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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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1
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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 Mod[n^x+1,x]=0 and if n is odd, smallest prime factor of n+1 (2) is a solution of Mod[n^x+1,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).
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LINKS
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Table of n, a(n) for n=1..96.
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FORMULA
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a[n_] := (For[k=2, Mod[n^k+1, k]>0, k++ ];k)
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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}]
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CROSSREFS
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Cf. A020639, A092028.
Sequence in context: A135679 A092028 A020639 * A214606 A079879 A071889
Adjacent sequences: A092064 A092065 A092066 * A092068 A092069 A092070
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KEYWORD
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nonn
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AUTHOR
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Farideh Firoozbakht, Mar 28 2004
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STATUS
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approved
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