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

1,2

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

Table of n, a(n) for n=1..29.

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: A063107 A015629 A341374 * A332349 A354529 A242636

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