OFFSET
0,4
COMMENTS
Equivalently the number of symmetries of the n-dimensional cross-polytope that fix exactly 2*k vertices.
If a facet of the hypercube is fixed, then the opposite facet must also be fixed.
LINKS
Peter Kagey, Rows n = 0..100, flattened
Wikipedia, Cross-polytope
Wikipedia, Hypercube
Wikipedia, Hyperoctahedral group
EXAMPLE
Table begins:
n\k | 0 1 2 3 4 5 6 7 8 9
----+--------------------------------------------------------------
0 | 1
1 | 1 1
2 | 5 2 1
3 | 29 15 3 1
4 | 233 116 30 4 1
5 | 2329 1165 290 50 5 1
6 | 27949 13974 3495 580 75 6 1
7 | 391285 195643 48909 8155 1015 105 7 1
8 | 6260561 3130280 782572 130424 16310 1624 140 8 1
9 | 112690097 56345049 14086260 2347716 293454 29358 2436 180 9 1
For the cube in n=2 dimensions (the square) there is
T(2,2) = 1 symmetry that fixes all 2*2 = 4 sides, namely the identity:
2
+---+
3| |1;
+---+
4
T(2,1) = 2 symmetries that fix 2*1 = 2 sides, namely horizonal/vertical flips:
4 2
+---+ +---+
3| |1 and 1| |3;
+---+ +---+
2 4
and T(2,0) = 5 symmetries that fix 2*0 = 0 sides, namely rotations or diagonal flips:
1 4 3 3 1
+---+ +---+ +---+ +---+ +---+
2| |4, 1| |3, 4| |2, 2| |4, and 4| |2.
+---+ +---+ +---+ +---+ +---+
3 2 1 1 3
PROG
(PARI) f(n) = sum(k=0, n, (-1)^(n+k)*binomial(n, k)*k!*2^k); \\ A000354
T(n, k) = f(n-k)*binomial(n, k); \\ Michel Marcus, Mar 10 2021
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Peter Kagey, Mar 09 2021
STATUS
approved