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A238289
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Least positive integer k such that prime(k*n), prime((k+1)*n) and prime((k+2)*n) form an arithmetic progression, or 0 if such a number k does not exist.
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2
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2, 2, 17, 4, 1, 1, 59, 3, 56, 1, 39, 10, 9, 130, 2, 18, 11, 7, 80, 67, 2, 19, 27, 17, 92, 73, 180, 65, 5, 110, 282, 4, 6, 8, 16, 2, 23, 198, 20, 3, 99, 83, 217, 13, 110, 28, 16, 6, 5, 3, 144, 31, 9, 187, 176, 145, 75, 11, 43, 424, 4, 54, 272, 8, 26, 131, 123, 107, 8, 4, 52, 9, 127, 84, 264, 33, 145, 663, 16, 285
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OFFSET
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1,1
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COMMENTS
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Conjecture: (i) We always have 0 < a(n) < 3*prime(n) + 9.
(ii) For any integer n > 3, there is a positive integer k < 5*n^3 such that prime(k*n), prime((k+1)*n), prime((k+2)*n), and prime((k+3)*n) form a 4-term arithmetic progression.
(iii) In general, for each m = 3, 4, ..., if n is sufficiently large then there is a positive integer k = O(n^(m-1)) such that prime((k+j)*n) (j = 0, ..., m-1) form an arithmetic progression.
The conjecture is a refinement of the Green-Tao theorem.
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LINKS
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EXAMPLE
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a(2) = 2 since prime(2*2) = 7, prime(3*2) = 13 and prime(4*2) = 19 form a 3-term arithmetic progression, but prime(1*2) = 3, prime(2*2) = 7 and prime(3*2) = 13 do not form a 3-term arithmetic progression.
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MATHEMATICA
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d[k_, n_]:=Prime[(k+1)*n]-Prime[k*n]
Do[Do[If[d[k, n]==d[k+1, n], Print[n, " ", k]; Goto[aa]], {k, 1, 3*Prime[n]+8}];
Print[n, " ", 0]; Label[aa]; Continue, {n, 1, 80}]
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PROG
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(PARI) okpr(p, q, r) = (q - p) == (r - q);
a(n) = {k = 1; while(! okpr(prime(k*n), prime((k+1)*n), prime((k+2)*n)), k++); k; } \\ Michel Marcus, Aug 28 2014
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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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