OFFSET
0,6
COMMENTS
Under a partial order based on the bitwise OR operation (see Wikipedia link) as a join, the set N_n = {0,1,2,...,n-1} is downward closed for all nonnegative integers n. Let N_n under the bitwise ordering be the n-th natural downset. E.g., for n = 0 to n = 9, the sets N_0 through N_9 under a bitwise ordering form the downward closed posets represented by the following Hasse diagrams:
7 7
/ | \ / | \
3 3 3 5 3 5 6 3 5 6 3 5 6
/ \ | \ | X \ | X X | | X X | | X X |
1 1 2 1 2 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 8
| \ / \ / \ | / \ | / \ | / \ | / \ \ / /
{} 0 0 0 0 0 0 0 0 0
The sequence {a(n)} lists the length of the orbit of the n-th natural downset under permutations of its atoms. Equivalently, given the smallest number k such that n <= 2^k, a(n) is the number of labeled downsets of the Boolean lattice of size 2^k which are isomorphic to the n-th natural downset (see examples for an illustration of n = 5).
Intuitively, this can be understood as the number of ways to internally rotate the n-th natural downset within this smallest Boolean lattice that it can fit while still being a downset.
More generally, for any nonnegative integer m, the number of labeled downsets of the Boolean lattice of size 2^m which are isomorphic to the n-th natural downset is given by a(n)*binomial(m,k). Thus, a(n) is the smallest (nonzero) orbit length, which obtains when binomial(m,k) = 1.
Note that k is the number of elements in the 1st rank of the posets, which is also the number of vertices in the corresponding simplicial complex, or k = ceiling(log_2(n)) for n > 0.
LINKS
Bruno L. O. Andreotti, Table of n, a(n) for n = 0..9999
Bruno L. O. Andreotti, Python program for n = 0 to 128
Wikipedia, Bitwise OR
FORMULA
Let k(n) = ceiling(log_2(n)) for n > 0, j = 2^k(n)-n, and k(j) = ceiling(log_2(j)) if j > 0, or k(j) = 0 if j = 0. Provably, a(n) = a(j)*binomial(k(n),k(j)).
EXAMPLE
For any nonnegative integer m the natural downset corresponding to N_2^m = {0,1,2,...,(2^m)-1} is a Boolean lattice. For n = 5 we have k = 3 which corresponds to the Boolean lattice N_2^k = N_8. We can illustrate a(5) = 3 under this definition based on the three downsets of N_8 which are isomorphic to N_5 (including N_5 itself):
7
/ | \
3 5 6 3 5 6
| X X | : | \ , / \ , / |
1 2 4 1 2 4 1 2 4 1 2 4
\ | / \ | / \ | / \ | /
0 0 0 0
Other examples:
a(0) = 1: N_0 = {} -> {}
a(1) = 1: N_1 = {0} -> {0}
a(2) = 1: N_2 = {0,1} -> {0,1}
a(3) = 1: N_3 = {0,1,2} -> {0,1,2}
a(4) = 1: N_4 = {0,1,2,3} -> {0,1,2,3}
a(5) = 3: N_5 = {0,1,2,3,4} -> {0,1,2,3,4}, {0,1,2,4,5}, {0,1,2,4,6}
a(6) = 3: N_6 = {0,1,2,3,4,5} -> {0,1,2,3,4,5}, {0,1,2,3,4,6}, {0,1,2,4,5,6}
a(7) = 1: N_7 = {0,1,2,3,4,5,6} -> {0,1,2,3,4,5,6}
a(8) = 1: N_8 = {0,1,2,3,4,5,6,7} -> {0,1,2,3,4,5,6,7}
a(9) = 4: N_9 = {0,1,2,3,4,5,6,7,8} -> {0,1,2,3,4,5,6,7,8}, {0,1,2,3,8,9,10,11}, {0,1,4,5,8,9,12,13}, {0,2,4,6,8,10,12,14}
PROG
(Python) # See Andreotti link.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno L. O. Andreotti, May 09 2023
STATUS
approved