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A246783
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Smallest number m such that all the n numbers np(m+k-1), 0 < k < n+1 are equal, where np(t) is number of primes p with prime(t) < p < prime(t)^(1+1/t).
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3
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1, 1, 1, 1, 67, 67, 67, 67, 67, 54412, 161342, 161342, 1214143, 9915018, 9915018, 68964006, 68964006, 810832784, 19867608968, 52415066804, 119937255921, 272007811177
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OFFSET
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1,5
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COMMENTS
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Conjecture: For every n, a(n) exists.
The np values for distinct terms of the sequence are: 1, 5, 11, 11, 16, 14, ... . It is interesting that for n = 1, 5, 11 & 14, np(a(n)) = n. What is the next term of this sequence (1, 5, 11, 14, ?, ... .)?
The sequence giving the np values of this sequence is A245101.
The sequence {1, 5, 11, 11, 16, 14} commented on above is expanded in A245098.
The next term of the sequence (1, 5, 11, 14, ?, ... .) commented on above is >22 and corresponds to t>10^12. - Robert Price, Nov 12 2014
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LINKS
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EXAMPLE
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a(9) = 67, since all the nine numbers np(67), np(67+1), np(67+2), ..., np(67+8) are equal and 67 is the smallest such number. Note that np(67) = 5.
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MATHEMATICA
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np[n_]:=Length[Select[Range[Prime[n]+1, Prime[n]^(1+1/n)], PrimeQ]]; a[n_]:=(For[m=1, Length[Union[Table[np[m+k-1], {k, n}]]]!=1, m++]; m); Do[Print[a[n]], {n, 15}]
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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