[seqfan] An Arrangement Of Partitions
Leroy Quet
q1qq2qqq3qqqq at yahoo.com
Tue Nov 3 20:53:22 CET 2009
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 partitions. 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 sequence.
%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 (1,1,3). But 17 (which is < 23) in binary is 10001, which has the run-lengths of (1,3,1). Since the multisets (1,1,3) and (1,3,1) are identical, then 23 is not in this sequence.
%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 together) in the binary representation of A175020(n).
%C A175022 a(n) gives the number of terms in row n of irregular tables A175023 and A175024.
%Y A175022 Cf. A175020,A175023,A175024
%K A175022 base,more,nonn
%O A175022 1,2
%I A175023
%S A175023 1,1,1,2,1,2,1,1,1,3,1,3,1,2,1,1,1,1,1,2,2,4,1,4,1,3,1,1,2,1,1,1,2,2,1,
%T A175023 1,1,1,1,2,3,5
%N A175023 Irregular table read by rows: Row n (of A175022(n) terms) contains the run-lengths in the binary representation of A175020(n), reading left to right.
%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 one 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, etc, where 1,2,3,5,7,... are the number of unrestricted partitions of 1,2,3,4,5,...
%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 order.)
As noted, this is another way to order the unrestricted partitions of the 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 position(index) 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 unrestricted 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?
Thanks,
Leroy Quet
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