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A339743
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a(n) is the least k > 0 such that 1+k, 1+2*k, ..., 1+n*k are pairwise coprime.
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3
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1, 1, 2, 4, 6, 6, 6, 30, 30, 60, 60, 210, 210, 210, 210, 210, 210, 210, 2310, 2310, 2310, 2310, 18480, 120120, 120120, 150150, 150150, 150150, 150150, 660660, 1531530, 2492490, 3063060, 3063060, 4594590, 38798760, 38798760, 38798760, 38798760, 38798760, 48498450, 193993800
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OFFSET
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1,3
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COMMENTS
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In other words, a(n) is the least k such that A339749(k) >= n.
If 2*p < n then p | a(n) for prime p. Proof: Suppose for such p, we don't have p | a(n). Then for some 1 <= k <= n/2 we have p | 1 + a(n)*k and so we have p | 1 + a(n) * (k + p) as well and gcd(1 + a(n) * k, 1 + a(n) * (k + p)) >= p > 1. Contradiction.
This sequence is weakly increasing. Proof: Let m > n. Then if a(m) < a(n) then by construction a(n) = a(m). Contradiction.
Let f(n) = Product_{i = 1..pi(n/2)} and let c(n) be a candidate for a(n) such that f(n) | c(n) where pi = A000720. If for some 1 + m*c(n) and 1 + (m + t)*c(n) we have gcd(1 + m * c(n), 1 + (m + t)*c) > 1 then n / 2 < t < n and t is prime. (End)
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LINKS
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EXAMPLE
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For n = 5:
- so a(5) = 6.
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PROG
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(PARI) { n = 1; for (k=1, 38798760, p = 1; for (m=1, oo, if (gcd(p, 1+m*k)>1, break, p *= 1+m*k; if (m==n, print1 (k ", "); n++)))) }
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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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