%I #10 Apr 15 2017 11:14:28
%S 1,1,1,1,5,2,1,19,62,8,1,65,1322,1976,48,1,211,24182,318392,140496,
%T 480,1,665,408842,42729464,260656752,19373280,5760,1,2059,6609302,
%U 5208402488,395975417424,485262187680,4125121920,92160,1,6305,103999562,600582229496
%N Square array A(n,k): number of integers having prime(n) as k-th factor when written as product of primes in nondecreasing order, in any interval of primorial(n)^k positive integers.
%C Square array read by descending antidiagonals: A(n,k) with rows n >= 1, columns k >= 1. Primorial(n) = A002110(n): product of first n primes.
%C Visualize the prime factorizations of the positive integers as a table with row headings giving each successive integer, and the primes of which the row heading is the product listed across the columns in nondecreasing order, repeated when necessary. Except for 1, which lacks prime factors, column 1 has the row heading's least prime factor, column 2 has a value for composite numbers but is blank for primes, and so on. This sequence measures precisely how frequently the various primes occur in each column. This is possible because any given prime occurs cyclically in any given column, for the reason following.
%C The occurrence pattern of up to k factors of prime(n) in such prime factorizations has a fundamental period over the positive integers of prime(n)^k. The least common period for up to k factors of each of the first n primes is Primorial(n)^k, and this covers everything that can affect the occurrence of prime(n) in the least k factors. Thus prime(n) is k-th least prime factor of integer m if and only if it is k-th least prime factor of m+Primorial(n)^k.
%C Intermediate values in the calculation of this sequence appear in A281891.
%C A(n,1) = A005867(n-1) in accordance with the comment on A005867 dated Jul 16 2006.
%C A(2,k) = A001047(k) = 3^k - 2^k.
%F A(n,k) = primorial(n-1) * A281891(n,k-1) - prime(n)^(k-1) * A281891(n-1,k).
%e Prime(2)=3 occurs as second least factor five times in the prime factorizations of every interval of 36=Primorial(2)^2 positive integers. See A014673. So A(2,2) = 5.
%Y A079474 re-read as a square array gives values of primorial(n)^k = A002110(n)^k.
%Y The values in the body of the factorization table described in the author's comments are in the irregular array A027746.
%Y Cf. A001047, A005867, A014673, A281891.
%K nonn,tabl
%O 1,5
%A _Peter Munn_, Feb 08 2017
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