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%I #33 May 30 2018 03:08:41
%S 0,1,2,3,2,4,2,5,5,3,6,5,3,5,6,7,3,7,6,2,8,4,8,9,5,3,6,2,6,14,5,8,3,
%T 11,3,9,7,6,8,8,3,13,2,6,3,14,15,5,3,7,9,3,11,8,9,9,3,9,6,3,13,16,7,3,
%U 6,16,8,13,3,6,9,10,9,9,6,8,11,6,12,14,4,14,2,10,7,8,11,6,4,6,16,10,6,13
%N Number of distinct prime factors of all composite numbers between n-th and (n+1)st primes.
%C From _Lei Zhou_, Mar 18 2014: (Start)
%C This is also the number of primes such that the (n+1)-th prime (mod i-th prime) is smaller than the (n+1)-th prime (mod n-th prime) for 1 <= i < n.
%C Proof: We denote the n-th prime number as P_n. Suppose P_(n+1) mod P_i = k; we can write P_(n+1) = m*P_i + k. Setting l = P_(n+1) - P_n, the composite numbers between P_n and P_(n+1) will be consecutively m*P_i + C, where C = k-l+1, k-l+2, ..., k-1. If k < l, there must be a value at which C equals zero since k-1 > 0 and k-l+1 <= 0, so P_i is a factor of a composite number between P_n and P_(n+1). If k >= l, all C values are greater than zero, thus P_i cannot be a factor of a composite number between P_n and P_(n+1). (End)
%H T. D. Noe, <a href="/A052297/b052297.txt">Table of n, a(n) for n = 1..10000</a>
%e n=30, p(30)=113, the next prime is 127. Between them are 13 composites: {114, 115, ..., 126}. Factorizing all and collecting prime factors, the set {2,3,5,7,11,13,17,19,23,29,31,41,59,61} is obtained, consisting of 14 primes, so a(30)=14.
%t Length[Union[Flatten[Table[Transpose[FactorInteger[n]][[1]],{n, First[#]+ 1, Last[#]-1}]]]]&/@Partition[Prime[Range[100]],2,1] (* _Harvey P. Dale_, Jan 19 2012 *)
%Y Cf. A052180, A052248, A061214, A077218.
%K nonn
%O 1,3
%A _Labos Elemer_, Feb 09 2000
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