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%I #39 Sep 08 2022 08:45:09
%S 1,3,7,29,69,287,683,2841,6761,28123,66927,278389,662509,2755767,
%T 6558163,27279281,64919121,270037043,642633047,2673091149,6361411349,
%U 26460874447,62971480443,261935653321,623353393081,2592895658763
%N Positive integer values of n such that 6n^2-5 is a square.
%C The corresponding sequence for which 6n^2-6 is a square is A001079.
%C Positive values of x (or y) satisfying x^2 - 10xy + y^2 + 20 = 0. - _Colin Barker_, Feb 09 2014
%H Vincenzo Librandi, <a href="/A080806/b080806.txt">Table of n, a(n) for n = 1..200</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PellEquation.html">Pell Equation.</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,10,0,-1).
%F a(n) = 10*a(n-2)-a(n-4).
%F G.f.: x*(1-x)*(1+4*x+x^2)/(1-10*x^2+x^4). - _Colin Barker_, Jun 13 2012
%F a(2*n+1) = ((6+r)*(5+2*r)^n+(6-r)*(5+2*r)^n)/12, a(2*n+2) = ((18+7*r)*(5+2*r)^n+(18-7*r)*(5-2*r)^n)/12, where r=sqrt(6) and n>=0. - _Paul Weisenhorn_, Sep 01 2012
%e 29 is a term of the sequence since 6*29^2 - 5 = 5041 = 71^2.
%t Do[ If[ IntegerQ[ Sqrt[6n^2 - 5]], Print[n]], {n, 1, 3*10^7}]
%t a[1]=1; a[2]=3; a[3]=7; a[4]=29; a[n_] := a[n]=10a[n-2]-a[n-4]
%t CoefficientList[Series[(1 - x) (1 + 4 x + x^2)/(1 - 10 x^2 + x^4), {x, 0, 30}], x] (* _Vincenzo Librandi_, Feb 10 2014 *)
%o (Magma) I:=[1,3,7,29]; [n le 4 select I[n] else 10*Self(n-2)-Self(n-4): n in [1..30]]; // _Vincenzo Librandi_, Feb 10 2014
%Y Cf. A001079, A077288.
%K nonn,easy
%O 1,2
%A _John W. Layman_, Mar 24 2003
%E Extended by _Robert G. Wilson v_, Apr 14 2003
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