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A280201
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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).
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1
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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, 3, 1, 1, 3, 3, 2, 4, 2, 2, 2, 0, 1, 2, 5, 1, 1, 2, 2
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OFFSET
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1,1
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COMMENTS
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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.
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
S(n) = 1.28 0.98 0.81
n=733158 was the largest available index when I analyzed a pool of primes <=10^9.
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
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LINKS
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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