A076563 -id:A076563 - cp's OEIS frontend

A356429 Smallest m such that A356428(m) = n, or -1 if there is no such m.

2, 8, 48, 315, 320, 6664, 135450, 273000, 518661, 519440, 519622, 148830266, 558797841, 558797968, 24900609294

Offset: 1

Jianing Song, Aug 07 2022

a(n) is the smallest m such that there are exactly n distinct gpf(x)'s in the iterations x -> x - gpf(x) starting at m and ending at 0, where gpf = A006530.
Conjecture: a(n) != -1 for all n. This would be true if A356428 is unbounded; otherwise, this sequence consists of entirely -1's after some point.
Since A356428(n) - A356428(n-gpf(n)) = 0 or 1, sequence is strictly increasing if no term equals -1.
If a(m) > -1 for m >= 15 then a(m) > 10^9. - David A. Corneth, Aug 09 2022

			In the following examples the numbers produced by the iterations are listed together with their GPFs.
320 (5) -> 315 (7) -> 308 (11) -> 297 (11) -> 286 (13) -> 273 (13) -> 260 (13) -> 247 (19) -> ... -> 19 (19) -> 0, the distinct gpf(x)'s are 5, 7, 11, 13, and 19. 320 is the smallest number such that the distinct gpf(x)'s in the iterations is 5, so a(5) = 320.
6664 (17) -> 6647 (23) -> 6624 (23) -> 6601 (41) -> 6560 (41) -> 6519 (53) -> 6466 (53) -> 6413 (53) -> 6360 (53) -> 6307 (53) -> 6254 (59) -> 6195 (59) -> 6136 (59) -> 6077 (103) -> ... -> 103 (103) -> 0, the distinct gpf(x)'s are 17, 23, 41, 53, 59, and 103. 6664 is the smallest number such that the distinct gpf(x)'s in the iterations is 6, so a(6) = 6664.

Cf. A006530 (gpf), A076563, A309892, A356428.

a(12) from Michael S. Branicky, Aug 08 2022
a(13)-a(14) from David A. Corneth, Aug 09 2022
a(15) from Jinyuan Wang, Jul 07 2025

A382986 a(n) is the number of iterations that n requires to reach 0 under the map k -> b(k) where b(k) = k+1 if k is even, and b(k) = k-gpf(k) if k is odd, where gpf(k) is the greatest prime dividing k.

Original entry on oeis.org

0, 1, 2, 1, 2, 1, 2, 1, 4, 3, 2, 1, 2, 1, 4, 3, 2, 1, 2, 1, 6, 5, 2, 1, 8, 7, 10, 9, 2, 1, 2, 1, 4, 3, 4, 3, 2, 1, 12, 11, 2, 1, 2, 1, 4, 3, 2, 1, 4, 3, 6, 5, 2, 1, 6, 5, 14, 13, 2, 1, 2, 1, 16, 15, 4, 3, 2, 1, 4, 3, 2, 1, 2, 1, 4, 3, 4, 3, 2, 1, 4, 3, 2, 1, 6, 5, 4, 3, 2, 1

Offset: 0

Jakub Buczak, Apr 11 2025

k -> 0 occurs when k is 1 or an odd prime and this is the only way to reach 0.
At 0, a loop 0 -> 1 -> 0 occurs and there are no other loops since either 1 or 2 steps is a decrease: b(k) < k for odd k, or b(b(k)) < k for even k >= 2.
Consecutive terms can form decreasing runs starting at an even term, such as: (4, 3, 2, 1) starting at the 8th term.

			a(0) = 0, since 0 already matches the ending condition.
a(2) = 2, since (b2)=3 and then b(3)=0.
a(3) = 1, since 3 - gpf(3) = 0.

Jakub Buczak, Table of n, a(n) for n = 0..10000

Cf. A006530, A076563.

gpf(n) = if (n==1, 1, vecmax(factor(n)[,1]));
b(m) = if (m % 2, m - gpf(m), m+1);
a(n)= my(nb=0); while (n>1, n=b(n); nb++); nb; \\ Michel Marcus, Apr 13 2025

from sympy import factorint
from itertools import count
def b(n): return n - max(factorint(n), default=0) if n&1 else n + 1
def a(n): return next(i for i in count(1) if not (n:=b(n))) if n > 1 else n
print([a(n) for n in range(90)]) # Michael S. Branicky, Apr 13 2025

a(p) = 1, for odd prime p.

a(2*n) = a(2*n+1) + 1.

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A356429 Smallest m such that A356428(m) = n, or -1 if there is no such m.

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A382986 a(n) is the number of iterations that n requires to reach 0 under the map k -> b(k) where b(k) = k+1 if k is even, and b(k) = k-gpf(k) if k is odd, where gpf(k) is the greatest prime dividing k.

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