cp's OEIS Frontend

A382250 Irregular 3-dimensional table, where layer n is an irregular 2D table with A000041(n) columns, each of which lists the n-bit binary numbers whose run lengths correspond to a given partition.

%I A382250 #16 Mar 26 2025 22:03:28
%S A382250 0,0,0,1,0,1,3,2,0,1,7,2,4,6,3,5,0,1,15,2,8,14,3,7,4,6,12,5,9,11,13,
%T A382250 10,0,1,31,2,16,30,3,15,4,6,8,14,24,28,5,17,23,29,7,9,11,13,19,25,27,
%U A382250 10,18,20,22,26,12,21,0,1,63,2,32,62,3,31,4,6,16,30,48,60,5,33,47,61,7,15,8,14
%N A382250 Irregular 3-dimensional table, where layer n is an irregular 2D table with A000041(n) columns, each of which lists the n-bit binary numbers whose run lengths correspond to a given partition.
%C A382250 The n-th layer contains the 2^(n-1) numbers from 0 to 2^(n-1)-1, each of which corresponds, through the run lengths of the digits when written with n bits, uniquely to one of the 2^(n-1) compositions of n. (For example, 000 <=> 3, 001 <=> 2+1; 010 <=> 1+1+1.) The numbers are grouped together in columns which correspond to the distinct partitions of n, so there are A000041(n) of these, where A000041 are the partition numbers.
%C A382250 Numbers that will never be in a top row are listed in A175021 = 6, 11, 13, 14, 20, 22, 23, 25, 26, 27, 28, 29, 30, 38, .... All other numbers will eventually be in a fixed position in the top row of all large enough layers.
%F A382250 If we denote by A(n, c, r) the r-th element of column number c in layer n, then
%F A382250 A(n, c, 1) = c-1 for 1 <= c <= min(n, 6); lim_{n -> oo} A(n, c+1, 1) = A175020(c).
%F A382250 A(n, 2, 2) = 2^(n-1) - 1 is the last element of column 2 for all n > 2.
%F A382250 A(n, 3, 2) = 2^(n-2), and A(n, 3, 3) = 2^(n-1) - 2 is the last element of column 3 for all n > 3.
%F A382250 A(n, 4, 2) = 2^(n-2) - 1 is the last element of column 4 for all n > 4.
%e A382250 The table starts as follows:
%e A382250 n = 0: There is A000041(0) = 1 partition of 0, the empty partition, which equals the
%e A382250        run lengths of the empty sequence of digits of 0 written with 0 binary digits.
%e A382250        So there is 1 column with just one number, 0:
%e A382250        0
%e A382250 n = 1: Again there is A000041(1) = 1 partition of 1, so there is 1 column, which
%e A382250        contains the number 0 written with n = 1 bits, so that run lengths are (1):
%e A382250        0
%e A382250 n = 2: There are A000041(2) = 2 columns for the two partitions of 2; each column
%e A382250        contains one number: 0 = 00 <=> partition (2), resp. 1 = 01 <=> (1,1):
%e A382250        0  1
%e A382250 n = 3: There are A000041(3) = 3 columns for the three partitions of 3,
%e A382250        corresponding to the 2^2 = 4 compositions which are the run lengths of
%e A382250        0 = 000 <=> partition (3) in column 1,
%e A382250        1 = 001 and 3 = 011 (partition 2+1 = 1+2) in column 2,
%e A382250        and 2 = 010 (partition (1,1,1) or 1+1+1) in column 3:
%e A382250        0  1  2
%e A382250           3
%e A382250 n = 4: Here are A000041(4) = 5 columns for the five partitions of 4, corresponding
%e A382250        to 2^3 = 8 compositions of 4 given as run lengths of the numbers 0, ..., 7
%e A382250        written with 4 bits: Column 1 holds the number 0 = 0000 <=> partition (4),
%e A382250        column 2 holds the numbers 1 = 0001 and 7 = 0111 <=> partition 3+1 = 1+3,
%e A382250        column 3 holds 2 = 0010, 4 = 0100 and 6 = 0110 for 2+1+1 = 1+1+2 = 1+2+1,
%e A382250        column 4 holds 3 = 0011 for 2+2, and column 5 holds 5 = 0101 for 1+1+1+1:
%e A382250        0  1  2  3  5
%e A382250           7  4
%e A382250              6
%e A382250 n = 5: 0   1   2   3   4   5  10
%e A382250           15   8   7   6   9
%e A382250               14      12  11
%e A382250                           13
%e A382250 n = 6: 0   1   2   3   4   5   7   9  10  12   21
%e A382250           31  16  15   6  17      11  18
%e A382250               30       8  23      13  20
%e A382250                       14  29      19  22
%e A382250                       24          25  26
%e A382250                       28          27
%o A382250 (PARI) {layer(n)=my(M=Map(), C=[], p, i); for(k=1, 2^max(n-1,0), mapisdefined(M, p=vecsort(A101211_row(2^n-k)), &i) || mapput(M, p, i=#C=concat(C,[[]])); C[i]=concat(C[i], k-1)); C}
%Y A382250 Cf. A000041 (partition numbers), A000079 (powers of 2), A007088 (binary numbers), A101211 and A318927 (run lengths of binary numbers).
%Y A382250 The number of columns of length 1 in layer n is A000005(n).
%Y A382250 Cf. A175021 (numbers never in the first row), A175020 (limit of the first rows without initial 0).
%K A382250 nonn
%O A382250 0,7
%A A382250 _Ali Sada_ and _M. F. Hasler_, Mar 24 2025