A214018 - OEIS

%I #33 May 05 2017 02:50:43

%S 1,1,1,1,1,1,2,1,1,2,1,1,2,1,1,3,2,1,1,2,1,1,3,3,4,2,1,2,3,3,1,3,2,3,

%T 2,1,1,1,5,3,2,3,1,1,2,3,1,1,2,3,3,2,3,3,5,5,2,1,1,2,1,3,4,2,1,3,1,7,

%U 2,3,3,3,1,3,3,1,5,1,3,3,2,1,2,3,4,3,3

%N Least k >= 1, such that prime(n) + k has the form 2^m * q, m >= 0, where q >= 2 is prime.

%C By the definition, a(n)<=p_(n+1)-p_n. It is well known that, for large n, p_(n+1)-p_n on average is approximately log(n). What is the average behavior of a(n)? By the Broughan-Qizhi inequality, A192869(n)>>n*(log(n))^2. Besides, they conjecture that A192869(n)=O(n*(log(n))^2). But in the case of this sequence, we have, on average, log(n)/2 possible odd values of k< p_(n+1)-p_n.

%C Therefore, we conjecture that, on average, a(n) is approximately c*log(n) with c in (0,1). Calculations up to 10^6 (_Peter J. C. Moses_) show that, most likely, c < 0.53 (cf. comment in A213892).

%H G. C. Greubel, <a href="/A214018/b214018.txt">Table of n, a(n) for n = 1..1000</a>

%H Kevin Broughan and Zhou Qizhi, <a href="https://doi.org/10.1017/S0004972710000067">Flat primes and thin primes</a>, Bulletin of the Australian Mathematical Society 82:2 (2010), pp. 282-292.

%e a(1)=1, since 2+1=3=2^0*3; a(2)=1, since 3+1=2^1*2.

%t Table[NestWhile[#+1&, 1, Not[Apply[Or, PrimeQ[(Prime[n]+#)/(2^Range[0,Floor[Log[Prime[n]]/Log[2]]])]]]&], {n, 100}] (* _Peter J. C. Moses_, Jul 09 2012 *)

%t Table[p = Prime[n]; k = 1; While[q = (p + k)/2^IntegerExponent[p + k, 2]; ! (q == 1 || PrimeQ[q]), k++]; k, {n, 100}] (* _T. D. Noe_, Jul 10 2012 *)

%Y Cf. A192869, A213982.

%K nonn

%O 1,7

%A _Vladimir Shevelev_, Jul 01 2012