

A085427


Least k such that k*2^n  1 is prime.


13



3, 2, 1, 1, 2, 1, 2, 1, 5, 7, 5, 3, 2, 1, 5, 4, 2, 1, 2, 1, 14, 7, 26, 13, 39, 22, 11, 16, 8, 4, 2, 1, 5, 6, 3, 24, 12, 6, 3, 25, 24, 12, 6, 3, 14, 7, 20, 10, 5, 19, 11, 21, 20, 10, 5, 3, 32, 16, 8, 4, 2, 1, 12, 6, 3, 67, 63, 43, 63, 40, 20, 10, 5, 15, 12, 6, 3, 55, 47, 30, 15, 30, 15, 64, 32, 16, 8
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OFFSET

0,1


COMMENTS

First few pairs (n,k) such that k > n are (1,2), (22,26), (24,39), (65,67), (110,150), (112,140), (135,150), (137,169), ... Also, for n=398 there is an interesting anomaly since k=893 which is > 2n.
Conjecture: for every n there exists a number k < 3n such that k*2^n  1 is prime. Comment from T. D. Noe: this fails at n=624, where a(n)=2163.
Define sumk = Sum_{n=1..N} k(n), and define sumn = Sum_{n=1..N} n, then as N increases the ratio sumk/sumn tends to log(2)/2 = 0.3465735.... so on average k(n) is about 0.35*n and seems to be always < 3.82*n or 11*log(2)/2.  Pierre CAMI, Feb 27 2009


LINKS



FORMULA



MATHEMATICA

k2np[n_]:=Module[{k=1, x=2^n}, While[!PrimeQ[k x1], k++]; k]; Array[ k2np, 90, 0] (* Harvey P. Dale, Nov 19 2011 *)


PROG

(PARI) lim=10^9; for(n=0, 200, k=1; i=0; while(k < lim, if(ispseudoprime(k*2^n1), print1(k, ", "); i++; break({1})); if(i==0 && k >= lim1, print1(">", lim, ", "); i=0); k++)) \\ Felix Fröhlich, Sep 20 2014


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



