A329065 Smallest m_0 such that A118106(m_0) = n; smallest m_0 such that if we write m_0 = Product_{i=1..t} p_i^e_i, then lcm_{1<=i,j<=t, i!=j} ord(p_i,p_j^e_j) = n, where ord(a,r) is the multiplicative order of a modulo r.

1, 6, 14, 10, 55, 18, 203, 34, 146, 22, 46, 26, 689, 86, 302, 51, 5759, 38, 955, 50, 98, 69, 94, 288, 505, 5462, 327, 58, 466, 77, 9305, 384, 5447, 309, 142, 74, 446, 2933, 158, 246, 3403, 129, 862, 115, 543, 141, 4702, 119, 5713, 453, 206, 106, 5671, 162, 605, 928, 687, 118

Offset: 1

Author

Jianing Song, Nov 03 2019

Keywords

• nonn

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.

Examples

			A118106(203) = 7; for any m < 203, A118106(m) is not equal to 7, so a(7) = 203.
		

Crossrefs

Cf. A118106, A112927, A212552.

Programs

PARI

a(n) = for(k=1, oo, if(A118106(k)==n, return(k))) \\ See A118106 for its program