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Numbers X such that (X^2-19)/57 is a square.
1

%I #20 Apr 07 2024 17:42:18

%S 38,11438,3454238,1043168438,315033414038,95139047871038,

%T 28731677423639438,8676871442891239238,2620386444075730610438,

%U 791348029239427753113038,238984484443863105709527038

%N Numbers X such that (X^2-19)/57 is a square.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (302,-1).

%F a(n) = 302*a(n-1) - a(n-2).

%F From _R. J. Mathar_, Nov 27 2011: (Start)

%F G.f.: -38*x*(-1+x) / ( 1-302*x+x^2 ).

%F a(n) = 38*A145123(n). (End)

%e a(1)=38 because 38^2 = 57*25+19.

%t LinearRecurrence[{302,-1},{38,11438},20] (* _Harvey P. Dale_, Jan 24 2013 *)

%K easy,nonn

%O 1,1

%A _Richard Choulet_, Oct 02 2008