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%I #28 Mar 03 2024 15:40:14
%S 2,4,19,29,114,172,667,1005,3890,5860,22675,34157,132162,199084,
%T 770299,1160349,4489634,6763012,26167507,39417725,152515410,229743340,
%U 888924955,1339042317,5181034322,7804510564,30197280979,45488021069,176002651554,265123615852
%N Indices k such that 6 plus the k-th triangular number is a perfect square.
%C In general, indices k such that A001109(2j) plus the k-th triangular number is a perfect square may be found as follows:
%C b(2n-1) = A001652(n+j-1) - A001653(n-j);
%C b(2n) = A001652(n-j-1) + A001653(n+j);
%C Indices k such that A001109(2j-1) plus the k-th triangular number is a perfect square may be found as follows:
%C b(2n-1) = A001652(n+j-1) - A001653(n-j+1);
%C b(2n) = A001652(n-j) + A001653(n+j). - _Charlie Marion_, Mar 10 2011
%H F. T. Adams-Watters, <a href="http://list.seqfan.eu/oldermail/seqfan/2009-October/002506.html">SeqFan Discussion</a>, Oct 2009.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1, 6, -6, -1, 1).
%F {k: 6+k*(k+1)/2 in A000290}.
%F a(2*n-1) = A001652(n) - A001653(n-1).
%F a(2*n) = A001652(n-2) + A001653(n+1).
%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*(2 +2*x +3*x^2 -2*x^3 -3*x^4)/((1-x)* (x^2-2*x-1)* (x^2+2*x-1))
%F G.f.: ( 6 + (-1 -4*x)/(x^2+2*x-1) + (6 +13*x)/(x^2-2*x-1) + 1/(x-1) )/2. (End)
%F a(1..4) = (2,4,19,29); a(n) = 6*a(n-2) - a(n-4) + 2, for n > 4. - _Ctibor O. Zizka_, Nov 10 2009
%e 2*(2+1)/2+6 = 3^2. 4*(4+1)/2+6 = 4^2. 19*(19+1)/2+6 = 14^2. 29*(29+1)/2+6 = 21^2.
%t LinearRecurrence[{1,6,-6,-1,1},{2,4,19,29,114},40] (* Following first conjecture *) (* _Harvey P. Dale_, Apr 11 2016 *)
%t Join[{2}, Select[Range[1, 1010], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 6 &] ] (* _G. C. Greubel_, Sep 03 2016 *)
%o (Magma) [2] cat [n: n in [0..2*10^7] | (Ceiling(Sqrt(n*(n + 1)/2)) )^2 - n*(n + 1)/2 eq 6]; // _Vincenzo Librandi_, Sep 03 2016
%Y Cf. A000217, A000290, A006451.
%K nonn
%O 1,1
%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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