The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A339457 Decimal expansion of the smallest positive number d such that numbers of the sequence floor(2^(n^d)) are distinct primes for all n>=1. 3
 1, 5, 0, 3, 9, 2, 8, 5, 2, 4, 0, 6, 9, 5, 2, 0, 6, 3, 3, 5, 2, 7, 6, 8, 9, 0, 6, 7, 8, 9, 7, 5, 8, 3, 1, 9, 9, 1, 9, 0, 7, 3, 8, 8, 4, 9, 5, 8, 1, 1, 3, 8, 4, 2, 9, 0, 0, 2, 9, 9, 9, 3, 5, 0, 6, 5, 7, 6, 5, 9, 5, 4, 7, 5, 6, 1, 6, 3, 0, 5, 7, 6, 4, 3, 1, 7, 1, 0, 1, 8, 9, 0, 8, 0, 8, 8, 6, 5, 2, 2, 4, 6, 8, 7, 4, 0, 1, 3, 0 (list; constant; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Assuming Cramer's conjecture on prime gaps, it can be proved that there exists at least one constant d such that all floor(2^(n^d)) are primes for n>=1 as large as required. The constant giving the smallest growth rate is d=1.503928524069520633527689067897583199190738... Algorithm to generate the smallest constant d and the associated prime number sequence a(n)=floor(2^(n^d)). 0.   n=1, a(1)=2, d=1 1.   n=n+1 2.   b=floor(2^(n^d)) 3.   p=smpr(b)     (smallest prime >= b) 4.   If p=b, then a(n)=p, go to 1. 5.   d=log(log(p)/log(2))/log(n) 6.   a(n)=p 7.   k=1 8.   b=floor(2^(k^d)) 9.   If b<>a(k) and b not prime, then p=smpr(b), n=k, go to 5. 10.  If b is prime, then a(k)=b 11.  If k

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified April 14 22:47 EDT 2021. Contains 342971 sequences. (Running on oeis4.)