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A329065
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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.
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0
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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A118106(203) = 7; for any m < 203, A118106(m) is not equal to 7, so a(7) = 203.
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PROG
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(PARI) a(n) = for(k=1, oo, if(A118106(k)==n, return(k))) \\ See A118106 for its program
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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