A154144 Indices k such that 13 plus the k-th triangular number is a perfect square.

2, 8, 23, 53, 138, 312, 807, 1821, 4706, 10616, 27431, 61877, 159882, 360648, 931863, 2102013, 5431298, 12251432, 31655927, 71406581, 184504266, 416188056, 1075369671, 2425721757, 6267713762, 14138142488, 36530912903, 82403133173, 212917763658, 480280656552

Offset: 1

Author

R. J. Mathar, Oct 18 2009

Keywords

• nonn

Examples

			2*(2+1)/2+13 = 4^2. 8*(8+1)/2+13 = 7^2. 23*(23+1)/2+13 = 17^2. 53*(53+1)/2+13 = 38^2.
		

Links

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

Crossrefs

Cf. A000217, A000290, A006451.

Programs

Mathematica

With[{nn=25000},Transpose[Select[Thread[{Range[nn],Accumulate[ Range[nn]]}], IntegerQ[Sqrt[#[[2]]+13]]&]][[1]]] (* Harvey P. Dale, Jan 13 2012 *)
Join[{2, 8}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 13 &]] (* G. C. Greubel, Sep 03 2016 *)

Formula

{k: 13+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*(2 +6*x +3*x^2 -6*x^3 -3*x^4)/((1-x) * (x^2-2*x-1) * (x^2+2*x-1))

G.f.: ( 6 + (-3-2*x)/(x^2+2*x-1) + 1/(x-1) + (8+19*x)/(x^2-2*x-1) )/2 . (End)

a(1..4) = (2,8,23,53); a(n) = 6*a(n-2) - a(n-4) + 2, for n>2. - Ctibor O. Zizka, Nov 10 2009

Extensions

a(16)-a(24) from Donovan Johnson, Nov 01 2010

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