

A328914


Smallest index m such that from the mth term on, the sequence {k^k mod A192135(n): k >= 0} enters into a cycle.


1



3, 3, 3, 3, 4, 7, 4, 7, 7, 4, 7, 11, 7, 11, 7, 11, 6, 11, 7, 15, 7, 15, 6, 15, 7, 15, 6, 19, 7, 19, 13, 6, 19, 19, 13, 8, 23, 6, 13, 23, 23, 8, 16, 11, 23, 16, 27, 11, 27, 8, 16, 27, 27, 16, 11, 8, 31, 16, 31, 11, 31, 16, 8, 31, 11, 22, 35, 35, 22, 8, 35, 16, 35, 22, 39, 8, 16, 25, 39, 39, 25, 12, 16, 39, 15, 25, 43
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

Let f(n) be the smallest index m such that from the mth term on, the sequence {k^k mod n: k >= 0} enters into a cycle, then:
(a) if gcd(n1,n2) = 1, then f(n1*n2) = max{f(n1), f(n2)}. For example, f(648) = max{f(8), f(81)} = 4;
(b) f(1) = 0; for primes p, f(p^e) = 1 if e <= p; f(p^e) is the largest s that is no greater than e and that s1 is divisible by p but not by p^2.
Proof: If we define b(k) = k^k mod p^e if gcd(k,p) = 1, 0 otherwise, it is easy to see that {b(k): k >= 0} is purely periodic. As a result, for every k >= f(p^e), we have either gcd(k,p) = 1 or p^e  k^k. In other words, let t be the largest number divisible by p such that p^e does not divide t^t, then f(p^e) = t+1.
If p^2 divides t, write t = r*p^2, then v(t^t,p) < v((t+p)^(t+p),p), where t(,p) is the padic valuation. This gives r = 0. As a result, f(p^e) is either 1 or a number s such that s1 is divisible by p but not by p^2. In the latter case, v((s1)^(s1),p) = s1, so s is the largest such number that is no greater than e.
The records in this sequence are {3, 4, 7, 11, 15, 19, 23, ...}


LINKS



FORMULA

a(n) = f(A192135(n)), where f is defined in the comment section.


EXAMPLE

A table for f(p^e):
p
e 2 3 5 7 11 13
1 1 1 1 1 1 1
2 1 1 1 1 1 1
3 3 1 1 1 1 1
4 3 4 1 1 1 1
5 3 4 1 1 1 1
6 3 4 6 1 1 1
7 7 7 6 1 1 1
8 7 7 6 8 1 1
9 7 7 6 8 1 1
10 7 7 6 8 1 1
11 11 7 11 8 1 1
12 11 7 11 8 12 1
13 11 13 11 8 12 1
14 11 13 11 8 12 14
15 11 13 11 15 12 14
16 11 16 16 15 12 14


PROG

(PARI) b(p, e) = if(!e, 0, if(e<=p, 1, forstep(k=e, p+1, 1, if(k%p==1&&k%(p^2)!=1, return(k)))))
L=List(); my(lim=12); forprime(p=2, lim, for(n=p+1, lim*log(lim)\log(p), listput(L, p^n))); listsort(L); L \\ generates all terms of A192135 below lim^lim
for(k=1, #L, my(p=factor(L[k])[1, 1], e=factor(L[k])[1, 2]); print1(b(p, e), ", "))


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



