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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.
2

%I #13 Jan 10 2018 20:25:14

%S 1,1,2,2,6,4,6,18,28,8,24,72,128,120,16,120,360,672,840,496,32,720,

%T 2160,4128,5760,5312,2016,64,5040,15120,29280,43200,47616,32928,8128,

%U 128,40320,120960,236160,360000,435264,387072,201728,32640,256,362880,1088640,2136960,3326400,4249920,4314240,3121152,1226880,130816,512

%N 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.

%C 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.

%e 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):

%e .....1

%e .....1........2

%e .....2........6........4

%e .....6.......18.......28........8

%e ....24.......72......128......120.......16

%e ...120......360......672......840......496.......32

%e ...720.....2160.....4128.....5760.....5312.....2016.......64

%e ..5040....15120....29280....43200....47616....32928.....8128......128

%e .40320...120960...236160...360000...435264...387072...201728....32640......256

%e 362880..1088640..2136960..3326400..4249920..4314240..3121152..1226880...130816......512

%o (Sage)

%o # Counting by exhaustive examination after a C program by Bartoletti.

%o def A249796_row(n):

%o def F(p, n):

%o for k in range(2, n):

%o a = mod(p[k] + 1, n)

%o b = mod(p[k] - 1, n)

%o fa, fb = false, false

%o for i in range(k):

%o if a == p[i] : fa = true

%o if b == p[i] : fb = true

%o if fa and fb:

%o counts[k] += 1

%o return

%o counts = [0]*n

%o for p in Permutations(n):

%o F(p, n)

%o for k in range(2, n):

%o counts[k] = counts[k] / (2*n)

%o return counts

%o for n in range(9): A249796_row(n) # _Peter Luschny_, Nov 11 2014

%K nonn,tabl

%O 1,3

%A _Tony Bartoletti_, Nov 05 2014