A154140 Indices k such that 6 plus the k-th triangular number is a perfect square.

2, 4, 19, 29, 114, 172, 667, 1005, 3890, 5860, 22675, 34157, 132162, 199084, 770299, 1160349, 4489634, 6763012, 26167507, 39417725, 152515410, 229743340, 888924955, 1339042317, 5181034322, 7804510564, 30197280979, 45488021069, 176002651554, 265123615852

Offset: 1

Author

R. J. Mathar, Oct 18 2009

Keywords

• nonn

Comments

In general, indices k such that A001109(2j) plus the k-th triangular number is a perfect square may be found as follows:

b(2n-1) = A001652(n+j-1) - A001653(n-j);

b(2n) = A001652(n-j-1) + A001653(n+j);

Indices k such that A001109(2j-1) plus the k-th triangular number is a perfect square may be found as follows:

b(2n-1) = A001652(n+j-1) - A001653(n-j+1);

b(2n) = A001652(n-j) + A001653(n+j). - Charlie Marion, Mar 10 2011

Examples

			2*(2+1)/2+6 = 3^2. 4*(4+1)/2+6 = 4^2. 19*(19+1)/2+6 = 14^2. 29*(29+1)/2+6 = 21^2.
		

Links

• F. T. Adams-Watters, SeqFan Discussion, Oct 2009.
• Index entries for linear recurrences with constant coefficients, signature (1, 6, -6, -1, 1).

Crossrefs

Cf. A000217, A000290, A006451.

Programs

Magma

[2] cat [n: n in [0..2*10^7] | (Ceiling(Sqrt(n*(n + 1)/2)) )^2 - n*(n + 1)/2 eq  6]; // Vincenzo Librandi, Sep 03 2016

Mathematica

LinearRecurrence[{1,6,-6,-1,1},{2,4,19,29,114},40] (* Following first conjecture *) (* Harvey P. Dale, Apr 11 2016 *)
Join[{2}, Select[Range[1, 1010], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 6 &] ] (* G. C. Greubel, Sep 03 2016 *)

Formula

{k: 6+k*(k+1)/2 in A000290}.

a(2*n-1) = A001652(n) - A001653(n-1).

a(2*n) = A001652(n-2) + A001653(n+1).

Conjectures: (Start)

a(n) = +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).

G.f.: x*(2 +2*x +3*x^2 -2*x^3 -3*x^4)/((1-x)* (x^2-2*x-1)* (x^2+2*x-1))

G.f.: ( 6 + (-1 -4*x)/(x^2+2*x-1) + (6 +13*x)/(x^2-2*x-1) + 1/(x-1) )/2. (End)

a(1..4) = (2,4,19,29); a(n) = 6*a(n-2) - a(n-4) + 2, for n > 4. - Ctibor O. Zizka, Nov 10 2009

Extensions

a(17)-a(24) from Donovan Johnson, Nov 01 2010

a(25)-a(30) from Lars Blomberg, Jul 07 2015