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%I #48 Sep 08 2022 08:46:04
%S 1,4,16,79,319,1576,6364,31441,126961,627244,2532856,12513439,
%T 50530159,249641536,1008070324,4980317281,20110876321,99356704084,
%U 401209456096,1982153764399,8004078245599,39543718583896,159680355455884,788892217913521
%N Numbers m such that 11*m^2 - 7 is a square.
%C See the first comment of A221762.
%C a(n) == 1 (mod 3).
%C a(n+1)/a(n) tends alternately to (2+sqrt(11))^2/7 and (5+sqrt(11))^2/14; a(n+2)/a(n) tends to A176395^2/2.
%C Positive values of x (or y) satisfying x^2 - 20xy + y^2 + 63 = 0. - _Colin Barker_, Feb 18 2014
%H Bruno Berselli, <a href="/A221763/b221763.txt">Table of n, a(n) for n = 1..500</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,20,0,-1).
%F G.f.: x*(1+4*x-4*x^2-x^3)/(1-20*x^2+x^4).
%F a(n) = ((11+2*t*(-1)^n)*(10-3*t)^floor(n/2)+(11-2*t*(-1)^n)*(10+3*t)^floor(n/2))/22, where t=sqrt(11).
%F a(n)*a(n-3)-a(n-1)*a(n-2) = (3/2)*(9+(-1)^n).
%p A221763:=proc(q)
%p local n;
%p for n from 1 to q do if type(sqrt(11*n^2-7), integer) then print(n);
%p fi; od; end:
%p A221763(1000); # _Paolo P. Lava_, Feb 19 2013
%t LinearRecurrence[{0, 20, 0, -1}, {1, 4, 16, 79}, 24]
%t CoefficientList[Series[(1 + 4 x - 4 x^2 - x^3)/(1 - 20 x^2 + x^4), {x, 0, 25}], x] (* _Vincenzo Librandi_, Aug 18 2013 *)
%o (Magma) m:=24; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+4*x-4*x^2-x^3)/(1-20*x^2+x^4)));
%o (Maxima) makelist(expand(((11+2*sqrt(11)*(-1)^n)*(10-3*sqrt(11))^floor(n/2)+(11-2*sqrt(11)*(-1)^n)*(10+3*sqrt(11))^floor(n/2))/22), n, 1, 24);
%o (Magma) I:=[1,4,16,79]; [n le 4 select I[n] else 20*Self(n-2)-Self(n-4): n in [1..25]]; // _Vincenzo Librandi_, Aug 18 2013
%Y Cf. A221762.
%K nonn,easy
%O 1,2
%A _Bruno Berselli_, Jan 24 2013
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