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%I #32 Jul 26 2024 08:58:11
%S 6,9,21,26,44,51,75,84,114,125,161,174,216,231,279,296,350,369,429,
%T 450,516,539,611,636,714,741,825,854,944,975,1071,1104,1206,1241,1349,
%U 1386,1500,1539,1659,1700,1826,1869,2001,2046,2184,2231,2375,2424,2574,2625
%N Integers of the form k*(5+k)/4.
%C Integers of the form k+k*(k+1)/4 = k+A000217(k)/2; for k see A014601, for A000217(k)/2 see A074378.
%C Are all terms composite?
%C Yes, because a(2*k) = k*(4*k+5) and a(2*k-1) = (k+1)*(4*k-1). - _Bruno Berselli_, Apr 07 2013
%C Numbers m such that 16*m + 25 is a square. - _Vincenzo Librandi_, Apr 07 2013
%H Vincenzo Librandi, <a href="/A165717/b165717.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F From _R. J. Mathar_, Sep 25 2009: (Start)
%F a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
%F G.f.: x*(-6-3*x+x^3)/( (1+x)^2 * (x-1)^3 ). (End)
%F Sum_{n>=1} 1/a(n) = 29/25 - Pi/5. - _Amiram Eldar_, Jul 26 2024
%e For k =1,2,3,.. the value of k*(k+5)/4 is 3/2, 7/2, 6, 9, 25/2, 33/2, 21, 26, 63/2, 75/2, 44, 51,.. and the integer values define the sequence.
%t q=2;s=0;lst={};Do[s+=((n+q)/q);If[IntegerQ[s],AppendTo[lst,s]],{n,6!}];lst
%t Select[Table[k*(5+k)/4,{k,100}],IntegerQ] (* or *) LinearRecurrence[ {1,2,-2,-1,1},{6,9,21,26,44},60] (* _Harvey P. Dale_, Aug 11 2011 *)
%t Select[Range[1, 3000], IntegerQ[Sqrt[16 # + 25]]&] (* _Vincenzo Librandi_, Apr 07 2013 *)
%o (Magma) [n: n in [1..3000] | IsSquare(16*n+25)]; // _Vincenzo Librandi_, Apr 07 2013
%Y Cf. A000217, A014601, A074378.
%K nonn,easy
%O 1,1
%A _Vladimir Joseph Stephan Orlovsky_, Sep 24 2009
%E Definition simplified by _R. J. Mathar_, Sep 25 2009