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A342381
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Triangle read by rows: T(n,k) is the number of symmetries of the n-dimensional hypercube that fix exactly 2*k facets; n,k >= 0
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1
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1, 1, 1, 5, 2, 1, 29, 15, 3, 1, 233, 116, 30, 4, 1, 2329, 1165, 290, 50, 5, 1, 27949, 13974, 3495, 580, 75, 6, 1, 391285, 195643, 48909, 8155, 1015, 105, 7, 1, 6260561, 3130280, 782572, 130424, 16310, 1624, 140, 8, 1, 112690097, 56345049, 14086260, 2347716, 293454, 29358, 2436, 180, 9, 1
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OFFSET
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0,4
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COMMENTS
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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.
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LINKS
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FORMULA
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T(n,k) = A000354(n-k)*binomial(n,k).
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EXAMPLE
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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
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PROG
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(PARI) f(n) = sum(k=0, n, (-1)^(n+k)*binomial(n, k)*k!*2^k); \\ A000354
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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