# twice partitioned numbers

Meeussen Wouter (bkarnd) wouter.meeussen at vandemoortele.com
Tue Aug 21 13:47:43 CEST 2001

```hi,

has anyone ever encountered a problem where the partitioning of an integer
was performed twice? A flippant example would be:

Take a search party of n people, and split it up into clusters according to
any partition of n:
if n=6,  a possible partitioning is  (3+3)

now, after a time, the organiser of the search decides to split 'm up even
further :
the possibilities are
( (3),(2+1),(1+1+1) )  *outer product* ( (3),(2+1),(1+1+1) )  equals

((3)    ,(3)),     ((3),(2+1)),     ((3),(1+1+1)),
((2+1)  ,(3)),   ((2+1),(2+1)),   ((2+1),(1+1+1)),
((1+1+1),(3)), ((1+1+1),(2+1)), ((1+1+1),(1+1+1))

*** How many different results can be obtained after the second split? ***
Remark that these are more diverse than the "plane partitions", since there
is no restriction on the second 'dimension'.

Can anyone come up with a better description or example?
Table[Length[Flatten[Flatten[Outer[f,Sequence@@(Partitions/@#)
,1]]&/@Partitions[i]]],{i,20}]

1, 3, 6, 15, 28, 66, 122, 266, 503, 1027, 1913, 3874, 7099, 13799, 25501,
48508, 88295, 165942, 299649, 554545

Although the Outer Product cooresponds better to the example above, a more
sensible
calculation uses the PartitionsP-count :
Table[ Plus@@    Apply[Times,Partitions[i]/.i_Integer:>PartitionsP[i],2]
,{i,36}]

{1, 3, 6, 15, 28, 66, 122, 266, 503, 1027, 1913, 3874, 7099,
13799, 25501, 48508, 88295, 165942, 299649, 554545, 997281, 1817984,
3245430, 5875438, 10410768, 18635587, 32885735, 58399350, 102381103,
180634057, 314957425, 551857780, 958031826, 1667918758, 2881852770,
4992747861}

is it easy to set up a generating function for such a thing?

PS.
A third split *seems* to produce (but I need to check):

Table[Plus@@((Apply[Plus, #/. i_Integer-> PartitionsP[i] ,{1}]/. f->Times)&
/@
Flatten[Flatten[Outer[f,Sequence@@(Partitions/@#)
,1]]&/@Partitions[w]]),{w,16}]

{1, 5, 14, 51, 125, 429, 1039, 3258, 8254, 23554, 58934,
168803, 412177, 1114550, 2795446, 7345875}

what would happen if we 'force' each integer >1 to split into at least 2
pieces,
effectively forcing towards a deeply nested list of all ones? Do we end onto
binary trees ?

Wouter Meeussen
tel +32 (0)51 33 21 24
fax +32 (0)51 33 21 75
wouter.meeussen at vandemoortele.com

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