A154138 Indices k such that 3 plus the k-th triangular number is a perfect square.

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

Offset: 1

Author

R. J. Mathar, Oct 18 2009

Keywords

• nonn

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

Examples

			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.
		

Links

• F. T. Adams-Watters, SeqFan Discussion, Oct 2009

Crossrefs

Cf. A000217, A000290, A006451.

Programs

Magma

[n: n in [0..2*10^7] | IsSquare(3+n*(n+1)/2)]; // Vincenzo Librandi, Sep 03 2016

Magma

[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

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 *)

PARI

for(n=0, 1e10, if(issquare(3+n*(n+1)/2), print1(n", "))) \\ Altug Alkan, Oct 16 2015

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]

Extensions

More terms from Zak Seidov, Oct 21 2009