A356008 -id:A356008 - cp's OEIS frontend

A356014 Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique e; the resulting list corresponds to the exponents in the prime factorization of a(n).

1, 2, 3, 4, 3, 2, 3, 8, 9, 10, 3, 12, 3, 10, 3, 16, 3, 18, 3, 20, 21, 10, 3, 24, 9, 10, 27, 20, 3, 2, 3, 32, 21, 10, 3, 4, 3, 10, 21, 40, 3, 10, 3, 20, 45, 10, 3, 48, 9, 50, 21, 20, 3, 54, 21, 40, 21, 10, 3, 12, 3, 10, 63, 64, 21, 10, 3, 20, 21, 10, 3, 72, 3

Offset: 1

Rémy Sigrist, Jul 23 2022

nonn

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

This sequence operates on prime exponents as A090079 and A337864 operate on binary and decimal digits, respectively.

			For n = 99:
- 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,
- the list of exponents is: 1 0 0 2 0,
- compressing consecutive values, we obtain: 1 0 2 0,
- so a(99) = 7^1 * 5^0 * 3^2 * 2^0 = 63.

Cf. A007814, A061742, A067255, A090079, A115964, A294674, A319521, A337864, A356008, A356021 (fixed points).

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

a(a(n)) = a(n).
a(n^k) = a(n)^k for any k >= 0.
a(n) = A319521(A356008(n)).
A007814(a(n)) = A007814(n).
a(n) = 3 iff n belongs to A294674 \ {1}.
a(n) = 4 iff n belongs to A061742 \ {1}.
a(n) = 8 iff n belongs to A115964.

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).

Original entry on oeis.org

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

Rémy Sigrist, Jul 23 2022

nonn

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.

			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.

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

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)) }

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.

cp's OEIS Frontend

A356014 Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique e; the resulting list corresponds to the exponents in the prime factorization of a(n).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula