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%I #27 Jan 02 2023 12:30:47
%S 0,10,13,63,80,370,469,2159,2736,12586,15949,73359,92960,427570,
%T 541813,2492063,3157920,14524810,18405709,84656799,107276336,
%U 493415986,625252309,2875839119,3644237520,16761618730,21240172813,97693873263,123796799360,569401620850
%N Indices k such that 9 plus the k-th triangular number is a perfect square.
%H Robert Israel, <a href="/A154142/b154142.txt">Table of n, a(n) for n = 1..2351</a>
%H F. T. Adams-Watters, <a href="http://list.seqfan.eu/oldermail/seqfan/2009-October/002504.html">SeqFan Discussion</a>, Oct 2009
%F {k: 9+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^2*(10 +3*x -10*x^2 -x^3)/((1-x) * (x^2-2*x-1) * (x^2+2*x-1))
%F G.f.: ( 2 + (-5+4*x)/(x^2+2*x-1) + (6+17*x)/(x^2-2*x-1) + 1/(x-1) )/2. (End)
%F a(1..4) = (0,10,13,63); a(n) = 6*a(n-2) - a(n-4) + 2, for n > 4. - _Ctibor O. Zizka_, Nov 10 2009
%F From _Robert Israel_, Jul 07 2015: (Start)
%F These conjectures follow from the theory of Pell-like equations.
%F a(2*k+1) = (8 * A001109(k) -7 * A001541(k) - 1)/2.
%F a(2*k) = (8 * A001109(k) -9 * A001541(k) - 1)/2. (End)
%e 0*(0+1)/2+9 = 3^2. 10*(10+1)/2+9 = 8^2. 13*(13+1)/2+9 = 10^2. 63*(63+1)/2+9 = 45^2.
%p seq(seq((8*orthopoly[U](k+j,3) - (8 - (-1)^j)*orthopoly[T](k+j,3)-1)/2, j=0..1),k=0..20); # _Robert Israel_, Jul 07 2015
%t Join[{0}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 9 &]] (* _G. C. Greubel_, Sep 03 2016 *)
%t Select[Range[0, 2 10^7], IntegerQ[Sqrt[9 + # (# + 1) / 2]] &] (* _Vincenzo Librandi_, Sep 03 2016 *)
%o (Magma) [n: n in [0..2*10^7] | IsSquare(9 + n*(n+1)/2)];
%o /* or */ [0] cat [n: n in [0..2*10^7] | (Ceiling(Sqrt(n*(n+ 1)/2)))^2-n*(n+1)/2 eq 9]; // _Vincenzo Librandi_, Sep 03 2016
%Y Cf. A000217, A000290, A006451, A001109, A001541.
%K nonn
%O 1,2
%A _R. J. Mathar_, Oct 18 2009
%E a(16)-a(24) from _Donovan Johnson_, Nov 01 2010
%E a(25)-a(30) from _Lars Blomberg_, Jul 07 2015
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