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A086566 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). 0
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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.

LINKS

Table of n, a(n) for n=1..39.

Carlos Rivera, Conjecture 30. The Firoozbakht Conjecture, The Prime Puzzles and Problems Connection.

FORMULA

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])

EXAMPLE

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).

MATHEMATICA

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}]

CROSSREFS

Cf. A000040.

Sequence in context: A237288 A293074 A005105 * A235213 A188552 A104892

Adjacent sequences:  A086563 A086564 A086565 * A086567 A086568 A086569

KEYWORD

nonn

AUTHOR

Farideh Firoozbakht, Sep 11 2003

STATUS

approved

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Last modified December 2 13:34 EST 2021. Contains 349444 sequences. (Running on oeis4.)