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
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Jul 01 2012
STATUS
approved