OFFSET

1,2

COMMENTS

All of the inequalities in the proof [of the maximum formula by Doslic] are strict inequalities (> instead of >=). So the partition which is found to be optimal is not just >= any other, it is >. And hence it is unique. - Franklin T. Adams-Watters [Copied from the SeqFan discussion by J.-F. Alcover, Sep 19 2019]

LINKS

Jean-François Alcover, Table of n, a(n) for n = 1..27908

Tomislav Doslic, Maximum product over partitions into distinct parts, J. of Integer Sequences, Vol. 8 (2005), Article 05.5.8.

SeqFan, Is a partition with distinct parts and maximum product unique ?, Discussion on SeqFan-mailing list, September 2019.

EXAMPLE

The partitions of 10 into distinct addenda are {{10}, {9, 1}, {8, 2}, {7, 3}, {7, 2, 1}, {6, 4}, {6, 3, 1}, {5, 4, 1}, {5, 3, 2}, {4, 3, 2, 1}}, then the maximal product is attained with 5*3*2 = 30, so row(10) is {5, 3, 2}.

Table begins:

1

2

3

4

3, 2

4, 2

4, 3

5, 3

4, 3, 2

5, 3, 2

...

MATHEMATICA

$RecursionLimit = 2000;

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]]]]];

A034893[n_] := b[n, n];

sol[n_, pro_] := Do[If[pro == Product[i, {i, j, m}]/k && n == (m - j + 1)*(j + m)/2 - k , Return[ {j, k, m}]], {j, 2, 3}, {m, Floor[Sqrt[2 n]], Ceiling[Sqrt[2 n]] + 1}, {k, j + 1, m}];

row[1] = {1}; row[4] = {4}; row[n_] := Module[{j, k, m}, {j, k, m} = sol[n, A034893[n]]; DeleteCases[Range[j, m], k] // Reverse];

Array[row, 100] // Flatten (* Jean-François Alcover, Sep 14 2019, after Alois P. Heinz in A034893 *)

CROSSREFS

KEYWORD

nonn,tabf

AUTHOR

Jean-François Alcover, Aug 20 2019

EXTENSIONS

b-file extended to 1000 rows by Jean-François Alcover, Sep 14 2019

STATUS

approved