A154141 Indices k such that 8 plus the k-th triangular number is a perfect square.

1, 7, 16, 46, 97, 271, 568, 1582, 3313, 9223, 19312, 53758, 112561, 313327, 656056, 1826206, 3823777, 10643911, 22286608, 62037262, 129895873, 361579663, 757088632, 2107440718, 4412635921, 12283064647, 25718726896, 71590947166, 149899725457, 417262618351

Offset: 1

Author

R. J. Mathar, Oct 18 2009

Keywords

• nonn

Comments

Also numbers n such that (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2 = 8. - Ctibor O. Zizka, Nov 10 2009

Examples

			1*(1+1)/2+8 = 3^2. 7*(7+1)/2+8 = 6^2. 16*(16+1)/2+8 = 12^2. 46*(46+1)/2+8 = 33^2.
		

Links

• F. T. Adams-Watters, SeqFan Discussion, Oct 2009.

Crossrefs

Cf. A000217, A000290, A006451.

Programs

Magma

[1] cat [n: n in [0..2*10^7] | (Ceiling(Sqrt(n*(n+ 1)/2)))^2-n*(n+1)/2 eq  8]; // Vincenzo Librandi, Sep 03 2016

Magma

[n: n in [0..2*10^7] | IsSquare(8+n*(n+1)/2)]; // Vincenzo Librandi, Sep 03 2016

Mathematica

Join[{1}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 8 &]] (* G. C. Greubel, Sep 03 2016 *)
Select[Range[0, 2 10^7], IntegerQ[Sqrt[8 + # (# + 1) / 2]] &] (* Vincenzo Librandi, Sep 03 2016 *)

PARI

isok(n) = issquare(8 + n*(n+1)/2); \\ Michel Marcus, Sep 03 2016

Formula

{k: 8+k*(k+1)/2 in A000290}

Conjectures: (Start)

a(n) = a(n-1) + 6*a(n-2) - 6*a(n-3) - a(n-4) + a(n-5).

G.f.: x*(1 +6*x +3*x^2 -6*x^3 -2*x^4)/((1-x) * (x^2-2*x-1) * (x^2+2*x-1)).

G.f.: ( 4 + 1/(x-1) - 3/(x^2+2*x-1) + (6+15*x)/(x^2-2*x-1) )/2. (End)

a(1..4) = (1,7,16,46); a(n) = 6*a(n-2) - a(n-4) + 2, for n>4. - Ctibor O. Zizka, Nov 10 2009

Extensions

a(17)-a(24) from Donovan Johnson, Nov 01 2010

a(25)-a(30) from Lars Blomberg, Jul 07 2015