Partitions Based On Permutations
Leroy Quet
q1qq2qqq3qqqq at yahoo.com
Tue Dec 4 21:31:43 CET 2007
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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