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A154141
Indices k such that 8 plus the k-th triangular number is a perfect square.
4
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
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
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