A240565 - OEIS

%I #14 Apr 16 2014 15:55:00

%S 13,19,23,29,31,37,41,47,53,59,67,71,79,83,89,97,101,109,113,127,131,

%T 139,149,157,163,173,181,191,199,211,223,227,239,251,257,263,277,283,

%U 293,307,313,331,337,347,353,367,379,389,397,409,419,431,439,457,461,479,487,499

%N Largest possible subsequence Q of the primes such that sum(1/p, p in Q and p <= n) >= log log log n, for odd primes n.

%C The question arose whether one can find a subsequence Q of the primes such that the sum of the reciprocals of the terms <= n is asymptotically equivalent to log log log n, as a variant of Mertens' theorem that the partial sums of the reciprocals of the primes behaves as log log n. Clearly, the present sequence has this behavior. The possibility of imposing the inequality to hold for prime n or for all n (n>2 else log log log n is not a real number) suggests that there are infinitely many such sequences. Another sequence with the same property would be the smallest subsequence of the primes such that the partial sums of reciprocals are always smaller than log log log n (for n>16).

%C The primes not in this sequence are (2, 3, 5, 7, 11, 17, 43, 61, 73, 103, 107, 137, 151, 167, 179, 193, 197, 229, 233, 241, 269, 271, 281, 311, 317, 349, 359, 373,...), cf A240565 (= complement of this sequence within the primes).

%H J. Lapuyade-Lahorgue et al., <a href="http://www.linkedin.com/groupItem?view=&amp;gid=4510047&amp;type=member&amp;item=5858850584894910467">Find a subset Q of the prime numbers such that sum_{k in Q,k<=n} 1/k ~ log(log(log(n)))</a>, on LinkedIn.com

%F sum_{k <= n} 1/a(k) ~ log(log(log(a(n)))).

%e 1/13 > log log log 13 = -0.06,

%e 1/13 + 1/19 > log log log 19 = 0,076,

%e etc.

%e One could try to start (17, 19, ...) but then 1/17+1/19+1/23 < log log log 23, and of course any larger starting term(s) would yield the same problem after at least 3 terms.

%o (PARI) A_vec(pMax=3000, a=[], p=1, s=0, f=p->log(abs(log(log(p)))))={ while( p<pMax, p=nextprime(p+1); s>= f(p) && next; while(s+1/p < f(p) || setsearch(a, p), p=precprime(p-1)); while( #a && p<=a[#a], s -= 1/a[#a]; a=vecextract(a,"^-1")); s+=1/p; a=concat(a,p)); a }

%K nonn

%O 1,1

%A _M. F. Hasler_, Apr 07 2014