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%I #26 Jan 02 2023 12:30:47
%S 1,7,16,46,97,271,568,1582,3313,9223,19312,53758,112561,313327,656056,
%T 1826206,3823777,10643911,22286608,62037262,129895873,361579663,
%U 757088632,2107440718,4412635921,12283064647,25718726896,71590947166,149899725457,417262618351
%N Indices k such that 8 plus the k-th triangular number is a perfect square.
%C Also numbers n such that (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2 = 8. - _Ctibor O. Zizka_, Nov 10 2009
%H F. T. Adams-Watters, <a href="http://list.seqfan.eu/oldermail/seqfan/2009-October/002504.html">SeqFan Discussion</a>, Oct 2009.
%F {k: 8+k*(k+1)/2 in A000290}
%F Conjectures: (Start)
%F a(n) = a(n-1) + 6*a(n-2) - 6*a(n-3) - a(n-4) + a(n-5).
%F 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)).
%F G.f.: ( 4 + 1/(x-1) - 3/(x^2+2*x-1) + (6+15*x)/(x^2-2*x-1) )/2. (End)
%F 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
%e 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.
%t Join[{1}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 8 &]] (* _G. C. Greubel_, Sep 03 2016 *)
%t Select[Range[0, 2 10^7], IntegerQ[Sqrt[8 + # (# + 1) / 2]] &] (* _Vincenzo Librandi_, Sep 03 2016 *)
%o (PARI) isok(n) = issquare(8 + n*(n+1)/2); \\ _Michel Marcus_, Sep 03 2016
%o (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
%Y Cf. A000217, A000290, A006451.
%K nonn
%O 1,2
%A _R. J. Mathar_, Oct 18 2009
%E a(17)-a(24) from _Donovan Johnson_, Nov 01 2010
%E a(25)-a(30) from _Lars Blomberg_, Jul 07 2015
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