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 1-3 of 3 results.

A361574 a(n) is the number of Fibonacci meanders of length m*n and central angle 360/m degrees where m = 3.

Original entry on oeis.org

1, 3, 8, 21, 68, 242, 861, 3151, 11874, 45192, 173496, 673042
Offset: 1

Views

Author

Peter Luschny, Mar 16 2023

Keywords

Comments

A binary curve C is a pair (m, S) such that
(a) S is a list with values in {L, R} that
(b) starts with an L, and
(c) m > 0 is an integer that divides the length of S.
Given a binary curve C = (m, S), we call m the modulus of the curve and S the steps of C. 'L' stands for a positively oriented arc (left turn) and 'R' for a negatively oriented arc (right turn).
Let SS denote a pair of consecutive steps in C. The direction d of a curve at n starts with d = 0, and for n > 0, d = d + 1 if SS = LL and d = d - 1 if SS = RR; otherwise, d remains unchanged.
A binary curve C = (m, S) is a meander if the values d mod m are assumed with equal frequency. A meander is a closed smooth curve in the plane, possibly self-overlapping (see the plots).
A binary curve 'changes direction' if two consecutive steps are of the same type, i.e., is a pair of steps of the form LL or RR.
A Fibonacci meander is a meander that does not change direction to the left except at the beginning of the curve, where any number of left turns can appear.

Examples

			For n = 4 the Fibonacci meanders with central angle 120 degrees and length 12, written as binary strings (L = 1, R = 0), are:
100000010001, 100010000001, 110000000001, 100000100100, 100100000100, 100010001000,
110000001000, 100100100000, 110001000000, 111000000000, 110100100101, 111001001001,
111100010001, 111110000001, 111010010010, 111100100100, 111110001000, 111111000000,
111111110001, 111111111000, 111111111111.
		

Crossrefs

Cf. A000071 (m=1), A201631 (m=2), A197657.

Programs

  • SageMath
    def isFibonacci(S: list[bool]) -> bool:
        dir, os = True, S[0]
        for s in S:
            if s and os:
                if not dir:
                    return False
                dir = True
            else:
                dir = False
            os = s
        return True
    def isMeander(n: int, S: list[bool]) -> bool:
        if S[0] != True:
            return False
        vec = [0] * n
        max = len(S) // n
        dir, ob = 0, True
        for b in S:
            if b and ob:
                dir += 1
            elif not b and not ob:
                dir -= 1
            dir = dir % n
            v = vec[dir] = vec[dir] + 1
            if v > max:
                return False
            ob = b
        return True
    def FibonacciMeanders(m: int, steps: int) -> int:
        size = m * steps
        count = 0
        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
                count += 1
            print(count)
        return count
    def FibonacciMeandersColumn(m: int, size: int) -> list[int]:
        return [FibonacciMeanders(m, n) for n in range(1, size + 1)]
    print([FibonacciMeandersColumn(3, n) for n in range(1, 7)])

A110198 Antidiagonal sums of number triangle A110197.

Original entry on oeis.org

1, 2, 4, 9, 20, 46, 109, 262, 638, 1569, 3886, 9680, 24225, 60856, 153368, 387573, 981742, 2491934, 6336721, 16139616, 41166912, 105139773, 268841100, 688157430, 1763206441, 4521749642, 11605580290, 29809644693, 76621733444, 197074591420, 507193044993
Offset: 0

Views

Author

Paul Barry, Jul 15 2005

Keywords

Comments

Partial sums of A051286.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/((1-x)*Sqrt[(1+x+x^2)*(1-3*x+x^2)]), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 08 2014 *)

Formula

G.f.: 1/((1-x)*sqrt((1+x+x^2)*(1-3x+x^2))); a(n) = sum{k=0..floor(n/2), sum{i=0..n-2k, binomial(i+k, k)^2}}.
a(n) = sum{i=0..2n, A202411(i)}. - Peter Luschny, Jan 16 2012
Conjecture: n*a(n) +(-3*n+1)*a(n-1) +n*a(n-2) +(-n+2)*a(n-3) +(3*n-5)*a(n-4) +(-n+2)*a(n-5)=0. - R. J. Mathar, Nov 15 2012
a(n) ~ sqrt(100+45*sqrt(5)) * ((sqrt(5)+3)/2)^n / (10*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 08 2014
Equivalently, a(n) ~ phi^(2*n + 3) / (2 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 07 2021

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 1-3 of 3 results.