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%I #24 Jan 02 2023 12:30:47
%S 1,3,12,22,73,131,428,766,2497,4467,14556,26038,84841,151763,494492,
%T 884542,2882113,5155491,16798188,30048406,97907017,175134947,
%U 570643916,1020761278,3325956481,5949432723,19385094972,34675835062,112984613353
%N Indices k such that 3 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 = 3. - _Ctibor O. Zizka_, Nov 10 2009
%C Note that 3 is 2nd triangular number A000217(2) = 2(2+1)/2, hence 2nd and n-th triangular numbers sum up to a square. - _Zak Seidov_, Oct 16 2015
%H F. T. Adams-Watters, <a href="http://list.seqfan.eu/oldermail/seqfan/2009-October/002511.html">SeqFan Discussion</a>, Oct 2009
%F {k: 3+k*(k+1)/2 in A000290}.
%F Conjectures:
%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 + 2*x + 3*x^2 - 2*x^3 - 2*x^4)/((1-x)*(x^2-2*x-1)*(x^2+2*x-1)). [Comment from _Zak Seidov_, Oct 21 2009: I believe both of these conjectures are correct.]
%F a(1..4)=(1,3,12,22); a(n>4)=6*a(n-2)-a(n-4)+2. [_Zak Seidov_, Oct 21 2009]
%e 1*(1+1)/2+3 = 2^2. 3*(3+1)/2+3 = 3^2. 12*(12+1)/2+3 = 9^2. 22*(22+1)/2+3 = 16^2.
%t a[1]=1;a[2]=3;a[3]=12;a[4]=22;a[n_]:=a[n]=6*a[n-2]-a[n-4]+2;Table[a[n],{n,35}] (* _Zak Seidov_, Oct 21 2009 *)
%t Select[Range[100], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 3 &] (* _G. C. Greubel_, Sep 02 2016 *)
%t Select[Range[0, 2 10^7], IntegerQ[Sqrt[3 + # (# + 1) / 2]] &] (* _Vincenzo Librandi_, Sep 03 2016 *)
%o (PARI) for(n=0, 1e10, if(issquare(3+n*(n+1)/2), print1(n", "))) \\ _Altug Alkan_, Oct 16 2015
%o (Magma) [n: n in [0..2*10^7] | IsSquare(3+n*(n+1)/2)]; /* or */ [1] cat [n: n in [0..2*10^7] | (Ceiling(Sqrt(n*(n+1)/2)))^2-n*(n+1)/2 eq 3]; // _Vincenzo Librandi_, Sep 03 2016
%Y Cf. A000217, A000290, A006451.
%K nonn
%O 1,2
%A _R. J. Mathar_, Oct 18 2009
%E More terms from _Zak Seidov_, Oct 21 2009
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