A353172 a(n) is the least k > 1 such that Omega(n) = Omega(n mod k), where Omega = A001222.

2, 3, 4, 5, 3, 7, 4, 9, 5, 6, 3, 13, 5, 5, 9, 17, 3, 10, 4, 12, 11, 6, 3, 25, 7, 10, 15, 10, 3, 11, 4, 33, 9, 5, 13, 20, 5, 8, 5, 24, 3, 15, 4, 9, 25, 6, 3, 49, 5, 14, 9, 11, 3, 19, 7, 20, 12, 6, 3, 22, 7, 7, 11, 65, 11, 18, 4, 10, 5, 25, 3, 40, 5, 5, 19, 16

Offset: 1

Author

Thomas Scheuerle, Apr 28 2022

Keywords

• nonn
• easy

Comments

It appears that a(m) = m*k/p if m = p*2^n ... . Are these formulas related to some well-known sequence of rational numbers?

Examples

			a(10) = 6 because 10 = 5*2 and 10 mod 6 = 4 = 2*2.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A001222, A003627, A029744, A061142, A077065, A095925.

Programs

PARI

a(n) = my(k=2); while(bigomega(n) != bigomega(max(n%k,1)), k++); k

Python

from itertools import count
from sympy.ntheory.factor_ import primeomega
def A353172(n):
    a = primeomega(n)
    for k in count(2):
        if (m := n % k) > 0 and primeomega(m) == a:
            return k # Chai Wah Wu, Jun 20 2022

Formula

a(A029744(n)) = A029744(n) + 1.

a(A003627(n)) = 3.

a(A000040(n)) = A095925(n).

a(A077065(n)) = 6. For n > 2.

If a(n) = 10, then n mod 10 is in most cases 8 and seldom 6.

a(m) = m*3/5 if m = 5*2^n or m = 15. This formula is valid for all positive n because (5*2^n) mod (5*2^n)*(3/5) = 2^(n+1). If the sequence of solutions does not create powers of two in the modulo operation, it will be of finite length. See next two formulas:

a(m) = m*3/11 if m = 11, 22, 33 or 66.

a(m) = m*4/43 if m = 43*2^n for n < 4. This series of solutions terminates because of the next formula which replaces the powers of two:

a(m) = m*41/(43*2^4) if m = 43*2^4*2^n. This formula is valid for all positive n.

a(m) = m*5/9 if m = 9*2^n or m = 27 or 45. This formula is valid for all positive n.

For each k = a(p) if k < p and gcd(k, p) = 1 such a formula, of the form a(m) = m*k/p, if m = p*2^n ..., can be developed.