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

1,2

Comments

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

Links

Table of n, a(n) for n=1..30.

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

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

Example

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.

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 *)

Prog

(PARI) isok(n) = issquare(8 + n*(n+1)/2); \\ Michel Marcus, Sep 03 2016
(Magma) [1] cat [n: n in [0..2*10^7] | (Ceiling(Sqrt(n*(n+ 1)/2)))^2-n*(n+1)/2 eq  8]; /* or */ [n: n in [0..2*10^7] | IsSquare(8+n*(n+1)/2)]; // Vincenzo Librandi, Sep 03 2016

Crossrefs

Cf. A000217, A000290, A006451.

Sequence in context: A066009 A304013 A037241 * A173661 A152530 A065099

Adjacent sequences: A154138 A154139 A154140 * A154142 A154143 A154144

Keyword

nonn

Author

R. J. Mathar, Oct 18 2009

Extensions

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

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

Status

approved