Partitions Based On Permutations

[Leroy Quet]

First, thanks to Joshua Zucker for doing the
work.

It occurs to me that maybe (unless for some
reason the solution is obvious) figuring out by
hand the one representation for n=55 or the one
for n=56 might make a good puzzle for those on
Seq.fan.
:)
(Maybe this would be too easy, since we know for
one thing that the permutation
(m(1),m(2),...m(j)) is its own inverse
permutation. Maybe figuring out a representation
for other n's would be a better puzzle.)

Thanks,
Leroy Quet


--- Joshua Zucker <joshua.zucker at gmail.com>
wrote:

> On Dec 4, 2007 10:31 AM, Leroy Quet
> <q1qq2qqq3qqqq at yahoo.com> wrote:
> > %I A000001
> > %S A000001
> > 1,0,0,1,1,0,0,0,0,1,2,0,2,1,0,0,0,0,0,1,3
> > %N A000001 a(n) = the total number of
> > permutations (m(1),m(2),m(3)...m(j)) of
> > (1,2,3,...,j) where n = 1*m(1) + 2*m(2) +
> 3*m(3)
> > + ...+j*m(j), where j is over all positive
> > integers.
> 
> With PURE brute force (checking all
> permutations for each j such that
> 1*j + 2*(j-1) + ... + j*1 is less than n), in
> about 10 seconds my
> program gives the following first 164 terms:
> 1 0 0 1 1 0 0 0 0 1 2 0 2 1 0 0 0 0 0 1 3 1 4 2
> 2 2 4 1 3 1 0 0 0 0 1
> 4 3 6 7 6 4 10 6 10 6 10 6 10 4 6 7 6 3 4 1 1 5
> 6 9 16 12 14 24 20 21
> 23 28 24 34 20 32 42 29 29 42 32 20 34 24 28 23
> 21 20 25 20 22 30 38
> 32 40 47 55 54 74 70 84 90 78 90 129 106 123
> 134 147 98 168 130 175
> 144 168 144 184 144 168 144 175 130 168 98 148
> 141 138 128 176 144 148
> 184 213 194 252 237 292 284 311 290 408 363 390
> 406 518 394 542 492
> 640 557 666 595 776 684 786 718 922 745 917 781
> 982 826 950 844 1066
> 845 936 845 1066
> 
>...




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