|
| |
|
|
A329697
|
|
a(n) is the number of iterations needed to reach a power of 2 starting at n and using the map k -> k-(k/p), where p is the largest prime factor of k.
|
|
64
|
|
|
|
0, 0, 1, 0, 1, 1, 2, 0, 2, 1, 2, 1, 2, 2, 2, 0, 1, 2, 3, 1, 3, 2, 3, 1, 2, 2, 3, 2, 3, 2, 3, 0, 3, 1, 3, 2, 3, 3, 3, 1, 2, 3, 4, 2, 3, 3, 4, 1, 4, 2, 2, 2, 3, 3, 3, 2, 4, 3, 4, 2, 3, 3, 4, 0, 3, 3, 4, 1, 4, 3, 4, 2, 3, 3, 3, 3, 4, 3, 4, 1, 4, 2, 3, 3, 2, 4, 4, 2, 3, 3, 4, 3, 4, 4, 4, 1, 2, 4, 4, 2
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,7
|
|
|
COMMENTS
|
Also the least number of iterations of nondeterministic map k -> k-(k/p) needed to reach a power of 2, when any prime factor p of k can be used. The minimal length path to the nearest power of 2 (= 2^A064415(n)) is realized whenever one uses any of the A005087(k) distinct odd prime factors of the current k, at any step of the process. For example, this could be done by iterating with the map k -> k-(k/A078701(k)), i.e., by using the least odd prime factor of k (instead of the largest prime).
Proof: Viewing the prime factorization of changing k as a multiset ("bag") of primes, we see that liquefying any odd prime p with step p -> (p-1) brings at least one more 2 to the bag, while applying p -> (p-1) to any 2 just removes it from the bag, but gives nothing back. Thus the largest (and thus also the nearest) power of 2 is reached by eliminating - step by step - all odd primes from the bag, but none of 2's, and it doesn't matter in which order this is done.
The above implies also that the sequence is totally additive, which also follows because both A064097 and A064415 are. That A064097(n) = A329697(n) + A054725(n) for all n > 1 can be also seen by comparing the initial conditions and the recursion formulas of these three sequences.
For any n, A333787(n) is either the nearest power of 2 reached (= 2^A064415(n)), or occurs on some of the paths from n to there.
(End)
|
|
|
LINKS
|
Michael De Vlieger, Annotated fan style binary tree labeling the index n, with a color code where black represents a(n) = 0, red a(n) = 1, and magenta the largest value in a(n) for n = 1..16383.
|
|
|
FORMULA
|
a(1) = a(2) = 0; and for n > 2, a(p) = 1 + a(p-1) if p is an odd prime and a(n*m) = a(n) + a(m) if m,n > 1. [This is otherwise equal to the definition of A064097, except here we have a different initial condition, with a(2) = 0].
a(p) = 1+a(p-1), for all odd primes p.
If A209229(n) == 1 [when n is a power of 2], a(n) = 0,
Equivalently, for non-powers of 2, a(n) = 1 + a(n-(n/A078701(n))),
or equivalently, for non-powers of 2, a(n) = 1 + Min a(n - n/p), for p prime and dividing n.
(End)
(End)
|
|
|
EXAMPLE
|
The trajectory of 15 is {12, 8}, taking 2 iterations to reach 8 = 2^3. So a(15) is 2.
Considering all possible paths from 15 to 1 nondeterministic map k -> k-(k/p), where p can be any prime factor of k, we obtain the following graph:
15
/ \
/ \
10 12
/ \ / \
/ \ / \
5 8 6
\__ | __/|
\_|_/ |
4 3
\ /
\ /
2
|
1.
It can be seen that there's also alternative route to 8 via 10 (with 10 = 15-(15/3), where 3 is not the largest prime factor of 15), but it's not any shorter than the route via 12.
(End)
|
|
|
MATHEMATICA
|
a[n_] := Length@ NestWhileList[# - #/FactorInteger[#][[-1, 1]] &, n, # != 2^IntegerExponent[#, 2] &] -1; Array[a, 100]
|
|
|
PROG
|
(PARI)
up_to = 2^24;
A329697list(up_to) = { my(v=vector(up_to)); v[1] = 0; for(n=2, up_to, v[n] = if(!bitand(n, n-1), 0, 1+vecmin(apply(p -> v[n-n/p], factor(n)[, 1]~)))); (v); };
v329697 = A329697list(up_to);
|
|
|
CROSSREFS
|
Cf. A000265, A003401, A005087, A052126, A053575, A054725, A064097, A064415, A078701, A087436, A147545, A171462, A209229, A333123, A333787, A333790, A334107, A334109, A335875, A334204, A335880, A335881, A336396, A336466, A336469 [= a(phi(n))], A336928 [= a(sigma(n))], A336470, A336477, A339970.
Cf. A334099 (a right inverse, positions of the first occurrence of each n).
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
