OFFSET
1,2
COMMENTS
Two closed meanders s and t with 2n points are equivalent iff their corresponding permutations s(1) s(2) ... s(2n) and t(1) t(2) ... t(2n) have the same absolute difference sequence, i.e. |s(i+1) - s(i)| = |t(i+1) - t(i)| for all i = 1,2,..,2n, where s(1) = t(1) = s(2n+1) = t(2n+1) = 1.
LINKS
M. De Biasi, Permutation Reconstruction from Differences, Electronic Journal of Combinatorics, Volume 21 No. 4 (2014), P4.3 (23 pages).
A. Panayotopoulos, On Meandric Colliers, Mathematics in Computer Science, (2018).
J. Sawada and R. Li, Stamp foldings, semi-meanders, and open meanders: fast generation algorithms, Electronic Journal of Combinatorics, Volume 19 No. 2 (2012), P#43 (16 pages).
FORMULA
Sum_{k >= 1} k*T(n,k) = A005315(n) (closed meandric numbers).
EXAMPLE
Triangle begins:
1;
2;
8;
42;
262;
1820, 4;
13756, 32;
110394, 280;
928790, 2328, 4;
8110104, 21294, 56;
73040142, 191396, 540, 24;
674775338, 1798624, 5214, 472;
6370633938, 17113152, 48240, 6482, 32;
61269105780, 168043112, 450616, 83804, 464, 32, 0, 4;
...
For n = 6 there exist four 2-element equivalence classes:
1st class consists of permutations (1, 2, 5, 6, 7, 4, 3, 8, 9, 12, 11, 10) and (1, 2, 5, 4, 3, 6, 7, 12, 11, 8, 9, 10) having difference sequence: (1, 3, 1, 1, 3, 1, 5, 1, 3, 1, 1, 9).
2nd class consists of permutations (1, 12, 9, 10, 11, 8, 7, 2, 3, 6, 5, 4) and (1, 12, 9, 8, 7, 10, 11, 6, 5, 2, 3, 4) having difference sequence: (11, 3, 1, 1, 3, 1, 5, 1, 3, 1, 1, 3).
3rd class consists of permutations (1, 10, 9, 8, 11, 12, 7, 6, 3, 4, 5, 2) and (1, 10, 11, 12, 9, 8, 3, 4, 7, 6, 5, 2) having difference sequence: (9, 1, 1, 3, 1, 5, 1, 3, 1, 1, 3, 1).
4th class consists of permutations (1, 4, 5, 6, 3, 2, 7, 8, 11, 10, 9, 12) and (1, 4, 3, 2, 5, 6, 11, 10, 7, 8, 9, 12) having difference sequence: (3, 1, 1, 3, 1, 5, 1, 3, 1, 1, 3, 11).
CROSSREFS
KEYWORD
tabf,nonn
AUTHOR
Gerasimos Pergaris, Dec 06 2020
STATUS
approved