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%I #10 Dec 29 2016 05:12:26
%S 3,15,13,3,19,5,4,0,1,8,8,13,0,4,2,2,1,5,0,2,0,1,0,1,0,1,1,4,5,1,1,8,
%T 3,1,1,3,3,2,4,2,2,2,0,1,2,5,1,1,2,2
%N Let the smallest of three successive primes p, p+d, p+2d be a so-called d-triple and b(n) the sequence of d-triples with d<>6. Then a(n) is the number of 6-triples between b(n) and b(n+1).
%C The sequence of all d-triples A122535(n) = (3), 47, 151, 167, (199), 251, 257, 367, 557, 587, 601, 647, 727, 941, 971, 1097, 1117, 1181, 1217, 1361, (1499), ... is the union of A047948(n) with 6-triples and b(n) with terms in brackets. There are three 6-triples between 3 and 199 and 15 6-triples between 199 and 1499. Thus a(1)=3 (see example) and a(2)=15.
%C The average of the first 10 terms is (3+15+13+3+19+5+4+0+1+8)/10 = 7.1. This means that, in this section, the 6-triples are more than 7 times as frequent as the other d-triples as a whole. Let us compare longer sections of a(n) with different magnitudes of n, for example (with S(n)=sum(a(k),k,1,n)/n): n <= 10000 100000 733158
%C S(n) = 1.28 0.98 0.81
%C n=733158 was the largest available index when I analyzed a pool of primes <=10^9.
%C Result: For small n, 6-triples are more frequent than the whole of other d-triples; for large n, the reverse is true. Does S(n) tend to zero? It seems so, see link "Tendency of a(n)". - _Gerhard Kirchner_, Dec 28 2016
%H Gerhard Kirchner, <a href="/A280201/b280201.txt">Table of n, a(n) for n = 1..10000</a>
%H Gerhard Kirchner, <a href="/A280201/a280201.pdf">Tendency of a(n)</a>
%e The first d-triples are 3 (,5,7, d=2); 47 (,53,59, d=6); 151 (,157,163, d=6); 167 (,173,179, d=6); 199 (,211,223, d=12). So there are three 6-triples between the 2-triple and the 12-triple: a(1)=3.
%Y Cf. A047948, A122535.
%K nonn
%O 1,1
%A _Gerhard Kirchner_, Dec 28 2016
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