A249796 Triangle T(n,k), n>=3, 3<=k<=n, read by rows. Number of ways to make n selections without replacement from a circular array of n unlabeled cells (ignoring rotations and reflection), such that the first selection of a cell adjacent to previously selected cells occurs on the k-th selection.

1, 1, 2, 2, 6, 4, 6, 18, 28, 8, 24, 72, 128, 120, 16, 120, 360, 672, 840, 496, 32, 720, 2160, 4128, 5760, 5312, 2016, 64, 5040, 15120, 29280, 43200, 47616, 32928, 8128, 128, 40320, 120960, 236160, 360000, 435264, 387072, 201728, 32640, 256, 362880, 1088640, 2136960, 3326400, 4249920, 4314240, 3121152, 1226880, 130816, 512

Offset: 1

Author

Tony Bartoletti, Nov 05 2014

Keywords

• nonn
• tabl

Comments

With m=n+3, T(m,3) = n!, T(m,m) = 2^n (easy proofs), and T(m,m-1) = A006516(n) = 2^(n-1) * (2^n - 1). Remaining supplied elements generated by exhaustive examination of permutations.

Examples

			T(3,3) = 1 since, given any permutation of <1,2,3>, the third element will be the first to be adjacent to previous elements (modulo 3), and these 6 permutations are indistinguishable given rotations and reflection. Sample table (left-justified):
.....1
.....1........2
.....2........6........4
.....6.......18.......28........8
....24.......72......128......120.......16
...120......360......672......840......496.......32
...720.....2160.....4128.....5760.....5312.....2016.......64
..5040....15120....29280....43200....47616....32928.....8128......128
.40320...120960...236160...360000...435264...387072...201728....32640......256
362880..1088640..2136960..3326400..4249920..4314240..3121152..1226880...130816......512
		

Programs

Sage

# Counting by exhaustive examination after a C program by Bartoletti.
def A249796_row(n):
    def F(p, n):
        for k in range(2, n):
            a = mod(p[k] + 1, n)
            b = mod(p[k] - 1, n)
            fa, fb = false, false
            for i in range(k):
                if a == p[i] : fa = true
                if b == p[i] : fb = true
            if fa and fb:
               counts[k] += 1
               return
    counts = [0]*n
    for p in Permutations(n):
        F(p, n)
    for k in range(2, n):
        counts[k] = counts[k] / (2*n)
    return counts
for n in range(9): A249796_row(n) # Peter Luschny, Nov 11 2014