A154146 Numbers k such that 16 plus the k-th triangular number is a perfect square.

0, 14, 17, 87, 104, 510, 609, 2975, 3552, 17342, 20705, 101079, 120680, 589134, 703377, 3433727, 4099584, 20013230, 23894129, 116645655, 139265192, 679860702, 811697025

Offset: 0

Author

R. J. Mathar, Oct 18 2009

Keywords

• nonn
• less

Comments

Numbers k such that x=2*k+1 satisfies the Pell-type equation x^2 = 8*y^2 - 127. - Robert Israel, Jul 18 2019

Examples

			0, 14, 17, and 87 are terms:
   0* (0+1)/2 + 16 =  4^2,
  14*(14+1)/2 + 16 = 11^2,
  17*(17+1)/2 + 16 = 13^2,
  87*(87+1)/2 + 16 = 62^2.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..2608
• F. T. Adams-Watters, SeqFan Discussion, Oct 2009
• Robert Israel, Proof of conjectured recurrence
• Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Crossrefs

Cf. A000217, A000290, A006451.

Programs

Maple

f:= gfun:-rectoproc({a(n+4)-6*a(n+2)+a(n)=2, a(0)=0, a(1)=14, a(2)=17, a(3)=87}, a(n), remember):
map(f, [$0..40]); # Robert Israel, Jul 18 2019

Mathematica

Join[{0}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 16 &]] (* G. C. Greubel, Sep 03 2016 *)

PARI

{for (n=0, 10^9, if ( issquare(n*(n+1)\2 + 16), print1(n, ", ") ) );}

Formula

{k: 16+k*(k+1)/2 in A000290}.

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*(-14-3*x+14*x^2+x^3)/((x-1) * (x^2-2*x-1) * (x^2+2*x-1)).

G.f.: ( 2 + (8+23*x)/(x^2-2*x-1) + 1/(x-1) + (-7+6*x)/(x^2+2*x-1) )/2. (End)

Conjectures confirmed: see link. - Robert Israel, Jul 18 2019