A202409 -id:A202409 - cp's OEIS frontend

A132812 Triangle read by rows, n>=1, 1<=k<=n, T(n,k) = k*binomial(n,k)^2/(n-k+1).

1, 2, 2, 3, 9, 3, 4, 24, 24, 4, 5, 50, 100, 50, 5, 6, 90, 300, 300, 90, 6, 7, 147, 735, 1225, 735, 147, 7, 8, 224, 1568, 3920, 3920, 1568, 224, 8, 9, 324, 3024, 10584, 15876, 10584, 3024, 324, 9, 10, 450, 5400, 25200, 52920, 52920, 25200, 5400, 450, 10

Offset: 1

Gary W. Adamson, Sep 01 2007

A127648 * A001263. (Original name by Gary W. Adamson.)
Let a meander be defined as in the link and m = 2. Then T(n,k) counts the invertible meanders of length m(n+1) built from arcs with central angle 360/m whose binary representation have mk '1's. - Peter Luschny, Dec 19 2011
Antidiagonal sums = A110320. - Philippe Deléham, Jun 08 2013

			First few rows of the triangle are:
  1;
  2, 2;
  3, 9, 3;
  4, 24, 24, 4;
  5, 50, 100, 50, 5;
  6, 90, 300, 300, 90, 6;
  ...
Row 4 = (4, 24, 24, 4) = 4 * (1, 6, 6, 1), where (1, 6, 6, 1) = row 4 of the Narayana triangle. - _Gary W. Adamson_
T(3,1) = 3 because the invertible meanders of length 8 and central angle 180 degree which have two '1's in their binary representation are {10000100, 10010000, 11000000}. - _Peter Luschny_, Dec 19 2011

Michael De Vlieger, Table of n, a(n) for n = 1..11325 (rows 1..150).
Peter Luschny, Meanders.

Row sums: A001791. Cf. A001263, A127648, A001791, A202409.

A132812 := (n,k) -> k*binomial(n,k)^2/(n-k+1);
seq(print(seq(A132812(n,k),k=0..n-1)),n=1..6); # Peter Luschny, Dec 19 2011

Table[k Binomial[n, k]^2/(n - k + 1), {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Nov 15 2017 *)

A127648 * A001263 as infinite lower triangular matrices.
a(n) = n * A001263(n,k).
T(n,k) = binomial(n,k)*binomial(n,k-1). - Philippe Deléham, Jun 08 2013
G.f.: x*d(N(x,y))/dx, where N(x,y) is g.f. for Narayana numbers A001263. - Vladimir Kruchinin, Oct 22 2021

New name from Peter Luschny, Dec 19 2011

a(53) corrected by Michael De Vlieger, Nov 15 2017

A201640 a(n) = Sum_{k=1..n} kbinomial(n, k)^3(n^2 + n - kn - k + k^2)/((n - k + 1)^2n).

Original entry on oeis.org

1, 8, 54, 368, 2550, 17952, 128086, 924128, 6729858, 49395440, 364979560, 2712343680, 20257516240, 151957919232, 1144281700110, 8646263301056, 65531851263978, 498047725561104, 3794627850238756, 28976634967413920, 221728252767064596, 1699859618636556608

Offset: 1

Peter Luschny, Dec 19 2011

nonn

Let a meander be defined as in the link and m = 3. Then a(n) counts the invertible meanders of length m*(n + 1) built from arcs with central angle 360/m.

			a(2) = 8 because (the binary representations of) the invertible meanders of length 9 and central angle 120 degree are {100100100, 110110110, 100011000, 111001110, 110001000, 111011100, 111000000, 111111000}.

Peter Luschny, Meanders.

Row sums of A202409.

Cf. A000079 (m=1), A001791 (m=2), this sequence (m=3).

A201640 := proc(n) add(k*binomial(n,k)^3*(n^2+n-k*n-k+k^2)/((n-k+1)^2*n),k=1..n) end; seq(A201640(n,k),n=1..22);
# Alternative:
h := (a, b, n) -> hypergeom([-n, -n, 1 - n], [a, b], -1):
a := n -> h(1, 1, n) + (n - 1)*h(1, 2, n):
seq(simplify(a(n)), n = 1..22); # Peter Luschny, Mar 16 2023
# Recurrence:
a := proc(n) option remember; if n < 3 then return [1, 8][n] fi;
(8*n*(3*n - 1)*(n - 1)*(n - 2)*a(n-2) + n*(21*n^3 - 25*n^2 - 2*n + 8)*a(n-1)) /
((n - 1)*(n + 1)^2*(3*n - 4)) end: seq(a(n), n = 1..22); # Peter Luschny, Mar 16 2023

Table[Sum[k*Binomial[n,k]^3*(n^2+n-k*n-k+k^2)/((n-k+1)^2*n),{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Oct 24 2012 *)

Recurrence: (n+1)^2*a(n) = n*(13*n+2)*a(n-1) - 6*(2*n-1)*(3*n-5)*a(n-2) - 2*(17*n^2-62*n+32)*a(n-3) + 16*(n-4)*(n-3)*a(n-4). - Vaclav Kotesovec, Oct 24 2012
a(n) ~ sqrt(3)*8^n/(Pi*n). - Vaclav Kotesovec, Oct 24 2012
From Peter Luschny, Mar 16 2023: (Start)
a(n) = h(1, 1, n) + (n - 1)*h(1, 2, n), where h(a, b, n) = hypergeom([-n, -n, 1 - n], [a, b], -1).
a(n) = (8*n*(3*n - 1)*(n - 1)*(n - 2)*a(n - 2) + n*(21*n^3 - 25*n^2 - 2*n + 8)* a(n-1)) / ((n - 1)*(n + 1)^2*(3*n - 4)) for n >= 3. (End)

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

Peter Luschny, Mar 20 2023

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?

			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.

Jean-Luc Baril, Sergey Kirgizov, Rémi Maréchal, and Vincent Vajnovszki, Enumeration of Dyck paths with air pockets, arXiv:2202.06893 [cs.DM], 2022-2023.
Peter Luschny, Fibonacci meanders.

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

# 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))

cp's OEIS Frontend

A132812 Triangle read by rows, n>=1, 1<=k<=n, T(n,k) = k*binomial(n,k)^2/(n-k+1).

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A201640 a(n) = Sum_{k=1..n} kbinomial(n, k)^3(n^2 + n - kn - k + k^2)/((n - k + 1)^2n).

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

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cp's OEIS Frontend

A132812 Triangle read by rows, n>=1, 1<=k<=n, T(n,k) = k*binomial(n,k)^2/(n-k+1).

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Author

Keywords

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Examples

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Crossrefs

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Maple

Mathematica

Formula

Extensions

A201640 a(n) = Sum_{k=1..n} k*binomial(n, k)^3*(n^2 + n - k*n - k + k^2)/((n - k + 1)^2*n).

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Maple

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Formula

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.

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Author

Keywords

Comments

Examples

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SageMath

A201640 a(n) = Sum_{k=1..n} kbinomial(n, k)^3(n^2 + n - kn - k + k^2)/((n - k + 1)^2n).

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