A095392 Numbers n such that more than half of the reduced-residue system modulo 210 consists of primes in the following sense: in {210n + R} more than 24 = phi(210)/2 primes occur, i.e., 25-33, 35, 46.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 20, 22, 23, 26, 27, 29, 31, 36, 41, 44, 46, 48, 56, 59, 70, 72, 74, 95, 109, 113, 114, 127, 132, 136, 148, 312, 321, 347, 428, 506, 538, 551, 1274, 1296, 1442, 2875, 4576, 5504, 6928, 7870, 12880, 15745, 17518
Offset: 1
Keywords
Examples
210n + r, where r runs through RRS of 210 corresponds to prime-difference patterns with several relatively small first prime differences.
n=18543: 210*18543 + r includes 26 primes with the following difference pattern: {2,4,2,4,30,18,2,10,6,12,2,18,6,10,2,12,12,4,6,8,6,6,4,2,10}.
Programs
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Mathematica
{k=0};Do[{m=0}; Do[s=210k+r; s1=210k+r+2;If[PrimeQ[s], m=m+1], {r, 1, 210}]; If[Greater[m, 24], Print[{m, k}]], {k, 0, 10000000}]
Formula
Solutions to A095390(x) > 24 = phi(210).