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A034893 Maximum of different products of partitions of n into distinct parts.

%I A034893 #61 Aug 04 2025 10:30:23
%S A034893 1,1,2,3,4,6,8,12,15,24,30,40,60,72,120,144,180,240,360,420,720,840,
%T A034893 1008,1260,1680,2520,2880,5040,5760,6720,8064,10080,13440,20160,22680,
%U A034893 40320,45360,51840,60480,72576,90720,120960,181440,201600,362880,403200
%N A034893 Maximum of different products of partitions of n into distinct parts.
%C A034893 For n > 1, to find the partition whose parts multiply to the maximum product, select from the partitions without "1" those with the most parts and from these the one with the smallest largest part. - _Felix Huber_, Apr 04 2025
%H A034893 Alois P. Heinz, <a href="/A034893/b034893.txt">Table of n, a(n) for n = 0..10000</a> (terms n = 1..1000 from T. D. Noe)
%H A034893 Tomislav Doslic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL8/Doslic/doslic15.html">Maximum product over partitions into distinct parts</a>, J. of Integer Sequences, Vol. 8 (2005), Article 05.5.8.
%H A034893 Andrew V. Sills and Robert Schneider, <a href="https://arxiv.org/abs/1904.08004">The product of parts or "norm" of a partition</a>, arXiv:1904.08004 [math.NT], 2019. Also in <a href="http://math.colgate.edu/~integers/u8/u8.Abstract.html">Integers</a>, (2020) Vol. 20A, Article #A13.
%F A034893 a(n) = (b + k)!/(b!*(b + k - r)) for n >= 2, where k = floor((sqrt(8*n + 9) - 1)/2), b = floor(d/(k - 1)) + 1, r = d mod (k - 1), d = n - k*(k + 1)/2 + 1. - _Felix Huber_, Apr 04 2025
%e A034893 The partitions of n = 4 into distinct parts are (4) and (1, 3) with the products of partitions being 4 and 3 respectively and the maximum product is a(4) = max(4, 3) = 4. - _David A. Corneth_, Apr 28 2020 [Corrected by _Felix Huber_, Apr 04 2025]
%e A034893 The partitions of n = 11 into distinct parts are (11), (1, 10), (2, 9), (3, 8), (4, 7), (5, 6), (1, 2, 8), (1, 3, 7), (1, 4, 6), (2, 3, 6), (2, 4, 5) and (1, 2, 3, 5) with the products of partitions being 11, 10, 18, 24, 28, 30, 16, 21, 24, 36, 40 and 30 respectively and the maximum product is a(11) = max(11, 10, 18, 24, 28, 30, 16, 21, 24, 36, 40, 30) = 40. Alternatively, find the length of the longest partitions that don't contain 1 (length 3) and from those the one with the smallest largest part: (2, 4, 5). - _Felix Huber_, Apr 04 2025 [Corrected and expanded by _Paul Bischof_, Jul 31 2025]
%p A034893 b:= proc(n, i) option remember; `if`(i*(i+1)/2<n, 0,
%p A034893      `if`(n=0, 1, max(b(n, i-1), i*b(n-i, min(n-i, i-1)))))
%p A034893     end:
%p A034893 a:= n-> b(n$2):
%p A034893 seq(a(n), n=0..50);  # _Alois P. Heinz_, Apr 19 2019
%p A034893 # Alternative:
%p A034893 A034893:=proc(n)
%p A034893     local b,k,r;
%p A034893     if n<2 then return 1 fi;
%p A034893     k:=floor((sqrt(8*n+9)-1)/2);
%p A034893     b:=iquo(n-k*(k+1)/2+1,k-1,r)+1;
%p A034893     return (b+k)!/(b!*(b+k-r))
%p A034893 end proc;
%p A034893 seq(A034893(n),n=0..45); # _Felix Huber_, Apr 04 2025
%t A034893 Table[Max[Times@@@Select[IntegerPartitions[n],Max[Tally[#][[All,2]]]<2&]],{n,50}] (* _Harvey P. Dale_, May 28 2017 *)
%t A034893 b[n_, i_] := b[n, i] = If[i(i+1)/2<n, 0, If[n==0, 1, Max[b[n, i-1], i b[n-i, Min[n-i, i-1]]]]];
%t A034893 a[n_] := b[n, n];
%t A034893 Table[a[n], {n, 0, 50}] (* _Jean-François Alcover_, Aug 21 2019, after _Alois P. Heinz_ *)
%Y A034893 Cf. A000792, A025147, A309859 (corresponding partitions).
%K A034893 nonn,nice
%O A034893 0,3
%A A034893 _Erich Friedman_
%E A034893 a(0)=1 prepended by _Alois P. Heinz_, Apr 19 2019