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Least k such that floor(2^n/k) is prime.
1

%I #12 Aug 31 2020 02:40:25

%S 1,2,3,3,6,9,11,11,7,9,5,10,19,11,5,10,9,11,22,35,39,9,5,10,20,27,11,

%T 19,9,18,36,25,29,27,5,10,20,40,61,13,21,42,29,27,39,9,17,29,58,49,27,

%U 25,50,11,22,44,39,11,22,44,29,58,116,53,19,38,76,152,237,139,5,10,20

%N Least k such that floor(2^n/k) is prime.

%H Amiram Eldar, <a href="/A074717/b074717.txt">Table of n, a(n) for n = 1..10000</a>

%F There is probably a constant c such that Sum_{i=1..n} a(i) is asymptotic to c*n^2 (0 < c < 1/2).

%t a[n_] := Module[{k = 1}, While[! PrimeQ @ Floor[2^n/k], k++]; k]; Array[a, 100] (* _Amiram Eldar_, Aug 31 2020 *)

%o (PARI) a(n)=if(n<0,0,k=1; while(isprime(floor(2^n/k)) == 0,k++); k)

%Y Cf. A013597, A013603, A035050, A085427.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, Sep 04 2002