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A338759 a(n+1) is the maximum number of groups which can be built from the terms in this sequence so far and using each term only once which result in a(n) as their product with a(1) = 1.

%I A338759 #35 Dec 13 2020 16:30:19
%S A338759 1,1,2,1,3,1,4,1,5,1,6,2,2,3,2,4,4,5,2,5,3,3,4,6,6,7,1,7,2,6,8,5,4,8,
%T A338759 7,3,5,5,6,10,7,4,9,3,6,12,13,1,8,9,5,7,5,8,10,8,11,1,9,6,13,2,7,6,14,
%U A338759 7,7,8,12,15,7,9,7,10,10,11,2,8,13,3,7,11,3,8,14
%N A338759 a(n+1) is the maximum number of groups which can be built from the terms in this sequence so far and using each term only once which result in a(n) as their product with a(1) = 1.
%C A338759 This sequence is a variant of A332518 without the requirement that all factors have to be consecutive numbers.
%H A338759 S. Brunner, <a href="/A338759/b338759.txt">Table of n, a(n) for n = 1..10000</a>
%H A338759 S. Brunner, <a href="https://pastebin.com/TdhhvXLk">List for n = 1..10000 together with the factorizations</a>.
%H A338759 S. Brunner, <a href="https://pastebin.com/cGZ4H6xr">Python program</a>.
%e A338759 To get a(n+1), count how many times a(n) appears in the sequence.
%e A338759 For 1 and primes, this is already a(n+1).
%e A338759 For prime squares, also count how many times the prime factor appears in the sequence, divide it by 2 and round it down.
%e A338759 For example, the next term after a(43) = 9 is 3, because 9 appeared 1 time (at a(43) itself) and 3 appeared 5 times, which can arranged in 2 groups of 3 X 3.
%e A338759 For semiprimes, count how many times the semiprime itself appears in the sequence. Then count how many times the 2 factors appear and add the smallest number.
%e A338759 For example, the next term after a(30) = 6 is 8, because 6 appeared 4 times and the factors 2 and 3 appeared 6 and 4 times. We can build 4 groups of 2 X 3 of them.
%Y A338759 Cf. A001222 (bigomega), A162247 (all factorizations of n).
%Y A338759 Cf. A332518.
%K A338759 nonn
%O A338759 1,3
%A A338759 _S. Brunner_, Nov 07 2020