A356016 Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique k; the resulting list corresponds to the exponents in the prime factorization of a(n).

1, 2, 6, 2, 12, 4, 24, 2, 6, 30, 48, 6, 96, 90, 18, 2, 192, 6, 384, 30, 210, 270, 768, 6, 12, 810, 6, 90, 1536, 8, 3072, 2, 1050, 2430, 36, 4, 6144, 7290, 5250, 30, 12288, 60, 24576, 270, 30, 21870, 49152, 6, 24, 30, 26250, 810, 98304, 6, 420, 90, 131250

Offset

1,2

Comments

We ignore the exponents (all 0's) for the prime numbers beyond the greatest prime factor of n.

There are only two fixed points: a(1) = 1 and a(2) = 2.

Iterating the sequence starting from any n > 1 will always eventually reach the fixed point 2.

Links

Table of n, a(n) for n=1..57.

Formula

a(n) = A319522(A356008(n)).

a(n^k) = a(n) for any k > 0.

a(n) = 2 iff n is a power of 2 > 1.

a(n) = 4 iff n is a power of 6 > 1.

a(n) = 2^k iff n is a power of A002110(k) > 1 (with k > 0).

a(prime(n)) = 3*2^(n-1) for any n > 1.

Example

For n = 99:
- 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,
- the list of exponents is: 1 0 0 2 0,
- the run lengths are: 1 2 1 1,
- so a(99) = 7^1 * 5^2 * 3^1 * 2^1 = 1050.

Prog

(PARI) a(n) = { my (v=1, e=-1, k=0, r=0); forprime (p=2, oo, if (n==1, return (v*if (r, prime(k++)^r, 1)), if (e!=e=valuation(n, p), if (r, v*=prime(k++)^r; r=0)); r++; n/=p^e)) }

Crossrefs

Cf. A002110, A067255, A318928, A319522, A356008, A356014.

Sequence in context: A243197 A319048 A066900 * A336852 A079462 A189218

Adjacent sequences: A356013 A356014 A356015 * A356017 A356018 A356019

Keyword

nonn

Author

Rémy Sigrist, Jul 23 2022

Status

approved