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A092067
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a(n) is the smallest number m such that m > 1 and m divides n^m + 1.
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4
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2, 3, 2, 5, 2, 7, 2, 3, 2, 11, 2, 13, 2, 3, 2, 17, 2, 19, 2, 3, 2, 23, 2, 5, 2, 3, 2, 29, 2, 31, 2, 3, 2, 5, 2, 37, 2, 3, 2, 41, 2, 43, 2, 3, 2, 47, 2, 7, 2, 3, 2, 53, 2, 5, 2, 3, 2, 59, 2, 61, 2, 3, 2, 5, 2, 67, 2, 3, 2, 71, 2, 73, 2, 3, 2, 7, 2, 79, 2, 3, 2, 83, 2, 5, 2, 3, 2, 89, 2, 7, 2, 3, 2, 5, 2, 97
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OFFSET
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1,1
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COMMENTS
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a(n)=2 iff n is odd. If n is even then every prime factor of n+1 is a solution of the equation (n^x + 1) mod x = 0, and if n is odd, the smallest prime factor of n+1 (2) is a solution of (n^x + 1) mod x = 0, so for each n, a(n) is not greater than the smallest prime factor of n+1.
Conjecture 1: All terms of this sequence are primes. We know if n is odd a(n) is the smallest prime factor of n+1.
Conjecture 2: For each n, a(n) is the smallest prime factor of n+1 or a(n)=A020639(n+1).
Proof: If a(n) is composite (kp, say) then n^(kp) == -1 (mod p), but then n^k is also congruent to -1 (mod p) by Fermat's little theorem, contradicting the assumption that a(n) was minimal. Thus, a(n) must be prime, and using Fermat's little theorem again shows that n^p == -1 (mod p) iff n == -1 (mod p), and A020639(n+1) gives the least p such that this is the case. (End)
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LINKS
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EXAMPLE
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a(6)=7 because 7 divides 6^7 + 1 and there doesn't exist m such that 1 < m < 7 and m divides 6^m + 1.
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MATHEMATICA
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a[n_] := (For[k=2, Mod[n^k+1, k]>0, k++ ]; k); Table[a[n], {n, 100}]
snm[n_]:=Module[{m=2}, While[PowerMod[n, m, m]!=m-1, m++]; m]; Array[snm, 100] (* Harvey P. Dale, Jul 31 2021 *)
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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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