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Look at all the different ways to factorize n as a product of numbers bigger than 1, and for each factorization write down the sum of the factors; a(n) = number of different sums.
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%I #23 Oct 21 2017 16:07:23

%S 1,1,1,1,1,2,1,2,2,2,1,3,1,2,2,3,1,4,1,3,2,2,1,5,2,2,3,3,1,5,1,4,2,2,

%T 2,7,1,2,2,5,1,5,1,3,4,2,1,8,2,4,2,3,1,7,2,5,2,2,1,9,1,2,4,6,2,5,1,3,

%U 2,5,1,10,1,2,4,3,2,5,1,8,5,2,1,8,2,2,2,5,1,10,2,3,2,2,2,12,1,4,4,7,1,5,1

%N Look at all the different ways to factorize n as a product of numbers bigger than 1, and for each factorization write down the sum of the factors; a(n) = number of different sums.

%D Amarnath Murthy, Generalization of Partition Function and Introducing Smarandache Factor Partitions, Smarandache Notions Journal, Vol. 11, 1-2-3. Spring 2000.

%H Antti Karttunen, <a href="/A069016/b069016.txt">Table of n, a(n) for n = 1..10000</a>

%H Antti Karttunen, <a href="/A069016/a069016.txt">Scheme-program for computing this sequence</a>

%F a(n) <= A001055(n). - _David A. Corneth_, Oct 21 2017

%e The factorizations of 12 are (2,2,3), (2,6), (3,4), and (12), which have three distinct sums 7, 8, and 12. Hence a(12) = 3. - _Antti Karttunen_, Oct 21 2017

%e The factorizations of 30 are (2,3,5), (2,15), (3,10), (5,6) and (30), which have the 5 distinct sums 10, 17, 13, 11 and 30. Hence a(30) = 5.

%Y Cf. A001055, A034891.

%K nonn

%O 1,6

%A _Amarnath Murthy_, Apr 01 2002

%E Edited by _David W. Wilson_, May 27 2002

%E Edited by _N. J. A. Sloane_, Apr 28 2013