)
%o A131750 (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
%Y A131750 Cf. A001921, A001075, A001353.
%Y A131750 Intersection of A001844 and A005448.
%K A131750 nonn
%O A131750 1,2
%A A131750 _Richard Choulet_, Sep 20 2007
%E A131750 More terms from _R. J. Mathar_, Oct 24 2007
%E A131750 Recurrences corrected by _Robert Israel_, Aug 26 2015
%E A131750 Name corrected by _Daniel Poveda Parrilla_, Sep 19 2016