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A072924 Least k such that floor((1+1/k)^n) is prime. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
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
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
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 *)
PROG
(PARI) a(n)=if(n<0, 0, s=1; while(isprime(floor((1+1/s)^n)) == 0, s++); s)
CROSSREFS
Sequence in context: A073130 A238267 A143526 * A247869 A036263 A307378
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, Aug 11 2002
STATUS
approved

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Last modified March 29 08:13 EDT 2024. Contains 371265 sequences. (Running on oeis4.)