A165717 Integers of the form k*(5+k)/4.

6, 9, 21, 26, 44, 51, 75, 84, 114, 125, 161, 174, 216, 231, 279, 296, 350, 369, 429, 450, 516, 539, 611, 636, 714, 741, 825, 854, 944, 975, 1071, 1104, 1206, 1241, 1349, 1386, 1500, 1539, 1659, 1700, 1826, 1869, 2001, 2046, 2184, 2231, 2375, 2424, 2574, 2625

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Sep 24 2009

Keywords

• nonn
• easy

Comments

Integers of the form k+k*(k+1)/4 = k+A000217(k)/2; for k see A014601, for A000217(k)/2 see A074378.

Are all terms composite?

Yes, because a(2*k) = k*(4*k+5) and a(2*k-1) = (k+1)*(4*k-1). - Bruno Berselli, Apr 07 2013

Numbers m such that 16*m + 25 is a square. - Vincenzo Librandi, Apr 07 2013

Examples

			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.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A000217, A014601, A074378.

Programs

Magma

[n: n in [1..3000] | IsSquare(16*n+25)]; // Vincenzo Librandi, Apr 07 2013

Mathematica

q=2;s=0;lst={};Do[s+=((n+q)/q);If[IntegerQ[s],AppendTo[lst,s]],{n,6!}];lst
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 *)
Select[Range[1, 3000], IntegerQ[Sqrt[16 # + 25]]&] (* Vincenzo Librandi, Apr 07 2013 *)

Formula

From R. J. Mathar, Sep 25 2009: (Start)

a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).

G.f.: x*(-6-3*x+x^3)/( (1+x)^2 * (x-1)^3 ). (End)

Sum_{n>=1} 1/a(n) = 29/25 - Pi/5. - Amiram Eldar, Jul 26 2024

Extensions

Definition simplified by R. J. Mathar, Sep 25 2009