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A086566
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a(1)=2; for n>1 a(n) is the largest prime number m such that a(n-1)^(1/(n-1))>m^(1/n).
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0
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2, 3, 5, 7, 11, 17, 23, 31, 47, 71, 107, 163, 241, 367, 557, 839, 1277, 1933, 2939, 4463, 6793, 10337, 15733, 23929, 36389, 55381, 84263, 128239, 195163, 297023, 452077, 688073, 1047271, 1593947, 2426041, 3692527, 5620159, 8554093, 13019651
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OFFSET
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1,1
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COMMENTS
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For this subsequence of prime numbers the sequence a(n)^(1/n) is decreasing and with this property a(n)-a(n-1) is maximal. Note that conjecture 30 of www.primepuzzles.net says A000040(n)^(1/n) is decreasing. But A000040(n)-A000040(n-1) is not maximal.
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LINKS
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FORMULA
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a[1]=2; a[n_] := a[n]=(For[m=PrimePi[a[n-1]]+1, (a[n-1]^(1/(n-1))>Prime[m]^(1/n)), m++ ]; Prime[m-1])
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EXAMPLE
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a(6)=17 because for n=2,...,6 a(n-1)^(1/(n-1))> a(n)^(1/n) and if the prime number p > 17 then a(5)^(1/5)< p^(1/6).
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MATHEMATICA
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a[1]=2; a[n_] := a[n]=(For[m=PrimePi[a[n-1]]+1, (a[n-1]^(1/(n-1))>Prime[m]^(1/n)), m++ ]; Prime[m-1]); v ={}; Do[v=Append[v, a[n]]; Print[v], {n, 47}]
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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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