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A238281
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a(n) = |{0 < k < n: the two intervals (k*n, (k+1)*n) and ((k+1)*n, (k+2)*n) contain the same number of primes}|.
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7
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0, 1, 2, 1, 2, 3, 3, 1, 5, 2, 4, 4, 8, 3, 7, 4, 4, 4, 2, 3, 7, 3, 10, 4, 12, 7, 7, 15, 7, 9, 8, 5, 8, 9, 11, 8, 8, 10, 8, 4, 10, 10, 10, 11, 7, 10, 8, 11, 8, 8, 9, 9, 8, 11, 7, 8, 13, 10, 8, 14, 13, 4, 14, 8, 11, 12, 14, 12, 8, 10, 16, 12, 16, 12, 14, 19, 11, 14, 8, 9
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OFFSET
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1,3
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 1. Moreover, if n > 1 is not equal to 8, then there is a positive integer k < n with 2*k + 1 prime such that the two intervals ((k-1)*n, k*n) and (k*n, (k+1)*n) contain the same number of primes.
(ii) For any integer n > 4, there is a positive integer k < prime(n) such that all the three intervals (k*n, (k+1)*n), ((k+1)*n, (k+2)*n), ((k+2)*n, (k+3)*n) contain the same number of primes, i.e., pi(k*n), pi((k+1)*n), pi((k+2)*n), pi((k+3)*n) form a 4-term arithmetic progression.
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LINKS
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EXAMPLE
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a(8) = 1 since each of the two intervals (7*8, 8*8) and (8*8, 9*8) contains exactly two primes.
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MATHEMATICA
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d[k_, n_]:=PrimePi[(k+1)*n]-PrimePi[k*n]
a[n_]:=Sum[If[d[k, n]==d[k+1, n], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 80}]
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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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