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A323135
a(n) is the least number of iterations that n requires to reach a power of a prime under the map x -> A070229(x), or -1 if we never reach a power of a prime.
2
0, 0, 0, 0, 0, 1, 0, 0, 0, 3, 0, 3, 0, 5, 2, 0, 0, 5, 0, 1, 4, 9, 0, 1, 0, 11, 0, 3, 0, 3, 0, 0, 8, 15, 2, 11, 0, 17, 10, 9, 0, 1, 0, 7, 8, 21, 0, 15, 0, 7, 14, 9, 0, 17, 6, 7, 16, 27, 0, 9, 0, 29, 6, 0, 8, 5, 0, 13, 20, 5, 0, 15, 0, 35, 14, 15, 4, 7, 0, 13, 0
OFFSET
1,10
COMMENTS
The powers of primes correspond to A000961.
Apparently, a(n) >= 0 for any n > 0.
For a given number n, while iterating A070229, we may encounter several prime increments (=several greatest prime factors). It is likely that the number of distinct increments before reaching a power of a prime is not bounded.
For k = 0..9, the least numbers with k distinct increments are:
k n Increments
- ------ ------------------------------------------
0 1 {}
1 6 {3}
2 12 {3, 5}
3 72 {3, 5, 17}
4 135 {5, 7, 11, 17}
5 686 {7, 11, 13, 19, 41}
6 12408 {47, 53, 59, 71, 89, 149}
7 35378 {19, 23, 67, 89, 101, 179, 211}
8 127581 {43, 53, 73, 103, 113, 227, 283, 457}
9 222111 {37, 79, 97, 191, 233, 239, 311, 359, 631}
LINKS
FORMULA
a(n) = 0 iff n belongs to A000961.
EXAMPLE
For n = 12:
- 12 = 2^2 * 3,
- A070229(12) = 12 + 3 = 15 = 3 * 5,
- A070229(15) = 15 + 5 = 20 = 2^2 * 5,
- A070229(20) = 20 + 5 = 25 = 5^5,
- hence we need at least 3 iterations of A070229 to reach a power of a prime,
- and a(12) = 3.
PROG
(PARI) a(n) = for (k=0, oo, if (omega(n) <= 1, return (k), my (f=factor(n)); n += f[#f~, 1]))
CROSSREFS
See A323136 for the corresponding powers of primes.
Sequence in context: A240923 A272727 A333791 * A356205 A100258 A045763
KEYWORD
nonn,look
AUTHOR
Rémy Sigrist, Jan 05 2019
STATUS
approved