A309434 a(n) = floor(n*Im(2*e^(i*Pi/5))/(Im(2*e^(i*Pi/5)) - 1)).

6, 13, 20, 26, 33, 40, 46, 53, 60, 66, 73, 80, 87, 93, 100, 107, 113, 120, 127, 133, 140, 147, 154, 160, 167, 174, 180, 187, 194, 200, 207, 214, 220, 227, 234, 241, 247, 254, 261, 267, 274, 281, 287, 294, 301, 308, 314, 321, 328, 334, 341

Offset: 1

Author

Karl V. Keller, Jr., Jun 06 2020

Keywords

• nonn

Comments

This is the Beatty sequence for Im(2*e^(i*Pi/5))/(Im(2*e^(i*Pi/5)) - 1).

This is the complement of A335137.

Im(2*e^(i*Pi/5))/(Im(2*e^(i*Pi/5)) - 1) = (5 + sqrt(5))/2 + sqrt(5 + 2*sqrt(5)) = 6.695717525925148250774877410... = 2 + phi + tan(2*Pi/5) = A296184 + A019970.

For n < 10, a(n) = A109235(n).

Re(2*e^(i*Pi/5))/(Re(2*e^(i*Pi/5)) - 1) = (3 + sqrt(5))/2 = 1 + phi = phi^2 = A104457.

Floor(n*Re(2*e^(i*Pi/5))/(Re(2*e^(i*Pi/5)) - 1)) is A001950 (floor(n*phi^2)).

Examples

			For n = 3, floor(3*6.69571) = 20.
		

Links

• Karl V. Keller, Jr., Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Beatty Sequence.
• Index entries for sequences related to Beatty sequences

Crossrefs

Cf. A001950, A019970, A104457, A109235, A296184, A335137.

Programs

Mathematica

a[n_] := Floor[n * Im[2 * Exp[I * Pi/5]]/(Im[2 * Exp[I * Pi/5]] - 1)]; Array[a, 100] (* Amiram Eldar, Jul 06 2020 *)

Python

from sympy import floor, im, exp, I, pi
for n in range(1, 101): print(floor(n*im(2*exp(I*pi/5))/(im(2*exp(I*pi/5)) - 1)), end=', ')

Python

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