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
===============================
This email is confidential and intended solely for the use of the individual to whom it is addressed.
If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited.
You are explicitly requested to notify the sender of this email that the intended recipient was not reached.
More information about the SeqFan
mailing list