A072924 Least k such that floor((1+1/k)^n) is prime.

1, 2, 2, 2, 2, 2, 2, 4, 3, 3, 3, 3, 6, 8, 7, 6, 6, 7, 5, 11, 2, 2, 9, 4, 6, 10, 5, 9, 5, 6, 4, 7, 10, 11, 7, 6, 4, 3, 10, 4, 4, 3, 5, 4, 17, 6, 11, 7, 5, 14, 12, 8, 6, 11, 4, 14, 8, 7, 3, 16, 4, 21, 8, 12, 7, 8, 7, 7, 18, 12, 8, 17, 10, 12, 28, 6, 24, 16, 12, 16, 18, 7, 6, 6, 7, 11, 8, 14, 24, 8

Offset: 1

Author

Benoit Cloitre, Aug 11 2002

Keywords

• easy
• nonn

Comments

a(n) = 2 for n in A070759. a(n) = 3 for n in A070762 but not in A070759. - Robert Israel, Jan 09 2018

References

• R. K. Guy, Unsolved Problems in Number Theory, E19

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Robert Israel, Scatter plot of a(n)/sqrt(n) for n=1..15000

Crossrefs

Cf. A070759, A070762

Programs

Maple

f:= proc(n) local k;
  for k from 1 do if isprime(floor((1+1/k)^n)) then return k fi od
end proc:
map(f, [$1..100]); # Robert Israel, Jan 09 2018

Mathematica

lkp[n_]:=Module[{k=1},While[!PrimeQ[Floor[(1+1/k)^n]],k++];k]; Array[ lkp,90] (* Harvey P. Dale, Dec 02 2018 *)

PARI

a(n)=if(n<0,0,s=1; while(isprime(floor((1+1/s)^n)) == 0,s++); s)

Formula

It seems that a(n)/sqrt(n) is bounded. More precisely for n large enough it seems that (1/2)*sqrt(n) < a(n) < 3*sqrt(n).

On the contrary, A.L. Whiteman conjectured that the sequence floor(r^n) for non-integer rational r > 1 always contains infinitely many primes. If this conjecture is true for some r=1+1/k, then lim inf_{n -> infinity} a(n) is finite. - Robert Israel, Jan 09 2018