Roger Ford - cp's OEIS frontend

A378944 Triangle read by rows: T(n,k) = number of stamp foldings with stamp #1 first, n stamps and stamp #2 covered by exactly one fold. k = the stamp number before the fold covering stamp #2 divided by 2. See examples.

2, 4, 8, 6, 20, 12, 48, 24, 28, 132, 60, 56, 348, 144, 112, 162, 1008, 396, 280, 324, 2812, 1044, 672, 648, 1076, 8420, 3024, 1848, 1620, 2152

Offset: 5

Roger Ford, Dec 11 2024

The conjectured formula for the numbers in T(n,k) involves two unsolved sequences, semi-meanders and meandric numbers.

			                          _____    __         ______________
Vertical lines = stamp#  |     |  |  |       |   __    __   |   __
Horizontal lines = folds 1  5  2  3  4       |  |  |  |  |  |  |  |
                            |  |__|  |       1  6  5  4  3  2  8  7
                            |________|          |  |__|  |__|     |
     fold 4-5 covers stamp #2  k = 4/2          |_________________|
                        Example: T(5,2)      fold 6-7 covers stamp #2 k = 6/2
                                                               Example: T(8,3)
Irregular triangle begins:
   n\k  (2)  (3)  (4)  (5)  (6)
    5:   2
    6:   4
    7:   8    6
    8:   20   12
    9:   48   24   28
   10:   132  60   56
   11:   348  144  112  162
   12:   1008 396  280  324
   13:   2812 1044 672  648  1076
   14:   8420 3024 1848 1620 2152

Cf. A000682, A005316, A077054, A301620.

T(n,k) = 2 * A000682(n+1-2*k) * A077054(k-1).

A366415 a(n) is the number of exterior top arches (no covering arch) for semi-meanders in generation n+1 that are generated by semi-meanders with n top arches and floor(n/2) exterior top arches using the exterior arch splitting algorithm.

Original entry on oeis.org

10, 34, 78, 222, 362, 938, 1326, 3246, 4242, 10002, 12438, 28566, 34330, 77338, 90654, 201246, 231458, 507938, 575526, 1251366, 1400874, 3022890, 3350574, 7184430, 7897138, 16842802, 18382902, 39026742, 42336314, 89522234, 96600126, 203554878

Offset: 4

Roger Ford, Oct 10 2023

b(n) = ((n-4)*2^floor((n-1)/2)+2)*floor(n/2) is the number of exterior top arches for all semi-meander solutions with n top arches and floor(n/2) exterior top arches. Conjecture: for n>=5, lim_{n->oo} a(n)/b(n) = 3.

			For n=5, the number of semi-meanders with 5 top arches and 2 exterior top arches is equal to A259689(5,2) = 6:
     __                                __
    //\\          __        ____      //\\        __          ____
   ///\\\    __  //\\      /  /\\    ///\\\      //\\  __    //\  \
/\////\\\\, //\\///\\\, /\//\//\\\, ////\\\\/\, ///\\\//\\, ///\\/\\/\
There are 12 exterior arches for the 6 solutions.
Solutions for generation n+1 using the exterior arch splitting algorithm:
      __
     //\\                        __                      ____
    ///\\\         __           //\\     __             /____\
   ////\\\\   __  //\\         ///\\\   //\\  __       //  __\\   __    __
/\/////\\\\\,//\\///\\\/\,/\/\////\\\\,///\\\//\\/\,/\///\//\\\\,//\\/\//\\/\
    __
   //\\                      __                        ____
  ///\\\         __         //\\               __     /____\
 ////\\\\       //\\  __   ///\\\         __  //\\   //__  \\       __    __
/////\\\\\/\,/\///\\\//\\,////\\\\/\/\,/\//\\///\\\,////\\/\\\/\,/\//\\/\//\\
These 12 solutions have 34 exterior arches. Therefore a(5) = 34.

Paolo Xausa, Table of n, a(n) for n = 4..1000
Michael LaCroix, Approaches to the Enumerative Theory of Meanders, 2003, pg. 31-31, Demonstrates arch splitting with semi-meander models.
Index entries for linear recurrences with constant coefficients, signature (1,7,-7,-18,18,20,-20,-8,8).

Cf. A259689, A365679.

LinearRecurrence[{1, 7, -7, -18, 18, 20, -20, -8, 8}, {10, 34, 78, 222, 362, 938, 1326, 3246, 4242}, 50] (* Paolo Xausa, May 28 2024 *)

For n>2:
a(2*n) = (3*n-1)*((2*n-4)*2^(n-1) + 2) - (3*n-3)*((2*n-5)*2^(n-1) + 2) + a(2*n-1);
a(2*n+1) = 3*n*((2*n-3)*2^n + 2) - 3*n*((2*n-4)*2^(n-1) + 2) + a(2*n).
G.f.: 2*x^4*(5 + 12*x - 13*x^2 - 12*x^3 + 6*x^4)/((1 - x)^2*(1 + x)*(1 - 2*x^2)^3). - Stefano Spezia, Nov 07 2023

A365679 a(n) is the number of exterior top arches (no covering arch) for semi-meanders in generation n+1 that are generated by semi-meanders with n top arches and floor((n+2)/2) exterior top arches using the exterior arch splitting algorithm.

Original entry on oeis.org

4, 10, 14, 32, 40, 88, 104, 224, 256, 544, 608, 1280, 1408, 2944, 3200, 6656, 7168, 14848, 15872, 32768, 34816, 71680, 75776, 155648, 163840, 335872, 352256, 720896, 753664, 1540096, 1605632, 3276800, 3407872, 6946816, 7208960, 14680064

Offset: 2

Roger Ford, Sep 15 2023

nonn

b(n) = the number of exterior top arches for all semi-meanders with n top arches and floor((n+2)/2) exterior top arches = (floor(n/2)+1) * 2^(floor((n-1)/2)). For n>=2, lim_{n->oo} a(n)/b(n) = 3.

			For n=4, the number of semi-meanders with 4 top arches and 3 exterior top arches is equal to A259689(4,3) = 2:
       /\    /\
/\ /\ //\\, //\\ /\ /\ = 6 exterior arches. These 6 arches will generate 6 solutions in the n+1 generation using the exterior arch splitting algorithm.
     _____           __                                    __           _____
    /   /\\    /\   //\\    _____                _____    //\\   /\    //\   \
/\ //\ //\\\, //\\ ///\\\, //\ /\\ /\ /\, /\ /\ //\ /\\, ///\\\ //\\, ///\\ /\\ /\
These 6 solutions have 14 exterior arches. Therefore a(4) = 14.

Index entries for linear recurrences with constant coefficients, signature (0, 4, 0, -4).

Cf. A259689.

a[n_]:=(2*n-Floor[(n-1)/2]) * 2^Floor[(n-1)/2]; Array[a,36,2] (* Stefano Spezia, Sep 16 2023 *)

a(n) = (2*n-floor((n-1)/2)) * 2^floor((n-1)/2).

a(30) corrected by Georg Fischer, Jun 03 2024

A363910 Triangular array read by rows: T(n,k) = the number of closed meanders with n top arches and k closed meanders in the reduction of the closed meander by the reverse of the exterior arch splitting algorithm.

Original entry on oeis.org

1, 0, 2, 0, 2, 6, 0, 6, 14, 22, 0, 28, 56, 86, 92, 0, 162, 298, 428, 518, 422, 0, 1076, 1868, 2562, 3096, 3144, 2074, 0, 7852, 13076, 17292, 20624, 21990, 19366, 10754

Offset: 1

Roger Ford, Jun 27 2023

The terms of this sequence can also be derived from sequences of consecutively numbered stamps folded with stamp 1 on top.

			n\k  1     2      3     4     5     6     7     8
1:   1
2:   0     2
3:   0     2      6
4:   0     6     14    22
5:   0    28     56    86    92
6:   0   162    298   428   518   422
7:   0  1076   1868  2562  3096  3144  2074
8:   0  7852  13076 17292 20624 21990 19366 10754
Closed meander:         Closed meander split with bottom rotated right
4 top arches            to form top of semi-meander with 8 arches
    ______                   ______
   / ____ \                 / ____ \
  / / __ \ \               / / __ \ \              __
 / / /  \ \ \             / / /  \ \ \            /  \
/ / / /\ \ \ \           / / / /\ \ \ \  /\  /\  / /\ \
\ \/ /  \/  \/           binary representation of semi-meander
 \__/                    1 1 1 1 0 0 0 0 1 0 1 0 1 1 0 0
                    Semi-meander top arches with no covering center arch  =  cm
                              START:          center |
Reduction of semi-meander:    1  1  1  1  0  0  0  0  1  0  1  0  1  1  0  0 cm(1)
Combine end of first arch     1  1  1  1  0  0  0  0e 1  0  1  0  1s 1  0  0
Oe with beginning of last        1  1  1  0  0  0  1  1  0  1  0  0  1  0
arch 1s.  0e...1s becomes        1  1  1  0  0  0e 1  1  0  1  0  0  1s 0
1...0 in the next line. The         1  1  0  0  1  1  1  0  1  0  0  0
starting 1 and ending 0             1  1  0  0e 1s 1  1  0  1  0  0  0
are removed in the next line           1  0  1  0  1  1  0  1  0  0
reducing number of arches.             1  0e 1  0  1s 1  0  1  0
by one.                                   1  1  0  0  1  0  1  0             cm(2)
                                          1  1  0  0e 1  0  1s 0
                                             1  0  1  1  0  0
                                             1  0e 1s 1  0  0
                                                1  0  1  0                   cm(3)
  Example: T(4,3) 4 starting top arches with 3 closed meanders in history.

Cf. A005315 (row sums), A001181, A005316, A000682.

T(n,n) = A001181(n).

T(n,2) = A005316(2*n-4)*2 for n > 1.

A354387 a(n) is the number of arch configuration solutions with n arches derived from 2 concentric arches using the exterior arch splitting algorithm.

Original entry on oeis.org

1, 1, 3, 6, 18, 42, 130, 332, 1048, 2836, 9078, 25578, 82730, 240124, 782956, 2324800

Offset: 2

Roger Ford, May 24 2022

Every a(n) arch configuration solution when put on top of n concentric arches has exactly 2 components (2 distinct loops).
If the starting arch configuration is /\, the exterior splitting algorithm will generate all the top arch configurations for semi-meanders. A000682(n) is the number of semi-meanders with n top arches.
For a(n) with n odd, n > 2 and a center arch of /\, a(n) = A000682(n-1).
There is an infinite number of the starting arch configurations with one exterior arch. They generate an infinite number of unique sequences.
Conjecture from Roger Ford, Aug 26 2022: (Start)
a(n) = a subset of semi-meanders A000682(n+1) with an arch of length 1 starting in the second top arch position.
Example: a(4) = 3, There are 10 semi-meanders with 5 top arches. 3 of those semi-meanders have an arch of length 1 starting in the second position.
Solutions:     /\                                 /\
              /  \           /\   /\        /\   //\\
             //\/\\ /\/\,   //\\ //\\ /\,  //\\ ///\\\
Nonsolutions:               /\          /\
         /\                //\\        //\\
        //\\   /\         ///\\\      ///\\\            /\   /\
       ///\\\ //\\,   /\ ////\\\\,   ////\\\\ /\,   /\ //\\ //\\
         /\               /\
        /  \             /  \             /\
       //\  \           /  /\\           /  \
      ///\\/\\ /\,  /\ //\//\\\,  /\ /\ //\/\\  (End)

			The splitting exterior arch algorithm involves splitting an exterior arch and moving the split ends to the first and last position of the arch configuration on the x axis. Moving the ends of the split arch will cause one arch to disappear and two new arches to appear. The example below shows one exterior arch being split in a generation.
                                              split
      split                        split       /\                       /\
       /\         split        /\   /\        //\\                /\   /  \
      //\\  =>  /\ /\ /\  =>  //\\ //\\  =>  ///\\\ /\ /\  => /\ //\\ //\/\\
arches 2           3             4                5                 6

Cf. A000682, A331499, A339179, A287548.

a(n) = A331499(n, 2).
Conjecture: For n >= 2, a(n) = Sum_{k = 2..floor((n+2)/2)} A339179 T(n,k)*(k-1).
a(n) = A287548(n, n-1) - A287548(n, n).

A347433 Irregular triangle read by rows: T(n,k) is the difference between the total arch lengths of a semi-meander multiplied by its number of exterior arches and total arch lengths of the semi-meanders with n + 1 top arches generated by the exterior arch splitting algorithm on the given semi-meander.

Original entry on oeis.org

4, 4, 4, 10, 4, 11, 4, 12, 20, 4, 13, 22, 4, 14, 24, 34, 4, 15, 26, 37, 4, 16, 28, 40, 52, 4, 17, 30, 43, 56, 4, 18, 32, 46, 60, 74, 4, 19, 34, 49, 64, 79, 4, 20, 36, 52, 68, 84, 100, 4, 21, 38, 55, 72, 89, 106, 4, 22, 40, 58, 76, 94, 112, 130, 4, 23, 42, 61, 80, 99

Offset: 2

Roger Ford, Sep 01 2021

			n = number of top arches, k = number of exterior top arches:
n\k  2   3   4   5   6
2:   4
3:   4
4:   4   10
5:   4   11
6:   4   12  20
7:   4   13  22
8:   4   14  24  34
9:   4   15  26  37
10:  4   16  28  40  52
Length of each arch = 1 + number of arches covered:
Top arches of a given semi-meander:       Arch splitting generated
n = 5, k = 2                              semi-meanders (6 top arches):
     1     1    = 2 exterior arches                /\
           /\                                     //\\
     /\   //\\                                   ///\\\
    //\\ ///\\\                           /\ /\ ////\\\\
    21   321    = 9 length of top arches  1  1  4321     = 12 length of top arches
                                            /\
                                           //\\   /\
                                          ///\\\ //\\ /\
                                          321    21   1  = 10 length of top arches
    T(5,2) = 4 + (5+2)(2-2) = 4 --------------------------- 4 = (12+10) - (2 * 9);
Top arches of given semi meander:
n = 5, k = 3                                    /\
    1   1    1   = 3 exterior arches           /  \
        /\   /\                               /    \
    /\ //\\ //\\                             //\  /\\
    1  21   21   = 7 length top arches   /\ ///\\//\\\
                                         1  521  21     = 12 length of top arches
                                                   /\
                                          /\      //\\
                                         //\\ /\ ///\\\
                                         21   1  321    = 10 length of top arches
                                            /\
                                           /  \
                                          /  /\\
                                         //\//\\\ /\ /\
                                         41 21    1  1  = 10 length of top arches
    T(5,3) = 4 + (5+2)(3-2) = 11 --------------------- 11 = (12+10+10) - (3 * 7).

Cf. A345747.

For n >= 2 and k = 2..floor((n+2)/2), T(n,k) = 4 + (n+2)*(k-2).

A346757 a(n) is the number of arches with a length of one and exactly one covering arch for semi-meanders with n top arches.

Original entry on oeis.org

0, 2, 2, 12, 20, 88, 180, 716, 1648, 6244, 15512, 57188, 149892, 543324, 1481636, 5310764, 4930380, 53102508, 152935148, 540918660, 1588618212, 5595773136, 16701973552

Offset: 2

Roger Ford, Aug 01 2021

The ratio of the number of semi-meanders in generation n+1 to the number of semi-meanders in generation n is equal to the ratio of the differences of these semi-meander subset arch sequences in successive generations.

A000682(n+1)/A000682(n) = (A260785(n+3) - a(n+1))/(A260785(n+2) - a(n)).

			n = 5, || indicates an arch of length one with exactly one covering arch.
           /\       /\                                                /\
     /\   //\\     //\\   /\        /\   /\     /\   /\              /  \
    //\\ ///\\\   ///\\\ //\\   /\ //\\ //\\   //\\ //\\ /\   /\ /\ //\/\\
     ||                   ||        ||   ||     ||   ||              ||||
       /\               /\
      /  \             /  \       /\
     //\  \           /  /\\     /  \
    ///\\/\\ /\   /\ //\//\\\   //\/\\ /\ /\
         ||           ||         ||||            a(5) = 12.

See A000682.

P. Di Francesco, O. Golinelli and E. Guitter, Meanders: exact asymptotics, arXiv:cond-mat/9910453 [cond-mat.stat-mech], 1999-2000; Nuclear Physics B, volume 570, issue 3, 27 March 2000, 699-712.

Cf. A000682, A260785.

For n >= 2, a(n) = A260785(n+2) - 2*A000682(n).

A339179 Irregular triangle read by rows: for n >= 2, 2 <= k <= floor(n/2) + 1, T(n,k) = the number of semi-meanders with n top arches, a first arch of length one and k arch groupings.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 4, 4, 2, 10, 10, 4, 24, 24, 14, 4, 66, 66, 34, 8, 174, 174, 106, 42, 8, 504, 504, 284, 98, 16, 1406, 1406, 878, 390, 114, 16, 4210, 4210, 2486, 1002, 258, 32, 12198, 12198, 7738, 3652, 1270, 290, 32, 37378, 37378, 22714, 9962, 3140, 642, 64, 111278, 111278, 71370, 34986, 13370, 3794, 706, 64

Offset: 2

Roger Ford, Nov 26 2020

			For n = 6:   /\ = arch of length one;
       /\             /\             /\             /\
      /  \           //\\           /  \           //\\       4 with 2 groupings
     /  /\\         //  \\         /    \         ///\\\
    /  /  \\       //  /\\\       //\  /\\       ////\\\\
/\ //\//\/\\\, /\ ///\//\\\\, /\ ///\\//\\\, /\ /////\\\\\,
         /\                                              /\
        //\\              /\         /\                 /  \  4 with 3 groupings
       ///\\\       /\   //\\       //\\   /\          //\  \
/\ /\ ////\\\\, /\ //\\ ///\\\, /\ ///\\\ //\\, /\ /\ ///\\/\\,
           /\                                                 2 with 4 groupings
          /  \       /\      /\
/\ /\ /\ //\/\\, /\ //\\ /\ //\\,             T(6,2) = 4, T(6,3) = 4, T(6,4) = 2;
Irregular triangle begins:
    n\k (2) (3) (4) (5) (6)
    2:   1
    3:   1
    4:   1   1
    5:   2   2
    6:   4   4   2
    7:   10  10  4
    8:   24  24  14  4
    9:   66  66  34  8
   10:   174 174 106 42  8
         ...

Cf. A259689, A301620, Row sums: A000682(n-1).

T(2,2) = T(3,2) = 1.
For n >= 4, T(n,2) = T(n,3) = A000682(n-2).
For n >= 6 and k >= 4, T(n,k) = Sum {x = k-1..floor(n/2)} (A259689(T(n-2,x))).
For n >= 4, A301620(n-3) = Sum {k = 4..floor((n+2)/2)} (T(n,k)).

A334615 a(n) is the number of semi-meanders with n top arches that has no arch of length 1 at the ends of the top arch configuration and no arch of length 1 adjacent to the center of the top arch configuration.

Original entry on oeis.org

0, 0, 0, 2, 0, 10, 6, 72, 86, 602, 982, 5426, 10558, 51246, 111602, 500076, 1177210, 5001518, 12462762, 51003906, 132711162, 528420604, 1422458280, 5547419160, 15347206464

Offset: 2

Roger Ford, Sep 08 2020

The number of semi-meanders with n top arches is A000682(n). If a formula for a(n) could be found without using the values for A000682(n) or A301620(n) then there would be a recursive formula for semi-meanders with n top arches.

			For n = 7: a(7) = 10.  11111000001100, 11110000111000, 11110000101100, 11101000110100, 11100011110000, 11100011100100, 11011000111000, 11010011101000, 11001111100000, 11001011110000.       /\
                                       /  \
                                      / /\ \
      11001011110000 -->   /\        / //\\ \    10 = arch length 1
                          //\\  /\  / ///\\\ \
                         end   center|      end
                         11       01  11    00    no 10 in designated positions.

Cf. A000682, A301620.

For n>= 4: a(n) = A301620(n) - 2*A301620(n-1) = A000682(n) - 4*A000682(n-1) + 4*A000682(n-2).

A337581 a(n) is the number of semi-meanders with n top arches that have both an arch of length 1 adjacent to the center of the top arch configuration and an arch of length 1 starting or ending the top arch configuration.

Original entry on oeis.org

1, 2, 4, 8, 16, 40, 96, 264, 696, 2016, 5624, 16840, 48792, 149512, 445112, 1387384, 4215496, 13312752, 41097864, 131146520, 410045672, 1319612232, 4169110888, 13511677040, 43060097728, 140383359392

Offset: 2

Roger Ford, Sep 01 2020

The number of semi-meanders with n top arches is A000682(n). The number of semi-meanders with n top arches that have a starting or ending arch of length 1 is 2*A000682(n-1). The number of semi-meanders with n top arches that have an arch adjacent to the center with length 1 is 2*A000682(n-1).

			For n = 5: (10) is a starting or ending arch of length 1, [10] is an arch adjacent to the center with length 1,  a(5) = 8.
(10)101[10]100, 110[10]010(10), (10)1[10]11000, 11100[10]0(10), (10)111[10]000,
111[10]000(10), (10)1[10]01100, 11001[10]0(10).
/\ = arch with length 1, (10) = arches with length 1 at ends or adjacent to center
        /\      /\                       /\
       /  \    /  \                     /  \
      / /\ \  / /\ \  /\               /    \
      1 10 0  1(10) 0(10)     /\  /\  / /\ /\\
         center|      end    (10) 10  1(10)100
                             end center|

Cf. A000682.

For n >= 4: a(n) = 4*A000682(n-2).

cp's OEIS Frontend

User: Roger Ford

A378944 Triangle read by rows: T(n,k) = number of stamp foldings with stamp #1 first, n stamps and stamp #2 covered by exactly one fold. k = the stamp number before the fold covering stamp #2 divided by 2. See examples.

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A366415 a(n) is the number of exterior top arches (no covering arch) for semi-meanders in generation n+1 that are generated by semi-meanders with n top arches and floor(n/2) exterior top arches using the exterior arch splitting algorithm.

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A365679 a(n) is the number of exterior top arches (no covering arch) for semi-meanders in generation n+1 that are generated by semi-meanders with n top arches and floor((n+2)/2) exterior top arches using the exterior arch splitting algorithm.

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A363910 Triangular array read by rows: T(n,k) = the number of closed meanders with n top arches and k closed meanders in the reduction of the closed meander by the reverse of the exterior arch splitting algorithm.

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A354387 a(n) is the number of arch configuration solutions with n arches derived from 2 concentric arches using the exterior arch splitting algorithm.

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A346757 a(n) is the number of arches with a length of one and exactly one covering arch for semi-meanders with n top arches.

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A339179 Irregular triangle read by rows: for n >= 2, 2 <= k <= floor(n/2) + 1, T(n,k) = the number of semi-meanders with n top arches, a first arch of length one and k arch groupings.

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A334615 a(n) is the number of semi-meanders with n top arches that has no arch of length 1 at the ends of the top arch configuration and no arch of length 1 adjacent to the center of the top arch configuration.

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