%I #53 Apr 27 2023 21:33:12
%S 1,1,1,1,1,2,1,1,2,3,1,2,2,3,5,1,3,2,3,5,7,2,5,3,4,6,7,11,3,8,5,6,6,8,
%T 11,15,7,13,9,9,9,10,12,16,22,11,20,15,17,14,15,16,18,24,30,18,30,26,
%U 28,22,27,21,25,27,33,42,36,45,43,46,38,44,33,43,36,44,47,60,46,66,64,70,63,72,61,69,60,63,58,69,80
%N Number of maximally refined partitions of n into distinct parts.
%C Let a_1,a_2,...,a_k be a partition of n into distinct parts. We say that this partition can be refined if one of the summands, say a_i can be replaced with two numbers whose sum is a_i and the resulting partition is a partition into distinct parts. For example, the partition 5+2 can be refined because 5 can be replaced by 4+1 to give 4+2+1. If a partition into distinct parts cannot be refined we say that it is maximally refined.
%C The value of a(0) is taken to be 1 as is often done when considering partitions (also, the empty partition cannot be refined).
%C This sequence was suggested by _Moshe Shmuel Newman_.
%H Massimo Lauria, <a href="/A179009/b179009.txt">Table of n, a(n) for n = 0..1500</a> (first 1000 terms by Fausto A. C. Cariboni)
%H Riccardo Aragona, Lorenzo Campioni, Roberto Civino, and Massimo Lauria, <a href="https://arxiv.org/abs/2111.11084">On the maximal part in unrefinable partitions of triangular numbers</a>, arXiv:2111.11084 [math.CO], 2021.
%H Riccardo Aragona, Roberto Civino, and Norberto Gavioli, <a href="https://arxiv.org/abs/2301.06347">A modular idealizer chain and unrefinability of partitions with repeated parts</a>, arXiv:2301.06347 [math.RA], 2023.
%H Riccardo Aragona, Roberto Civino, Norberto Gavioli and Carlo Maria Scoppola, <a href="https://arxiv.org/abs/2107.04666">Unrefinable partitions into distinct parts in a normalizer chain</a>, arXiv:2107.04666 [math.CO], 2021.
%H Riccardo Aragona, Lorenzo Campioni, Roberto Civino and Massimo Lauria, <a href="https://arxiv.org/abs/2112.15096">Verification and generation of unrefinable partitions</a>, arXiv:2112.15096 [math.CO], 2021.
%H Joerg Arndt, <a href="http://www.jjj.de/fxt/demo/seq/#A179009">C++ program</a> to compute such partitions.
%e a(11)=2 because there are two partitions of 11 which are maximally refined, namely 6+4+1 and 5+3+2+1.
%e From _Joerg Arndt_, Apr 23 2023: (Start)
%e The 15 maximally refined partitions of 35 are:
%e 1: [ 1 2 3 4 5 6 14 ]
%e 2: [ 1 2 3 4 5 7 13 ]
%e 3: [ 1 2 3 4 5 8 12 ]
%e 4: [ 1 2 3 4 5 9 11 ]
%e 5: [ 1 2 3 4 6 7 12 ]
%e 6: [ 1 2 3 4 6 8 11 ]
%e 7: [ 1 2 3 4 6 9 10 ]
%e 8: [ 1 2 3 4 7 8 10 ]
%e 9: [ 1 2 3 5 6 7 11 ]
%e 10: [ 1 2 3 5 6 8 10 ]
%e 11: [ 1 2 3 5 7 8 9 ]
%e 12: [ 1 2 4 5 6 7 10 ]
%e 13: [ 1 2 4 5 6 8 9 ]
%e 14: [ 1 3 4 5 6 7 9 ]
%e 15: [ 2 3 4 5 6 7 8 ]
%e (End)
%K nonn
%O 0,6
%A _David S. Newman_, Jan 03 2011
%E More terms from _Joerg Arndt_, Jan 04 2011