A154145 Indices k such that 15 plus the k-th triangular number is a perfect square.

1, 4, 6, 11, 20, 33, 43, 70, 121, 196, 254, 411, 708, 1145, 1483, 2398, 4129, 6676, 8646, 13979, 24068, 38913, 50395, 81478, 140281, 226804, 293726, 474891, 817620, 1321913, 1711963, 2767870, 4765441, 7704676, 9978054, 16132331, 27775028

Offset: 1

Author

R. J. Mathar, Oct 18 2009

Keywords

• nonn

Comments

a(1..4)=(1,4,6,11); a(n>4)=6*a(n-2)-a(n-4)+2. [From Ctibor O. Zizka, Nov 13 2009]

Examples

			1*(1+1)/2+15 = 4^2. 4*(4+1)/2+15 = 5^2. 6*(6+1)/2+15 = 6^2. 11*(11+1)/2+15 = 9^2.
		

Links

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

Crossrefs

Cf. A000217, A000290, A006451.

Programs

Mathematica

Flatten[Position[Accumulate[Range[28000000]],?(IntegerQ[Sqrt[#+15]]&)]] (* This program will take a long time to run. *) (* _Harvey P. Dale, Jun 09 2014 *)
Join[{1, 4, 6}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 15 &]] (* G. C. Greubel, Sep 03 2016 *)

Formula

{k: 15+k*(k+1)/2 in A000290}.

Conjectures: (Start)

a(n)= +a(n-1) +6*a(n-4) -6*a(n-5) -a(n-8) +a(n-9).

G.f.: x*(-1-3*x-2*x^2-5*x^3-3*x^4+5*x^5+2*x^6+3*x^7+2*x^8)/((x-1) * (x^4+2*x^2-1) * (x^4-2*x^2-1)).

G.f.: ( 4 + (7+4*x+16*x^2+11*x^3)/(x^4-2*x^2-1) + 1/(x-1) + (-4-7*x-3*x^2-2*x^3)/(x^4+2*x^2-1) )/2. (End)

a(1..4) = (1,4,6,11); a(n) = 6*a(n-2) - a(n-4) + 2, for n>4. - Ctibor O. Zizka, Nov 13 2009

Extensions

a(32)-a(37) from Donovan Johnson, Nov 01 2010