A357005 Smallest k that is cyclically equivalent (see Comment for definition) to n.

1, 2, 3, 4, 5, 5, 7, 8, 9, 10, 11, 9, 11, 11, 15, 16, 17, 17, 19, 17, 19, 19, 23, 17, 19, 19, 23, 19, 23, 23, 31, 32, 33, 34, 35, 36, 37, 37, 39, 34, 37, 42, 43, 37, 45, 43, 47, 33, 35, 37, 39, 37, 43, 45, 47, 35, 39, 43, 47, 39, 47, 47, 63, 64, 65, 65, 67, 65

Offset: 1

Author

Pontus von Brömssen, Sep 08 2022

Keywords

• nonn

Comments

Two positive integers k and n are cyclically equivalent (as defined by Hladnik, Marušič, and Pisanski, 2002) if they have the same number m of binary digits, and there exist integers s and t such that gcd(s,m) = 1 and the map x -> s*x+t mod m maps the set of exponents of 2 occurring in the binary expansion of n bijectively to the corresponding set for k. (In particular, A000120(k) = A000120(n).) For example, k = 17 = 2^0 + 2^4 and n = 18 = 2^1 + 2^4 are cyclically equivalent, because the map x -> 2*x+2 mod 5 maps {1,4} to {0,4}.

The fixed points are the terms of A357006.

The number of fixed points n in the interval 2^(m-1) <= n < 2^m equals A002729(m)-1.

Links

• Pontus von Brömssen, Table of n, a(n) for n = 1..10000
• Milan Hladnik, Dragan Marušič, and Tomaž Pisanski, Cyclic Haar graphs, Discrete Mathematics 244 (2002), 137-152.

Crossrefs

Cf. A000120, A002729, A163382, A267508, A357004, A357006.

Programs

Python

from math import gcd
def A357005(n):
    p=[int(d) for d in format(n,'b')]
    m=len(p)
    p0=min([p[(k*i+j)%m] for i in range(m)] for k in range(1,m+1) if gcd(k,m)==1 for j in range(m) if p[j])
    return sum(p0[i]*2**(m-1-i) for i in range(m))

Formula

a(a(n)) = a(n).

a(n) = A357004(n) for n <= 146, but a(147) = 147 > 141 = A357004(147).

A357004(n) <= a(n) <= A163382(n).