A323785 Iteratively swap prime factors of n with their exponent (unless the exponent is 1), until a cycle is reached, and then record the largest term in the cycle.

1, 2, 3, 4, 5, 6, 7, 9, 9, 10, 11, 12, 13, 14, 15, 16, 17, 16, 19, 20, 21, 22, 23, 27, 32, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 32, 37, 38, 39, 45, 41, 42, 43, 44, 45, 46, 47, 48, 128, 32, 51, 52, 53, 54, 55, 63, 57, 58, 59, 60, 61, 62, 63, 32, 65, 66, 67

Offset: 1

Author

Gabriel Antonius, Aug 31 2019

Keywords

• nonn

Examples

			For n=196, the iterative procedure evolves as follows: 196 = 2^2 * 7^2 --> 2^2 * 2^7 = 2^9 --> 9^2 = 3^4 --> 4^3 = 2^6 --> 6^2 = 3^2 * 2^2 --> 2^3 * 2^2 = 2^5 --> 5^2 --> 2^5 = 32.
Since the iterative procedure stabilizes in the cycle 2^5=32 <--> 5^2=25, of length 2, the largest ,member of which is 32, we get a(196)=32.
		

Links

• Gabriel Antonius, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A008477.

Programs

Mathematica

p[x_, y_] := If[y > 1, y^x, x^y]; f[n_] := Times @@ (p[First[#], Last[#]] & /@ FactorInteger[n]); a[n_] := Module[{k = n, s = {k}}, While[! MemberQ[s, (k = f[k])], AppendTo[s, k]]; ind = Position[s, ?(# == k &)][[1, 1]]; Max[s[[ind ;; -1]]]]; Array[a, 57] (* _Amiram Eldar, Sep 02 2019 *)

Python

def a(n):
    np = swap_prime_factors_with_exponents(n)
    npp = swap_prime_factors_with_exponents(np)
    return max(n, np) if n == npp else a(npp)
def swap_prime_factors_with_exponents(n):
    np = 1
    for p in primes:
        q = 0
        while n % p == 0:
            q += 1
            n = n // p
        if q > 1: np *= q ** p
        if q == 1: np *= p
    if n > 1: np *= n
    return np
N = 200
primes = []
for n in range(2, int(N ** (1/2)) + 1):
    if all(n % p > 0 for p in primes):
        primes.append(n)
for n in range(1,N+1):
    print(n, a(n))