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%I #13 Jan 02 2023 12:30:47
%S 1,4,6,11,20,33,43,70,121,196,254,411,708,1145,1483,2398,4129,6676,
%T 8646,13979,24068,38913,50395,81478,140281,226804,293726,474891,
%U 817620,1321913,1711963,2767870,4765441,7704676,9978054,16132331,27775028
%N Indices k such that 15 plus the k-th triangular number is a perfect square.
%C a(1..4)=(1,4,6,11); a(n>4)=6*a(n-2)-a(n-4)+2. [From _Ctibor O. Zizka_, Nov 13 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: 15+k*(k+1)/2 in A000290}.
%F Conjectures: (Start)
%F a(n)= +a(n-1) +6*a(n-4) -6*a(n-5) -a(n-8) +a(n-9).
%F G.f.: x*(-1-3*x-2*x^2-5*x^3-3*x^4+5*x^5+2*x^6+3*x^7+2*x^8)/((x-1) * (x^4+2*x^2-1) * (x^4-2*x^2-1)).
%F G.f.: ( 4 + (7+4*x+16*x^2+11*x^3)/(x^4-2*x^2-1) + 1/(x-1) + (-4-7*x-3*x^2-2*x^3)/(x^4+2*x^2-1) )/2. (End)
%F a(1..4) = (1,4,6,11); a(n) = 6*a(n-2) - a(n-4) + 2, for n>4. - _Ctibor O. Zizka_, Nov 13 2009
%e 1*(1+1)/2+15 = 4^2. 4*(4+1)/2+15 = 5^2. 6*(6+1)/2+15 = 6^2. 11*(11+1)/2+15 = 9^2.
%t Flatten[Position[Accumulate[Range[28000000]],_?(IntegerQ[Sqrt[#+15]]&)]] (* This program will take a long time to run. *) (* _Harvey P. Dale_, Jun 09 2014 *)
%t Join[{1, 4, 6}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 15 &]] (* _G. C. Greubel_, Sep 03 2016 *)
%Y Cf. A000217, A000290, A006451.
%K nonn
%O 1,2
%A _R. J. Mathar_, Oct 18 2009
%E a(32)-a(37) from _Donovan Johnson_, Nov 01 2010
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