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Numbers k such that 2*k^2 + 17 is a square.
2

%I #32 Jul 25 2021 13:57:24

%S 2,4,16,26,94,152,548,886,3194,5164,18616,30098,108502,175424,632396,

%T 1022446,3685874,5959252,21482848,34733066,125211214,202439144,

%U 729784436,1179901798,4253495402,6876971644,24791187976,40081928066,144493632454,233614596752

%N Numbers k such that 2*k^2 + 17 is a square.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,0,-1).

%F G.f.: 2*x*(1+x)*(1+x+x^2) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - _R. J. Mathar_, Nov 27 2011

%F a(n) = 2*A077241(n-1). - _R. J. Mathar_, Nov 27 2011

%F a(n) = 6*a(n-2) - a(n-4). - _Colin Barker_, Oct 20 2014

%e a(1)=2 because 2*4 + 17 = 25 = 5^2.

%t Select[Range[6000000],IntegerQ[Sqrt[2#^2+17]]&] (* _Harvey P. Dale_, Aug 18 2012 *)

%t LinearRecurrence[{0, 6, 0, -1}, 2{1, 2, 8, 13}, 30] (* _Robert G. Wilson v_, Dec 02 2014 *)

%o (PARI) Vec(2*x*(1+x)*(1+x+x^2) / ((x^2+2*x-1)*(x^2-2*x-1)) + O(x^50)) \\ _Colin Barker_, Oct 20 2014

%Y Cf. A133301.

%K nonn,easy

%O 1,1

%A _Richard Choulet_, Sep 21 2008

%E Corrected by _R. J. Mathar_, Nov 27 2011

%E Editing and more terms from _Colin Barker_, Oct 20 2014