

A125584


Maximum number of divisors of Product(a_i) + Product(b_j) over all (disjoint) partitions of {1..n} into {a_i} and {b_j}.


1



2, 2, 2, 2, 4, 4, 12, 20, 16, 24, 64, 96, 144, 128, 320, 384, 512, 1008, 1296, 1024, 2700, 2592, 4800
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,1


COMMENTS

Answering a question asked by Leroy Quet in rec.puzzles on 20070105.
The terms were also calculated by Peter Pein and J. K. Haugland.


LINKS

Table of n, a(n) for n=0..22.
L. Quet, MultiplyThenAdd "Game", USENET post to rec.puzzles.


EXAMPLE

a(1) = 2 because the product over the empty set is defined here as 1. So we have a(1) = number of divisors of (1+1).
For n = 6 the maximum number of divisors occurs when S = 1*3*4*5 + 2*6 = 72. (This 12divisor solution is not unique.) So a(6) is the number of positive divisors of 72, which is 12.
a(7) = 20 because of the partition 3*4 + 2*5*6*7 = 432, which has 20 divisors (and no other partition yields more).


MAPLE

A125584 := proc(n) local bc, a, b, c, i, j, bL, S, bsiz ; a := 0 ; bc := {seq(i, i=1..n)} ; for bsiz from 0 to floor(n/2) do bL := combinat[choose](bc, bsiz) ; for i from 1 to nops(bL) do b := convert(op(i, bL), set) ; c := bc minus b ; if nops(b) = 0 then b := 1; else b := mul(j, j=b) ; fi ; if nops(c) = 0 then c := 1; else c := mul(j, j=c) ; fi ; S := numtheory[tau](c+b) ; a := max(a, S) ; od: od: RETURN(a) ; end: for n from 1 do A125584(n) ; od; # R. J. Mathar, Nov 11 2007


PROG

(MAGMA) [ n lt 3 select 2 else Max([NumberOfDivisors(x + (Factorial(n) div x)) where x is &*s : s in Subsets({3..n}) ] : n in [0..20] ];


CROSSREFS

Sequence in context: A010578 A240674 A005866 * A230447 A029078 A131799
Adjacent sequences: A125581 A125582 A125583 * A125585 A125586 A125587


KEYWORD

nonn


AUTHOR

Geoff Bailey (geoff(AT)maths.usyd.edu.au), Jan 04 2007


EXTENSIONS

2 more terms from R. J. Mathar, Nov 11 2007
Edited by N. J. A. Sloane, Jul 03 2008 at the suggestion of R. J. Mathar


STATUS

approved



