A263574 Beatty sequence for 1/sqrt(3) - log(phi)/3575 where phi is the golden ratio, A001622.

0, 0, 1, 1, 2, 2, 3, 4, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 10, 11, 12, 12, 13, 13, 14, 15, 15, 16, 16, 17, 17, 18, 19, 19, 20, 20, 21, 21, 22, 23, 23, 24, 24, 25, 25, 26, 27, 27, 28, 28, 29, 30, 30, 31, 31, 32, 32, 33, 34, 34, 35, 35, 36, 36, 37, 38, 38, 39, 39, 40, 40, 41, 42, 42, 43

Offset: 0

Author

Karl V. Keller, Jr., Oct 21 2015

Keywords

• nonn

Comments

The number 1/sqrt(3) - log(phi)/3575 (=0.577215664483...) is an approximation to Euler's constant (A001620) (=0.577215664901...).

M. Hudson found a similar Euler-Mascheroni constant approximation (see link), 1/sqrt(3)-1/7429 (=0.57721566157...).

Examples

			For n=9, floor(9*(0.577215664483)) = floor(5.194940980347) = 5.
		

Links

• Karl V. Keller, Jr., Table of n, a(n) for n = 0..100000
• Xavier Gourdon and Pascal Sebah, Collection of formulas for Euler's constant,Euler's constant.
• Eric Weisstein's World of Mathematics, Beatty Sequence.
• Eric Weisstein's World of Mathematics, Euler-Mascheroni Constant.
• Eric Weisstein's World of Mathematics, Euler-Mascheroni Constant Approximations.

Crossrefs

Cf. A001620, A020760 (1/sqrt(3)), A038128 (Beatty sequence for Euler's constant), A097337 (Beatty sequence for 1/sqrt(3)).

Programs

Magma

phi:= (1+Sqrt(5))/2; [Floor(n*(1/Sqrt(3) - Log(phi)/3575)): n in [0..100]]; // G. C. Greubel, Sep 05 2018

Mathematica

Table[Floor[n (1/Sqrt@ 3 - Log[GoldenRatio]/3575)], {n, 0, 75}] (* Michael De Vlieger, Nov 12 2015 *)

PARI

{phi = (1+sqrt(5))/2}; vector(100, n, n--; floor(n*(1/sqrt(3) - log(phi)/3575))) \\ G. C. Greubel, Sep 05 2018

Python

from sympy import floor, log, sqrt
for n in range(0,101):print(floor(n*(1/sqrt(3)-log(1/2+sqrt(5)/2)/3575)),end=', ')

Formula

a(n) = floor(n*(1/sqrt(3) - log(phi)/3575)).

a(n) = A038128(n) for n < 58628.