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 A154138 Indices k such that 3 plus the k-th triangular number is a perfect square. 6
 1, 3, 12, 22, 73, 131, 428, 766, 2497, 4467, 14556, 26038, 84841, 151763, 494492, 884542, 2882113, 5155491, 16798188, 30048406, 97907017, 175134947, 570643916, 1020761278, 3325956481, 5949432723, 19385094972, 34675835062, 112984613353 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Also numbers n such that (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2 = 3. - Ctibor O. Zizka, Nov 10 2009 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 LINKS F. T. Adams-Watters, SeqFan Discussion, Oct 2009 FORMULA {k: 3+k*(k+1)/2 in A000290}. Conjectures: a(n) = +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5); 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.] a(1..4)=(1,3,12,22); a(n>4)=6*a(n-2)-a(n-4)+2. [Zak Seidov, Oct 21 2009] EXAMPLE 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. MATHEMATICA 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 *) Select[Range[100], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 3 &] (* G. C. Greubel, Sep 02 2016 *) Select[Range[0, 2 10^7], IntegerQ[Sqrt[3 + # (# + 1) / 2]] &] (* Vincenzo Librandi, Sep 03 2016 *) PROG (PARI) for(n=0, 1e10, if(issquare(3+n*(n+1)/2), print1(n", "))) \\ Altug Alkan, Oct 16 2015 (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 CROSSREFS Cf. A000217, A000290, A006451. Sequence in context: A063598 A063107 A015629 * A242636 A012469 A103449 Adjacent sequences:  A154135 A154136 A154137 * A154139 A154140 A154141 KEYWORD nonn AUTHOR R. J. Mathar, Oct 18 2009 EXTENSIONS More terms from Zak Seidov, Oct 21 2009 STATUS approved

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