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Number of primes p such that sqrt(q) - sqrt(p) > 1/n, where q is the prime after p.
3

%I #25 Aug 06 2017 03:16:42

%S 0,6,22,41,75,132,186,258,330,416,511,613,724,860,1001,1163,1372,1563,

%T 1751,1965,2179,2412,2685,2945,3258,3581,3885,4194,4525,4857,5246,

%U 5644,6024,6402,6767,7229,7695,8177,8666,9156,9674,10185,10740,11283,11824

%N Number of primes p such that sqrt(q) - sqrt(p) > 1/n, where q is the prime after p.

%C Also, the number of terms by Andrica ranking which are greater than 1/n.

%H T. D. Noe, <a href="/A218015/b218015.txt">Table of n, a(n) for n = 1..200</a>

%H Marek Wolf, <a href="http://arxiv.org/abs/1010.3945">A Note on the Andrica Conjecture</a>, arXiv:1010.3945 [math.NT], 2010.

%e a(1) = 6 because only the primes 3, 7, 13, 23, 31 and 113 satisfy the criterion.

%e As an example, - sqrt(3) + sqrt(5) ~= 0.50401717 which is greater than 1/2.

%t lst = {}; p = 2; q = 3; While[p < 10^8, If[ Sqrt[q] - Sqrt[p] > 1/50, AppendTo[lst, {p, Sqrt[q] - Sqrt[p]}]]; p = q; q = NextPrime[q]]; Table[ Length@ Select[ lst, #[[2]] > 1/n &], {n, 50}]

%t nn = 50; t = Table[0, {nn}]; p = 2; q = 3; While[p < 10^8, n = Floor[1/(Sqrt[q] - Sqrt[p])]; If[n <= nn, t[[n]]++]; p = q; q = NextPrime[q]]; Join[{0}, Accumulate[t]] (* _T. D. Noe_, Oct 18 2012 *)

%Y Cf. A079296, A218012, A218014.

%K nonn

%O 1,2

%A Marek Wolf and _Robert G. Wilson v_, Oct 18 2012