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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

%I

%S 2,3,5,7,11,17,23,31,47,71,107,163,241,367,557,839,1277,1933,2939,

%T 4463,6793,10337,15733,23929,36389,55381,84263,128239,195163,297023,

%U 452077,688073,1047271,1593947,2426041,3692527,5620159,8554093,13019651

%N 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).

%C 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.

%H Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_037.htm">Conjecture 30. The Firoozbakht Conjecture</a>, The Prime Puzzles and Problems Connection.

%F a=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])

%e 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).

%t a=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}]

%Y Cf. A000040.

%K nonn

%O 1,1

%A _Farideh Firoozbakht_, Sep 11 2003

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Last modified January 27 08:21 EST 2022. Contains 350606 sequences. (Running on oeis4.)