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
Keywords
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))