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

1,1

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

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).

Formula

a(n)= a(n-1) +2*a(n-2) -2*a(n-3) -a(n-4) +a(n-5). - R. J. Mathar, Sep 25 2009

G.f.: x*(-6-3*x+x^3)/( (1+x)^2 * (x-1)^3 ). - R. J. Mathar, Sep 25 2009

Example

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.

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 *)

Prog

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

Crossrefs

Sequence in context: A357840 A355484 A358222 * A239874 A043103 A242756

Adjacent sequences: A165714 A165715 A165716 * A165718 A165719 A165720

Keyword

nonn,easy

Author

Vladimir Joseph Stephan Orlovsky, Sep 24 2009

Extensions

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

Status

approved