A214018 Least k >= 1, such that prime(n) + k has the form 2^m * q, m >= 0, where q >= 2 is prime.

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, 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, 2, 3, 3, 3, 1, 3, 3, 1, 5, 1, 3, 3, 2, 1, 2, 3, 4, 3, 3

Offset

1,7

Comments

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.

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).

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Kevin Broughan and Zhou Qizhi, Flat primes and thin primes, Bulletin of the Australian Mathematical Society 82:2 (2010), pp. 282-292.

Example

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

Mathematica

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 *)
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 *)

Crossrefs

Cf. A192869, A213982.

Sequence in context: A230404 A082647 A367627 * A304869 A161071 A161110

Adjacent sequences: A214015 A214016 A214017 * A214019 A214020 A214021

Keyword

nonn

Author

Vladimir Shevelev, Jul 01 2012

Status

approved