A327788 a(n) is the smallest nonnegative integer m such that the integer part of tan(m) is equal to n.

0, 1, 20, 17, 105, 237, 303, 14, 58, 80, 124, 146, 11151, 168, 46318, 190, 46695, 212, 23997, 58432, 234, 13014, 38574, 61649, 82949, 256, 16586, 33271, 48891, 63091, 76581, 89361, 278, 8088, 18738, 28678, 37908, 46783, 54948, 63113, 70568, 77668, 84768, 91158, 97193, 300, 4915, 10240, 15565, 20535, 25150

Offset: 0

Author

James Carruthers, Sep 25 2019

Keywords

• nonn
• look
• easy

Examples

			tan(0) = 0, so a(0) = 0.
tan(1) = 1.557407724654902230506974807... so a(1) = 1.
For m = 2, 3, 4, ... , 13, 15, 16, 18, 19, tan(m) < 2, tan (14) = 7.24460661609480..., tan(17) = 3.493915645474... and tan(20) = 2.2371609442247422652871732477... so a(2) = 20.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..5000

Crossrefs

Cf. A000503 (floor(tan(n))).

Programs

Magma

a:=[]; for n in [0..50] do m:=0; while Floor(Tan(m)) ne n do m:=m+1; end while; Append(~a,m); end for; a; // Marius A. Burtea, Oct 05 2019

Mathematica

Array[Block[{m = 0}, While[IntegerPart@ Tan@ m != #, m++]; m] &, 40, 0] (* Michael De Vlieger, Oct 05 2019 *)

PARI

a(n) = my(k=0); while (floor(tan(k)) != n, k++); k; \\ Michel Marcus, Oct 05 2019

Python

import numpy as np
import math as m
n = 1
i = 0
inp = np.zeros(1)
out = inp
while n < 10001:
    k=m.trunc(m.tan(i))
    if k==n:
        inp = np.append(inp,int(n))
        out = np.append(out,int(i))
        print(n,i)
        n += 1
        i = 0
        continue
    else:
        i+=1