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

1, 1, 2, 3, 4, 6, 8, 12, 15, 24, 30, 40, 60, 72, 120, 144, 180, 240, 360, 420, 720, 840, 1008, 1260, 1680, 2520, 2880, 5040, 5760, 6720, 8064, 10080, 13440, 20160, 22680, 40320, 45360, 51840, 60480, 72576, 90720, 120960, 181440, 201600, 362880, 403200

Offset: 0

Author

Erich Friedman

Keywords

• nonn
• nice

Comments

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

Examples

			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]
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]
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000 (terms n = 1..1000 from T. D. Noe)
• Tomislav Doslic, Maximum product over partitions into distinct parts, J. of Integer Sequences, Vol. 8 (2005), Article 05.5.8.
• Andrew V. Sills and Robert Schneider, The product of parts or "norm" of a partition, arXiv:1904.08004 [math.NT], 2019. Also in Integers, (2020) Vol. 20A, Article #A13.

Crossrefs

Cf. A000792, A025147, A309859 (corresponding partitions).

Programs

Maple

b:= proc(n, i) option remember; `if`(i*(i+1)/2 b(n$2):
seq(a(n), n=0..50);  # Alois P. Heinz, Apr 19 2019
# Alternative:
A034893:=proc(n)
    local b,k,r;
    if n<2 then return 1 fi;
    k:=floor((sqrt(8*n+9)-1)/2);
    b:=iquo(n-k*(k+1)/2+1,k-1,r)+1;
    return (b+k)!/(b!*(b+k-r))
end proc;
seq(A034893(n),n=0..45); # Felix Huber, Apr 04 2025

Mathematica

Table[Max[Times@@@Select[IntegerPartitions[n],Max[Tally[#][[All,2]]]<2&]],{n,50}] (* Harvey P. Dale, May 28 2017 *)
b[n_, i_] := b[n, i] = If[i(i+1)/2Jean-François Alcover, Aug 21 2019, after Alois P. Heinz *)

Formula

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

Extensions

a(0)=1 prepended by Alois P. Heinz, Apr 19 2019