A028231 From hexagons in a circle problem.

1, 22, 313, 4366, 60817, 847078, 11798281, 164328862, 2288805793, 31878952246, 444016525657, 6184352406958, 86136917171761, 1199732487997702, 16710117914796073, 232741918319147326, 3241676738553266497, 45150732421426583638, 628868577161418904441

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Numbers k such that (k^2 + k + 1)/3 is a square. - Arkadiusz Wesolowski, Feb 10 2012

Given by the numerators of the convergents to the continued fraction [1,(1,2)^i,3,(1,2)^{i-1},1]. - Jeffrey Shallit, Dec 11 2017

References

• J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.
• T. Nagell, Des équations indéterminées x^2 + x + 1 = y^n et x^2 + x + 1 = 3*y^n, Norsk Mat. Forenings Skrifter, Ser. I, (1921).

Links

• Michel Marcus, Table of n, a(n) for n = 0..100
• Kevin A. Broughan, An explicit bound for aliquot cycles of repdigits, #A15 INTEGERS Vol 12 (2012) p. 4.
• Index entries for linear recurrences with constant coefficients, signature (15,-15,1)

Crossrefs

Cf. A001570, which gives the corresponding values of y in 3y^2 = n^2 + n + 1. - Jeffrey Shallit, Dec 11 2017

Programs

Maple

f:= gfun:-rectoproc({a(n) = 15*a(n-1)-15*a(n-2)+a(n-3),a(0)=1,a(1)=22,a(2)=313},a(n),remember):
map(f, [$0..30]); # Robert Israel, Dec 12 2017

Mathematica

With[{k = Sqrt@ 3}, Simplify@ Array[k ((2 + k)^(2 # + 1) - (2 - k)^(2 # + 1))/4 - 1/2 &, 19, 0]] (* Michael De Vlieger, Dec 11 2017 *)

PARI

a(n) = {w = quadgen(12);w*((2+w)^(2*n+1) - (2-w)^(2*n+1))/4 - 1/2;} /* Michel Marcus, Jul 28 2012 */

Formula

a(n) = sqrt(3)*((2+sqrt(3))^(2*n+1) - (2-sqrt(3))^(2*n+1))/4 - 1/2 (see Kevin A. Broughan paper). - Michel Marcus, Jul 28 2012

a(n) = 15*a(n-1)-15*a(n-2)+a(n-3). G.f.: (1+7*x-2*x^2)/((1-x)*(1-14*x+x^2)). - conjectured by Colin Barker, Apr 10 2012; these follow easily from the formula.

Extensions

Edited by Robert Israel, Dec 12 2017