|
|
A109468
|
|
a(n) is the number of permutations of (1,2,3,...,n) written in binary such that no adjacent elements share a common 1-bit.
|
|
1
|
|
|
1, 2, 0, 4, 2, 0, 0, 0, 8, 32, 0, 8, 0, 0, 0, 0, 0, 64, 0, 1968, 508, 0, 0, 0, 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1024, 0, 0, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
In other words, if b(m) and b(m+1) are adjacent elements written in binary, then (b(m) AND b(m+1)) = 0 for 1 <= m <= n-1. (If a logical AND is applied to each pair of adjacent terms, the result is zero.)
Let 2^k be the largest power of 2 <= n. Note that element 2^k-1 can be adjacent only to 2^k. So 2^k-1 must be at the beginning or the end of the permutation while 2^k must be next to 2^k-1. The elements 2^k-1-2^i (i=1,...,k-1) can be adjacent only to 2^i, 2^k and 2^k+2^i implying that n must be >=2^k+2^(k-3) to yield a nonzero number of permutations.
Let 2^k be the smallest power of 2 that is greater than n. Note that for 0 <= i <= k-1, the element 2^k-1-2^i can only be adjacent to 2^i, so it must be at the beginning or the end of the permutation. If n >= 2^k-1-2^(k-3) and k >= 3, there are at least three such elements <= n, which is impossible, so a(n) = 0. Together with the preceding comment, this means that a(n) = 0 when 2^k-2^(k-3)-1 <= n <= 2^k+2^(k-3)-1, k >= 3, i.e., when n is in one of the intervals 6-8, 13-17, 27-35, 55-71, ... . - Pontus von Brömssen, Aug 15 2022
|
|
LINKS
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,more
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|