cp's OEIS Frontend

For n > 1, the number of permutations of n letters without overlaps [Sade, 1949]. - N. J. A. Sloane, Jul 05 2015

Number of ways to fold a strip of n labeled stamps with leaf 1 on top. [Clarified by Stéphane Legendre, Apr 09 2013]

From Roger Ford, Jul 04 2014: (Start)

The number of semi-meander solutions for n (a(n)) is equal to the number of n top arch solutions in the intersection of A001263 (with no intersecting top arches) and A244312 (arches forming a complete loop).

The top and bottom arches for semi-meanders pass through vertices 1-2n on a straight line with the arches below the line forming a rainbow pattern.

The number of total arches going from an odd vertex to a higher even vertex must be exactly 2 greater than the number of arches going from an even vertex to a higher odd vertex to form a single complete loop with no intersections.

The arch solutions in the intersection of A001263 (T(n,k)) and A244312 (F(n,k)) occur when the number of top arches going from an odd vertex to a higher even vertex (k) meets the condition that k = ceiling((n+1)/2).

Example: semi-meanders a(5)=10.

(A244312) F(5,3)=16 { 10 common solutions: [12,34,5 10,67,89] [16,23,45,78,9 10] [12,36,45,7 10,89] [14,23,58,67,9 10] [12,3 10,49,58,67] [18,27,36,45,9 10] [12,3 10,45,69,78] [18,25,34,67,9 10] [14,23,5 10,69,78] [16,25,34,7 10,89] } + [18,27,34,5 10,69] [16,25,3 10,49,78] [18,25,36,49,7 10] [14,27,3 10,58,69] [14,27,36,5 10,89] [16,23,49,58,7 10]

(A001263) T(5,3)=20 { 10 common solutions } + [12,38,45,67,9 10] [1 10,29,38,47,56] [1 10,25,34,69,78] [14,23,56,7 10,89] [12,3 10,47,56,89] [18,23,47,56,9 10] [1 10,29,36,45,78] [1 10,29,34,58,67] [1 10,27,34,56,89] [1 10,23,49,56,78].

(End)

From Roger Ford, Feb 23 2018: (Start)

For n>1, the number of semi-meanders with n top arches and k concentric starting arcs is a(n,k)= A000682(n-k).

/\ /\

Examples: a(5,1)=4 //\\ / \ /\

A000682(5-1)=4 ///\\\ / /\\ / \ /\ /\

/\////\\\\, /\//\//\\\, /\/\//\/\\, /\ //\\//\\

a(5,2)=2 /\ a(5,3)=1 /\

A000682(5-2)=2 /\ //\\ /\ /\ A000682(5-3)=1 //\\ /\

//\\///\\\, //\\//\\/\ ///\\\//\\

a(5,4)=1 /\

A000682(5-4)=1 //\\

///\\\

////\\\\/\. (End)

For n >= 4, 4*a(n-2) is the number of stamp foldings with leaf 1 on top, with leaf 2 in the second or n-th position, and with leaf n and leaf n-1 adjacent. Example for n = 5, 4*a(5-2) = 8: 12345, 12354, 12453, 12543, 13452, 13542, 14532, 15432. - Roger Ford, Aug 05 2019

From Martin Philp, Mar 25 2021: (Start)

The condition of having leaf n and leaf n-1 adjacent is the same as having one fewer leaf, and then counting each element twice. So the above comment is equivalent to saying:

For n >= 3, 2*a(n-1) is the number of stamp foldings with leaf 1 on top and leaf 2 in the second or n-th position. Example for n = 4, 2*a(4-1) = 4: 1234, 1243, 1342, 1432. Furthermore the number of stamp foldings with leaf 1 on top and leaf 2 in the n-th position is the same as the number of stamp foldings with leaf 1 on top and leaf 2 in the second position, as a cyclic rotation of 1 and mirroring the sequence maps one to the other. 1234, 1243 <-rot-> 2341, 2431 <-mirror-> 1432, 1342.

Hence, for n >= 2, a(n-1) is the number of stamp foldings having 1 and 2 (in this order) on top.

Not only is a(n) the number of stamp foldings with 1 on top, it is the number of stamp foldings with any particular leaf on top. This explains why A000136(n)= n*a(n).

(End)

The number of semi-meanders that in the first exterior top arch has exactly one arch of length one = Sum_{k=1..n-1} a(k). Example: for n = 5, Sum_{k=1..4} A000682(k) = 8, 10 = arch of length one, *start and end of first exterior top arch*; *10*11001100, *10*11110000, *10*11011000, *10*10110100, *1100*111000, *1100*110010, *111000*1100, *11110000*10. - Roger Ford, Jul 12 2020

This uses a too broad notion of equivalence. Besides the obvious reflection in a plane perpendicular to the straight line, if the end of the curve is in a free region of the plane, it is extended to infinity and the direction of the curve can then be reversed. A000560 uses a better definition of equivalence.

The 12th row of the triangle given in the Sade reference is incorrect, since the first column of this triangle is known (it is A000560).

The 12th row of the triangle (as given in the reference) is definitely wrong, since the first column of this triangle is known (it is A000560). The row sums are also known - see A000682.

From Roger Ford, Jul 06 2016: (Start)

To determine the first increasing run of the permutation 176852943 start on the left and move to the right counting the consecutive integers.

(1)7685(2)94(3). This permutation a has a first run of (3-1)=2. The permutation 123465 has a first run of (5-1)=4. (1)(2)(3)(4)6(5). (End)