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A224894
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a(1) = 1, a(n+1) = smallest prime divisor of 1 + product of all the primes p <= a(n).
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0
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OFFSET
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1,2
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COMMENTS
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Following Euclid's proof that there are infinitely many primes.
For example, 211 is 2*3*5*7 + 1 and 1051 is the smallest prime divisor of 2*3*5*...*211 + 1. This differs from the Euclid-Mullin sequence (A000945) because all the primes between a(n-1) and a(n) are used in calculating a(n+1).
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LINKS
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EXAMPLE
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a(5) = 2*3*5*7 + 1 = 211.
a(6) = 1051 because 1051 is the smallest prime divisor of 2*3*5*...*211 + 1.
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MATHEMATICA
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a[1] = 1; a[n_] := a[n] = Block[{pr = 1 + Product[Prime[k], {k, PrimePi@a[n - 1]}], p = NextPrime@a[n - 1]}, While[Mod[pr, p] > 0, p = NextPrime@p]; p]; Array[a, 7] (* Giovanni Resta, Jul 24 2013 *)
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CROSSREFS
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KEYWORD
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nonn,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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