OFFSET
2,2
COMMENTS
It appears that a(p)= p-1 (for prime p) and a(2p) = p-1 (for odd prime p), but this is only speculation. I (or someone else) need to calculate more values of a(n) to settle this conjecture.
From Lars Blomberg, Oct 27 2015: (Start)
It also appears that:
When all factors of n have multiplicity=1 then a(n)=lcm(all factors - 1).
When n=p^2 then a(n)=n-p+1.
When n=2^k, k>2 then a(n)=2^(k-2)+k-1.
These conjectures hold for n <= 10000. (End)
LINKS
Lars Blomberg, Table of n, a(n) for n = 2..10000
Charles Burnette, On power maps over weakly periodic rings, arXiv:2207.14283 [math.RA], 2022. See p. 11 for a proof of the conjectures.
FORMULA
EXAMPLE
m mod 4 = {0,1,2,3,0,1,2,3,...}; m^2 mod 4 = {0,1,0,1,...}; m^3 mod 4 = {0,1,0,3,0,1,0,3,...}; but m^4 mod 4 = {0,1,0,1} = m^2 mod 4; so for higher powers of m these sequences cycle and are not distinct, thus for mod 4 we have {0,1,2,3}, {0,1} and {0,1,0,3} as distinct, so a(4) = 3.
CROSSREFS
KEYWORD
nonn
AUTHOR
Bruce Corrigan (scentman(AT)myfamily.com), Aug 10 2005
EXTENSIONS
More terms from Lars Blomberg, Oct 27 2015
STATUS
approved