OFFSET
1,1
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)).
LINKS
Karl V. Keller, Jr., Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Beatty Sequence.
EXAMPLE
For n = 3, floor(3*6.69571) = 20.
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 *)
PROG
(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=', ')
CROSSREFS
KEYWORD
nonn
AUTHOR
Karl V. Keller, Jr., Jun 06 2020
STATUS
approved