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%I #18 Dec 02 2018 19:01:13
%S 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,
%T 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,
%U 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
%N Least k such that floor((1+1/k)^n) is prime.
%C a(n) = 2 for n in A070759. a(n) = 3 for n in A070762 but not in A070759. - _Robert Israel_, Jan 09 2018
%D R. K. Guy, Unsolved Problems in Number Theory, E19
%H Robert Israel, <a href="/A072924/b072924.txt">Table of n, a(n) for n = 1..10000</a>
%H Robert Israel, <a href="/A072924/a072924.png">Scatter plot of a(n)/sqrt(n) for n=1..15000</a>
%F 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).
%F 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
%p f:= proc(n) local k;
%p for k from 1 do if isprime(floor((1+1/k)^n)) then return k fi od
%p end proc:
%p map(f, [$1..100]); # _Robert Israel_, Jan 09 2018
%t lkp[n_]:=Module[{k=1},While[!PrimeQ[Floor[(1+1/k)^n]],k++];k]; Array[ lkp,90] (* _Harvey P. Dale_, Dec 02 2018 *)
%o (PARI) a(n)=if(n<0,0,s=1; while(isprime(floor((1+1/s)^n)) == 0,s++); s)
%Y Cf. A070759, A070762
%K easy,nonn
%O 1,2
%A _Benoit Cloitre_, Aug 11 2002
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