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Sorted integers m = (prime(n+1)^2 - prime(n)^2)/24, where prime(n) is A000040(n), with duplicates removed.
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%I #12 Mar 07 2015 15:27:15

%S 1,2,3,5,7,10,12,13,15,17,18,23,25,27,28,30,32,33,35,37,38,40,43,45,

%T 47,52,55,58,62,65,67,70,72,75,77,80,85,87,88,93,95,100,103,105,107,

%U 110,117,118,120,127,130,133,135,137,138,140,143,147

%N Sorted integers m = (prime(n+1)^2 - prime(n)^2)/24, where prime(n) is A000040(n), with duplicates removed.

%C A069482 gives the values of (prime(n+1)^2 - prime(n)^2), in order, with duplicates.

%C For n>=3 (prime(n+1)^2 - prime(n)^2)/24 is an integer.

%C The list here is sorted with duplicates removed to examine the nature and scope coverage over the integers of these ratios.

%C a(n) values have increasing differences on average, and approximately fit a curve for the n-th distinct value, given by (1/3)*n*log(n) + (3/10)*n*log(log(n))^3 for the first 10,000 values.

%C The differences between adjacent a(n) values, examined over the first 100,000 values, indicates all integers are covered (i.e., for any integer k there is at least one n where k = a(n+1) - a(n)).

%C Prime factorization of a(n) indicates every prime will appear as a factor for at least one a(n) value.

%t DeleteDuplicates[Sort[Table[(Prime[n + 1]^2 - Prime[n]^2)/24, {n, 3, 300}]]]

%Y Cf. A069482, A000040.

%K nonn

%O 1,2

%A _Richard R. Forberg_, Feb 19 2015