OFFSET

1,2

COMMENTS

For n != 1, 6, a(n) <= 2*A112927(n): suppose n != 1, 6, by Zsigmondy's theorem, 2^n - 1 has at least one primitive factor p. Here a primitive factor p means that ord(2,p) = n, where ord(a,r) is the multiplicative order of a modulo r. So we have A118106(2p) = lcm(ord(p,2),ord(2,p)) = lcm(1,n) = n. Specially, we have A118106(2*A112927(n)) = n for n != 1, 6.

There is another way to construct m such that A118106(m) = n > 1 (and usually this way generates smaller m's than the way above): let q be any prime factor of n, again, by Zsigmondy's theorem, q^n - 1 has at least one primitive factor p unless (n,q) = (6,2). Note that q^(p-1) == 1 (mod p), so q^gcd(p-1,n) == 1 (mod p). But n is the smallest positive number such that q^n == 1 (mod p), so gcd(p-1,n) = n. So we have A118106(pq) = lcm(ord(p,q),ord(q,p)) = lcm(1,n) = n. For example, if n = 5, then q = 5, p = 11, m = 55 (the way above gives A118106(62) = 5); if n = 7, then q = 7, p = 29, m = 203 (the way above gives A118106(254) = 7); if n = 13, then q = 13, p = 53, m = 689 (the way above gives A118106(16382) = 13). This gives a(q) <= q*A212552(q) for primes q.

CROSSREFS

KEYWORD

nonn

AUTHOR

Jianing Song, Nov 03 2019

STATUS

approved