[seqfan] Re: An Arrangement Of Partitions

franktaw at netscape.net franktaw at netscape.net
Tue Nov 3 21:14:20 CET 2009

A175020 can also be described as:

Numbers in whose binary representation, the runs of 1's are in 
increasing order of length, the runs of 0's are in decreasing order of 
length, and all runs of 0's are at least as long as any run of 1's.

Yes, the position of [1^m] in the partitions of m will be P(m-1).  It 
is the last partition in the list with a part of size 1; anything with 
a part of size 2 or more will start 100... in the binary 
representation, while this partition starts 101...; and any partition 
that does not have a part of size 1 will start 11....  Removing one 
part of size 1 from the partitions of size m that have such a part 
gives each partition of m-1 uniquely.  This relationship is expressed 
by the second formula of A002865.

Franklin T. Adams-Watters

-----Original Message-----
From: Leroy Quet <q1qq2qqq3qqqq at yahoo.com>

Here are 3 sequences I just submitted:

%I A175020
%S A175020 1,2,3,4,5,7,8,9,10,12,15,16,17,18,19,21,24,31,32
%N A175020 An integer n is included if n is the smallest positive 
integer with
its particular multiset of run-lengths (of either 0 or 1 considered 
together) in
its binary representation.
%C A175020 This sequence gives a way to enumerate the unrestricted 
The number of terms of this sequence that are each >= 2^(k-1) and <= 
2^k -1 is
equal to the number of unrestricted partitions of k.
%C A175020 A175021 contains those positive integers not in this 
%e A175020 9 in binary is 1001. The run lengths form the multiset 
(1,2,1). Since
no positive integer < 9 has this same multiset of run lengths, then 9 
is in this
sequence. On the other hand, 23 in binary is 10111. The run-lengths are 
But 17 (which is < 23) in binary is 10001, which has the run-lengths of 
Since the multisets (1,1,3) and (1,3,1) are identical, then 23 is not 
in this
%Y A175020 A175021,A175022,A175023,A175024
%K A175020 base,more,nonn
%O A175020 1,2

%I A175022
%S A175022 1,2,1,2,3,1,2,3,4,2,1,2,3,4,3,5,2,1
%N A175022 a(n) = the number of runs (those of 0 and of 1 considered 
in the binary representation of A175020(n).
%C A175022 a(n) gives the number of terms in row n of irregular tables 
and A175024.
%Y A175022 Cf. A175020,A175023,A175024
%K A175022 base,more,nonn
%O A175022 1,2

%I A175023
%S A175023 
%T A175023 1,1,1,1,2,3,5
%N A175023 Irregular table read by rows: Row n (of A175022(n) terms) 
the run-lengths in the binary representation of A175020(n), reading 
left to
%C A175023 This table lists the parts of the partitions of the positive
integers. Each partition is represented exactly once in this table. If 
n is such
that 2^(m-1) <= A175020(n) <= 2^m -1, then row n of this table gives 
partition of m.
%e A175023 Table to start:
%e A175023 1
%e A175023 1,1
%e A175023 2
%e A175023 1,2
%e A175023 1,1,1
%e A175023 3
%e A175023 1,3
%e A175023 1,2,1
%e A175023 1,1,1,1
%e A175023 2,2
%e A175023 4
%e A175023 1,4
%e A175023 1,3,1
%e A175023 1,2,1,1
%e A175023 1,2,2
%e A175023 1,1,1,1,1
%e A175023 2,3
%e A175023 5
%e A175023 Note there are: 1 row that sums to 1, two rows that sum to 
2, three
rows that sum to 3, five rows that sum to 4, seven rows that sum to 5, 
where 1,2,3,5,7,... are the number of unrestricted partitions of 
%Y A175023 Cf. A175020,A175022,A175024
%K A175023 base,more,nonn,tabf
%O A175023 1,4

(A175021 contains those positive integers not in A175020. And A175024 
is the
table A175023 with the terms of each row arranged in non-descending 

As noted, this is another way to order the unrestricted partitions of 
positive integers.

Let say we group the rows of A175023 based on their row-sums.

What I wonder is, among those rows with row sum m, what is the 
of the row made up of (1,1,1,....,1), m 1's?

It seems, just looking at the small example, to be the number of 
partitions of m-1.
But all I can easily prove is that the index (among those rows summing 
to m) of
the all-1 row is <= the number of unrestricted partitions of m-1.

Any other comments?

Leroy Quet

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