A073044 -id:A073044 - cp's OEIS frontend

A274228 Triangle read by rows: T(n,k) (n>=3, 0<=k<=n-3) = number of n-sequences of 0's and 1's with exactly one pair of adjacent 0's and exactly k pairs of adjacent 1's.

2, 3, 2, 4, 4, 2, 5, 8, 5, 2, 6, 12, 12, 6, 2, 7, 18, 21, 16, 7, 2, 8, 24, 36, 32, 20, 8, 2, 9, 32, 54, 60, 45, 24, 9, 2, 10, 40, 80, 100, 90, 60, 28, 10, 2, 11, 50, 110, 160, 165, 126, 77, 32, 11, 2, 12, 60, 150, 240, 280, 252, 168, 96, 36, 12, 2, 13, 72, 195, 350, 455, 448, 364, 216, 117, 40, 13, 2

Offset: 3

Jeremy Dover, Jun 14 2016

			n=3 => 100, 001 -> T(3,0) = 2.
n=4 => 0010, 0100, 1001 -> T(4,0) = 3; 0011, 1100 -> T(4,1) = 2.
Triangle starts:
2,
3, 2,
4, 4, 2,
5, 8, 5, 2,
6, 12, 12, 6, 2,
7, 18, 21, 16, 7, 2,
8, 24, 36, 32, 20, 8, 2,
9, 32, 54, 60, 45, 24, 9, 2,
10, 40, 80, 100, 90, 60, 28, 10, 2,
11, 50, 110, 160, 165, 126, 77, 32, 11, 2,
12, 60, 150, 240, 280, 252, 168, 96, 36, 12, 2,
13, 72, 195, 350, 455, 448, 364, 216, 117, 40, 13, 2,
...

Row sums give A001629.
Cf. A073044.
Columns of table:
  T(n,0)=A000027(n-1)
  T(n,1)=A007590(n-1)
  T(n,2)=A080838(n-1)
  T(n,3)=A032091(n)

Table[(k + 1) (Binomial[Floor[(n + k - 2)/2], k + 1] + Binomial[Floor[(n + k - 3)/2], k + 1]) + 2 Binomial[Floor[(n + k - 3)/2], k], {n, 3, 14}, {k, 0, n - 3}] // Flatten (* Michael De Vlieger, Jun 16 2016 *)

T(n,k) = (k+1)*(binomial((n+k-2)\2,k+1)+binomial((n+k-3)\2,k+1))+2*binomial((n+k-3)\2,k); \\ Michel Marcus, Jun 17 2016

T(n,k) = (k+1)*(binomial(floor((n+k-2)/2),k+1)+binomial(floor((n+k-3)/2),k+1))+2*binomial(floor((n+k-3)/2),k).
T(n,k) = (k+1)*A073044(n-2,k+1) + 2*A046854(n-3,k).
T(n,k) = A274742(n,k)+A274742(n-1,k)+A046854(n-3,k).

A345123 Number T(n,k) of ordered subsequences of {1,...,n} containing at least k elements and such that the first differences contain only odd numbers; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 2, 1, 4, 3, 1, 7, 6, 3, 1, 12, 11, 7, 3, 1, 20, 19, 14, 8, 3, 1, 33, 32, 26, 17, 9, 3, 1, 54, 53, 46, 34, 20, 10, 3, 1, 88, 87, 79, 63, 43, 23, 11, 3, 1, 143, 142, 133, 113, 83, 53, 26, 12, 3, 1, 232, 231, 221, 196, 156, 106, 64, 29, 13, 3, 1, 376, 375, 364, 334, 279, 209, 132, 76, 32, 14, 3, 1

Offset: 0

Alois P. Heinz, Jun 08 2021

The sequence of column k satisfies a linear recurrence with constant coefficients of order 2k if k >= 2 and of order 3 for k in {0, 1}.

			T(0,0) = 1: [].
T(1,1) = 1: [1].
T(2,2) = 1: [1,2].
T(3,1) = 6: [1], [2], [3], [1,2], [2,3], [1,2,3].
T(4,0) = 12: [], [1], [2], [3], [4], [1,2], [1,4], [2,3], [3,4], [1,2,3], [2,3,4], [1,2,3,4].
T(6,3) = 17: [1,2,3], [1,2,5], [1,4,5], [2,3,4], [2,3,6], [2,5,6], [3,4,5], [4,5,6], [1,2,3,4], [1,2,3,6], [1,2,5,6], [1,4,5,6], [2,3,4,5], [3,4,5,6], [1,2,3,4,5], [2,3,4,5,6], [1,2,3,4,5,6].
Triangle T(n,k) begins:
    1;
    2,   1;
    4,   3,   1;
    7,   6,   3,   1;
   12,  11,   7,   3,   1;
   20,  19,  14,   8,   3,   1;
   33,  32,  26,  17,   9,   3,  1;
   54,  53,  46,  34,  20,  10,  3,  1;
   88,  87,  79,  63,  43,  23, 11,  3,  1;
  143, 142, 133, 113,  83,  53, 26, 12,  3, 1;
  232, 231, 221, 196, 156, 106, 64, 29, 13, 3, 1;
  ...

Chu, Hung Viet, Various Sequences from Counting Subsets, Fib. Quart., 59:2 (May 2021), 150-157.

Alois P. Heinz, Rows n = 0..140, flattened
Chu, Hung Viet, Various Sequences from Counting Subsets, arXiv:2005.10081 [math.CO], 2021.

Columns k=0-3 give: A000071(n+3), A001911, A001924(n-1), A344004.
T(2n,n) give A340766.
Cf. A073044, A105438.

b:= proc(n, l, t) option remember; `if`(n=0, `if`(t=0, 1, 0), `if`(0
      in [l, irem(1+l-n, 2)], b(n-1, n, max(0, t-1)), 0)+b(n-1, l, t))
    end:
T:= (n, k)-> b(n, 0, k):
seq(seq(T(n, k), k=0..n), n=0..10);
# second Maple program:
g:= proc(n, k) option remember; `if`(k>n, 0,
     `if`(k in [0, 1], n^k, g(n-1, k-1)+g(n-2, k)))
    end:
T:= proc(n, k) option remember;
     `if`(k>n, 0, g(n, k)+T(n, k+1))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);
# third Maple program:
T:= proc(n, k) option remember; `if`(k>n, 0, binomial(iquo(n+k, 2), k)+
      `if`(k>0, binomial(iquo(n+k-1, 2), k), 0)+T(n, k+1))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);

T[n_, k_] := T[n, k] = If[k > n, 0, Binomial[Quotient[n+k, 2], k] +
     If[k > 0, Binomial[Quotient[n+k-1, 2], k], 0] + T[n, k+1]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 11}] // Flatten (* Jean-François Alcover, Nov 06 2021, after 3rd Maple program *)

A129714 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k runs (0<=k<=n). A Fibonacci binary word is a binary word having no 00 subword. A run is a maximal sequence of consecutive identical letters.

Original entry on oeis.org

1, 0, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 3, 2, 0, 1, 2, 4, 4, 2, 0, 1, 2, 5, 6, 5, 2, 0, 1, 2, 6, 8, 9, 6, 2, 0, 1, 2, 7, 10, 14, 12, 7, 2, 0, 1, 2, 8, 12, 20, 20, 16, 8, 2, 0, 1, 2, 9, 14, 27, 30, 30, 20, 9, 2, 0, 1, 2, 10, 16, 35, 42, 50, 40, 25, 10, 2, 0, 1, 2, 11, 18, 44, 56, 77, 70, 55, 30, 11, 2

Offset: 0

Emeric Deutsch, May 12 2007

Row sums are the Fibonacci numbers (A000045).

			T(5,3)=4 because we have 10111, 11011, 11101 and 01110.
Triangle starts:
  1;
  0,2;
  0,1,2;
  0,1,2,2;
  0,1,2,3,2;
  0,1,2,4,4,2;

M. Henk, J. Richter-Gebert and G. M. Ziegler, Basic properties of convex polytopes, Discrete and Computational Geometry, J.E. Goodman and J. O'Rourke eds., CRC Press, Boca Raton, Florida, 1997, Second Edition, 2004, 355-383.

Cf. A000045, A065941, A073044, A129715.

T:=proc(n,k): if k<0 then 0 elif k=0 and n=0 then 1 elif k=0 then 0 elif n=1 and k=1 then 2 elif n=2 and k=1 then 1 elif n=2 and k=2 then 2 elif k>n then 0 else T(n-1,k)+T(n-2,k-2) fi end: for n from 0 to 14 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[n_, k_] := T[n, k] = Which[k < 0, 0, k == 0 && n == 0, 1, k == 0, 0, n == 1 && k == 1, 2, n == 2 && k == 1, 1, n == 2 && k == 2, 2, k > n, 0, True, T[n-1, k] + T[n-2, k-2]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 23 2024, after Maple program *)

T(n,k) = A073044(n,n-k) (since in each Fibonacci binary word of length n the number of runs plus the number of 11's is equal to n).
Sum_{k=0..n} k*T(n,k) = A129715(n).
G.f.: G(t,z)=(1+tz)(1-z+tz)/(1-z-t^2*z^2).
T(n,k) = T(n-1,k)+T(n-2,k-2) for n>=3, k>=1 (see the Maple program).
For n >=1, T(n+1,k+1) = binomial(n-floor((k+1)/2),floor(k/2)) + binomial(n-1-floor(k/2),floor((k-1)/2)) = A065941(n,k) + A065941(n-1,k-1). T(n+1,2k) = 2*binomial(n-k,k-1) and T(n+1,2k+1) = n/(n-k)*binomial(n-k,k). For 0 <= k < n and n >=1, T(n+1,k+1) equals the number of facets of the k-dimensional cyclic polytope C_k(n), defined as the convex hull of the n points (1,1^2,...,1^k),...,(n,n^2,...n^k) in R^k [see Henk et al., p.11]. [Peter Bala, Sep 25 2008]

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A274228 Triangle read by rows: T(n,k) (n>=3, 0<=k<=n-3) = number of n-sequences of 0's and 1's with exactly one pair of adjacent 0's and exactly k pairs of adjacent 1's.

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A345123 Number T(n,k) of ordered subsequences of {1,...,n} containing at least k elements and such that the first differences contain only odd numbers; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A129714 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k runs (0<=k<=n). A Fibonacci binary word is a binary word having no 00 subword. A run is a maximal sequence of consecutive identical letters.

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