cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A201631 a(n) is the number of Fibonacci meanders of length m*n and central angle 360/m degrees where m = 2.

Original entry on oeis.org

1, 3, 6, 13, 30, 70, 167, 405, 992, 2450, 6090, 15214, 38165, 96069, 242530, 613811, 1556856, 3956316, 10070871, 25674210, 65541142, 167517654, 428635032, 1097874434, 2814611701, 7221917871, 18544968768, 47655572191, 122544150258, 315313433594, 811792614547
Offset: 1

Views

Author

Peter Luschny, Jan 15 2012

Keywords

Comments

Empirically the partial sums of A051291. - Sean A. Irvine, Jul 13 2022
The above conjecture was proved by Baril et al., which also give a formal definition of the Fibonacci meanders and describe a bijection with a certain class of peakless grand Motzkin paths of length n. - Peter Luschny, Mar 16 2023

Examples

			a(3) = 6 = card({100001, 100100, 110000, 111001, 111100, 111111}).
		

Crossrefs

Programs

  • Maple
    A201631 := n -> add(A202411(k),k=0..2*n-1): seq(A201631(i),i=1..9);
    # Alternative, using the g.f. of Baril et al.:
    S := (x^2 - x + 1 - R)/((x - 1)*(x^2 - x - 1 + R)*R):
    R := (((x - 3)*x + 1)*(x^2 + x + 1))^(1/2): ser := series(S, x, 33):
    seq(coeff(ser, x, n), n = 1..31); # Peter Luschny, Mar 16 2023
    # Using a recurrence:
    a := proc(n) option remember; if n < 5 then return [0, 1, 3, 6, 13][n + 1] fi;
    (n*(2*n - 1)*(2*n - 3)*(n - 5)*a(n - 5) - (n - 4)*(2*n - 1)^2*(3*n - 5)*a(n - 4) + (2*n - 5)*(n - 3)*(2*n^2 - 3*n + 2)*a(n - 3) - (2*n - 3)*(n - 2)*(2*n^2 - 3*n + 5)*a(n - 2) + (3*n - 4)*(2*n - 1)*(2*n - 5)*(n - 1)*a(n - 1))/(n*(2*n - 3)*(2*n - 5)*(n - 1)) end: seq(a(n), n = 1..31); # Peter Luschny, Mar 16 2023
  • Mathematica
    a[n_] := Sum[A202411[k], {k, 0, 2 n - 1}];
    Array[a, 31] (* Jean-François Alcover, Jun 29 2019 *)

Formula

a(n) = Sum_{k=0..2n-1} A202411(k).
a(n) = [x^n] (x^2 - x + 1 - R)/((x - 1)*(x^2 - x - 1 + R) * R), where R = (((x - 3)*x + 1)*(x^2 + x + 1))^(1/2). (This is Theorem 21 in Baril et al.) - Peter Luschny, Mar 16 2023

A361681 Triangle read by rows. T(n, k) is the number of Fibonacci meanders with a central angle of 360/m degrees that make m*k left turns and whose length is m*n, where m = 3.

Original entry on oeis.org

1, 2, 1, 5, 2, 1, 10, 8, 2, 1, 17, 40, 8, 2, 1, 26, 161, 44, 8, 2, 1, 37, 506, 263, 44, 8, 2, 1, 50, 1312, 1466, 268, 44, 8, 2, 1, 65, 2948, 6812, 1726, 268, 44, 8, 2, 1, 82, 5945, 26048, 11062, 1732, 268, 44, 8, 2, 1, 101, 11026, 84149, 64548, 11617, 1732, 268, 44, 8, 2, 1
Offset: 1

Views

Author

Peter Luschny, Mar 20 2023

Keywords

Comments

For an overview of the terms used see A361574, which gives the row sums of this triangle. The corresponding sequence counting meanders without the requirement of being Fibonacci is A202409.
The diagonals, starting from the main diagonal, converge to A141147?

Examples

			Triangle T(n, k) starts:
  [ 1]   1;
  [ 2]   2,     1;
  [ 3]   5,     2,     1;
  [ 4]  10,     8,     2,     1;
  [ 5]  17,    40,     8,     2,     1;
  [ 6]  26,   161,    44,     8,     2,    1;
  [ 7]  37,   506,   263,    44,     8,    2,   1;
  [ 8]  50,  1312,  1466,   268,    44,    8,   2,  1;
  [ 9]  65,  2948,  6812,  1726,   268,   44,   8,  2, 1;
  [10]  82,  5945, 26048, 11062,  1732,  268,  44,  8, 2, 1;
  [11] 101, 11026, 84149, 64548, 11617, 1732, 268, 44, 8, 2, 1.
.
T(4, 2) = 8 counts the Fibonacci meanders with central angle 120 degrees and length 12 that make 6 left turns. Written as binary strings (L = 1, R = 0):
110100100101, 111001001001, 111100010001, 111110000001, 111010010010,
111100100100, 111110001000, 111111000000.
		

Crossrefs

Cf. A361574 (row sums), A202409, A141147.

Programs

  • SageMath
    # using functions 'isMeander' and 'isFibonacci' from A361574.
    def FibonacciMeandersByLeftTurns(m: int, n: int) -> list[int]:
        size = m * n; A = [0] * n; k = -1
        for a in range(0, size + 1, m):
            S = [i < a for i in range(size)]
            for c in Permutations(S):
                if c[0] == 0: break
                if not isFibonacci(c): continue
                if not isMeander(m, c): continue
                A[k] += 1
            k += 1
        return A
    for n in range(1, 12):
        print(FibonacciMeandersByLeftTurns(3, n))

A361575 Number of Fibonacci meanders of length n.

Original entry on oeis.org

1, 3, 5, 11, 13, 30, 34, 71, 97, 177, 233, 481, 610, 1157, 1677, 3027, 4181, 8016, 10946, 20379, 29534, 52461, 75025, 140748, 196778, 355979, 526123, 933044, 1346269, 2469992, 3524578, 6342729, 9400985, 16487211
Offset: 1

Views

Author

Peter Luschny, Mar 16 2023

Keywords

Comments

For an overview of the terms and functions used, compare A361574. The corresponding sequence counting meanders without the requirement to be Fibonacci is A199932.

Examples

			Fibonacci meanders with length 6 can have the central angle 360/m, where m is in divisors(6) = {1, 2, 3, 6}. In total there are a(6) = 30 such meanders, the list shows their binary representation together with the multiplicity with which they appear.
100000 x 1, 100001 x 2, 100010 x 1, 100100 x 2, 100101 x 1, 101000 x 1,
101001 x 1, 101010 x 1, 110000 x 2, 110001 x 2, 110010 x 1, 110100 x 1,
110101 x 1, 111000 x 2, 111001 x 2, 111010 x 1, 111100 x 2, 111101 x 1,
111110 x 1, 111111 x 4.
		

Crossrefs

Programs

  • Maple
    # The list A was computed with the functions given in A361574. They correspond to the columns in the table shown in the reference.
    A := [[1, 2, 4, 7, 12, 20, 33, 54, 88, 143, 232, 376, 609, 986, 1596, 2583, 4180, 6764, 10945, 17710, 28656, 46367, 75024, 121392, 196417, 317810, 514228, 832039, 1346268, 2178308, 3524577, 5702886, 9227464, 14930351], [1, 3, 6, 13, 30, 70, 167, 405, 992, 2450, 6090, 15214, 38165, 96069, 242530, 613811, 1556856], [1, 3, 8, 21, 68, 242, 861, 3151, 11874, 45192, 173496], [1, 3, 10, 35, 154, 858, 4723, 25625], [1, 3, 12, 61, 360, 3058], [1, 3, 14, 111, 878], [1, 3, 16, 209], [1, 3, 18, 403], [1, 3, 20], [1, 3, 22], [1, 3, 24], [1, 3], [1, 3], [1, 3], [1, 3], [1, 3], [1, 3], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1], [1]];
    with(LinearAlgebra):  # a(n) is the sum of row n of this table.
    row := k -> [seq(`if`(irem(n, k) <> 0, 0, A[k][n/k]), n = 1..34)]:
    M := Transpose(Matrix([seq(row(n), n = 1..34)])):
    seq(add(m, m = Row(M, n)), n = 1..34);

A361894 Triangle read by rows. T(n, k) is the number of Fibonacci meanders with a central angle of 360/m degrees that make m*k left turns and whose length is m*n, where m = 2.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 6, 2, 1, 5, 16, 6, 2, 1, 6, 35, 20, 6, 2, 1, 7, 66, 65, 20, 6, 2, 1, 8, 112, 186, 70, 20, 6, 2, 1, 9, 176, 462, 246, 70, 20, 6, 2, 1, 10, 261, 1016, 812, 252, 70, 20, 6, 2, 1, 11, 370, 2025, 2416, 917, 252, 70, 20, 6, 2, 1, 12, 506, 3730, 6435, 3256, 924, 252, 70, 20, 6, 2, 1
Offset: 1

Views

Author

Peter Luschny, Mar 31 2023

Keywords

Comments

For an overview of the terms used see A361574. A201631 gives the row sums of this triangle.
The corresponding sequence counting meanders without the requirement of being Fibonacci is A103371 (for which in turn A103327 is a termwise majorant counting permutations of the same type).
The diagonals, starting from the main diagonal, apparently converge to A000984.

Examples

			Triangle T(n, k) starts:
  [ 1]  1;
  [ 2]  2,   1;
  [ 3]  3,   2,    1;
  [ 4]  4,   6,    2,    1;
  [ 5]  5,  16,    6,    2,    1;
  [ 6]  6,  35,   20,    6,    2,   1;
  [ 7]  7,  66,   65,   20,    6,   2,   1;
  [ 8]  8, 112,  186,   70,   20,   6,   2,  1;
  [ 9]  9, 176,  462,  246,   70,  20,   6,  2,  1;
  [10] 10, 261, 1016,  812,  252,  70,  20,  6,  2, 1;
  [11] 11, 370, 2025, 2416,  917, 252,  70, 20,  6, 2, 1;
  [12] 12, 506, 3730, 6435, 3256, 924, 252, 70, 20, 6, 2, 1.
.
T(4, k) counts Fibonacci meanders with central angle 180 degrees and length 8 that make k left turns. Written as binary strings (L = 1, R = 0):
k = 1: 11000000, 10010000, 10000100, 10000001;
k = 2: 11110000, 11100100, 11100001, 11010010, 11001001, 10100101;
k = 3: 11111100, 11111001;
k = 4: 11111111.
		

Crossrefs

Cf. A201631 (row sums), A361681 (m=3), A132812, A361574, A103371, A000984.

Programs

  • SageMath
    # using function 'FibonacciMeandersByLeftTurns' from A361681.
    for n in range(1, 12):
        print(FibonacciMeandersByLeftTurns(2, n))
Showing 1-4 of 4 results.