A059576 - cp's OEIS frontend

A059576 Summatory Pascal triangle T(n,k) (0 <= k <= n) read by rows. Top entry is 1. Each entry is the sum of the parallelogram above it.

1, 1, 1, 2, 3, 2, 4, 8, 8, 4, 8, 20, 26, 20, 8, 16, 48, 76, 76, 48, 16, 32, 112, 208, 252, 208, 112, 32, 64, 256, 544, 768, 768, 544, 256, 64, 128, 576, 1376, 2208, 2568, 2208, 1376, 576, 128, 256, 1280, 3392, 6080, 8016, 8016, 6080, 3392, 1280, 256

Offset: 0

Floor van Lamoen, Jan 23 2001

We may also relabel the entries as U(0,0), U(1,0), U(0,1), U(2,0), U(1,1), U(0,2), U(3,0), ... [That is, T(n,k) = U(n-k, k) for 0 <= k <= n and U(m,s) = T(m+s, s) for m,s >= 0.]
From Petros Hadjicostas, Jul 16 2020: (Start)
We explain the parallelogram definition of T(n,k).
    T(0,0) *
           |\
           | \
           |  * T(k,k)
  T(n-k,0) *  |
            \ |
             \|
              * T(n,k)
The definition implies that T(n,k) is the sum of all T(i,j) such that (i,j) has integer coordinates over the set
{(i,j): a(1,0) + b(1,1), 0 <= a <= n-k, 0 <= b <= k} - {(n,k)}.
The parallelogram can sometimes be degenerate; e.g., when k = 0 or n = k. (End)
T(n,k) is the number of 2-compositions of n having sum of the entries of the first row equal to k (0 <= k <= n). A 2-composition of n is a nonnegative matrix with two rows, such that each column has at least one nonzero entry and whose entries sum up to n. - Emeric Deutsch, Oct 12 2010
From Michel Marcus and Petros Hadjicostas, Jul 16 2020: (Start)
Robeva and Sun (2020) let A(m,n) = U(m-1, n-1) be the number of subdivisions of a 2-row grid with m points on the top and n points at the bottom (and such that the lower left point is the origin).
The authors proved that A(m,n) = 2*(A(m,n-1) + A(m-1,n) - A(m-1,n-1)) for m, n >= 2 (with (m,n) <> (2,2)), which is equivalent to a similar recurrence for U(n,k) given in the Formula section below. (They did not explicitly specify the value of A(1,1) = U(0,0) because they did not care about the number of subdivisions of a degenerate polygon with only one side.)
They also proved that, for (m,n) <> (1,1), A(m,n) = (2^(m-2)/(n-1)!) * Q_n(m) =
= (2^(m-2)/(n-1)!) * Sum_{k=1..n} A336244(n,k) * m^(n-k), where Q_n(m) is a polynomial in m of degree n-1. (End)
With the square array notation of Petros Hadjicostas, Jul 16 2020 below, U(i,j) is the number of lattice paths from (0,0) to (i,j) whose steps move north or east or have positive slope. For example, representing a path by its successive lattice points rather than its steps, U(1,2) = 8 counts {(0,0),(1,2)}, {(0,0),(0,1),(1,2)}, {(0,0),(0,2),(1,2)}, {(0,0),(1,0),(1,2)}, {(0,0),(1,1),(1,2)}, {(0,0),(0,1),(0,2),(1,2)}, {(0,0),(0,1),(1,1),(1,2)}, {(0,0),(1,0),(1,1),(1,2)}. If north (vertical) steps are excluded, the resulting paths are counted by A049600.  - David Callan, Nov 25 2021

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins
[0]   1;
[1]   1,   1;
[2]   2,   3,   2;
[3]   4,   8,   8,   4;
[4]   8,  20,  26,  20,   8;
[5]  16,  48,  76,  76,  48,  16;
[6]  32, 112, 208, 252, 208, 112, 32;
  ...
T(5,2) = 76 is the sum of the elements above it in the parallelogram bordered by T(0,0), T(5-2,0) = T(3,0), T(2,2) and T(5,2). We of course exclude T(5,2) from the summation. Thus
T(5,2) = Sum_{a=0..5-2, b=0..2, (a,b) <> (5-2,2)} T(a(1,0) + b(1,1)) =
= (1 + 1 + 2) + (1 + 3 + 8) + (2 + 8 + 26) + (4 + 20) = 76. [Edited by _Petros Hadjicostas_, Jul 16 2020]
From _Petros Hadjicostas_, Jul 16 2020: (Start)
Square array U(n,k) (with rows n >= 0 and columns k >= 0) begins
   1,   1,   2,    4,    8, ...
   1,   3,   8,   20,   48, ...
   2,   8,  26,   76,  208, ...
   4,  20,  76,  252,  768, ...
   8,  48, 208,  768, 2568, ...
  16, 112, 544, 2208, 8016, ...
  ...
Consider the following 2-row grid with n = 3 points at the top and k = 2 points at the bottom:
   A  B  C
   *--*--*
   |    /
   |   /
   *--*
   D  E
The sets of the dividing internal lines of the A(3,2) = U(3-1, 2-1) = 8 subdivisions of the above 2-row grid are as follows: { }, {DC}, {DB}, {EB}, {EA}, {DB, DC}, {DB, EB}, and {EA, EB}. See Robeva and Sun (2020).
These are the 2-compositions of n = 3 with sum of first row entries equal to k = 1:
[1; 2], [0,1; 2,0], [0,1; 1,1], [1,0; 0,2], [1,0; 1,1], [0,0,1; 1,1,0], [0,1,0; 1,0,1], and [1,0,0; 0,1,1]. We have T(3,2) = 8 such matrices. See _Emeric Deutsch_'s contribution above. See also Section 2 in Castiglione et al. (2007). (End)

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
G. Castiglione, A. Frosini, E. Munarini, A. Restivo and S. Rinaldi, Combinatorial aspects of L-convex polyominoes, European Journal of Combinatorics, 28(6) (2007), 1724-1741; see Fig. 5, p. 1729.
Fang, E., Jenkins, J., Lee, Z., Li, D., Lu, E., Miller, S. J., ... & Siktar, J. (2019). Central Limit Theorems for Compound Paths on the 2-Dimensional Lattice, arXiv preprint arXiv:1906.10645. Also Fib. Q., 58:1 (2020), 208-225.
Elina Robeva and Melinda Sun, Bimonotone Subdivisions of Point Configurations in the Plane, arXiv:2007.00877 [math.CO], 2020.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A000079, A001792, A003480 (row sums), A052141 (main diagonal).
Cf. A059474, A059473, A008288, A035002, A059226, A059283, A181292, A336244.
Cf. A011782, A049611, A049612, A055589, A055852, A055853, A181306.

a059576 n k = a059576_tabl !! n !! k
a059576_row n = a059576_tabl !! n
a059576_tabl = [1] : map fst (iterate f ([1,1], [2,3,2])) where
   f (us, vs) = (vs, map (* 2) ws) where
     ws = zipWith (-) (zipWith (+) ([0] ++ vs) (vs ++ [0]))
                      ([0] ++ us ++ [0])
-- Reinhard Zumkeller, Dec 03 2012

A011782:= func< n | n eq 0 select 1 else 2^(n-1) >;
function T(n,k) // T = A059576
  if k eq 0 or k eq n then return A011782(n);
  else return 2*T(n-1, k-1) + 2*T(n-1, k) - (2 - 0^(n-2))*T(n-2, k-1);
  end if; return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 02 2022

A059576 := proc(n,k) local b,t1; t1 := min(n+k-2,n,k); add( (-1)^b * 2^(n+k-b-2) * (n+k-b-2)! * (1/(b! * (n-b)! * (k-b)!)) * (-2 * n-2 * k+2 * k^2+b^2-3 * k * b+2 * n^2+5 * n * k-3 * n * b), b=0..t1); end;
T := proc (n, k) if k <= n then add((-1)^j*2^(n-j-1)*binomial(k, j)*binomial(n-j, k), j = 0 .. min(k, n-k)) fi end proc: 1; for n to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form # Emeric Deutsch, Oct 12 2010
T := (n, k) -> `if`(n=0, 1, 2^(n-1)*binomial(n, k)*hypergeom([-k, k - n], [-n], 1/2)): seq(seq(simplify(T(n, k)), k=0..n), n=0..10); # Peter Luschny, Nov 26 2021

T[0, 0] = 1; T[n_, k_] := 2^(n-k-1)*n!*Hypergeometric2F1[ -k, -k, -n, -1 ] / (k!*(n-k)!); Flatten[ Table[ T[n, k], {n, 0, 9}, {k, 0, n}]] (* Jean-François Alcover, Feb 01 2012, after Robert Israel *)

def T(n,k): # T = A059576
    if (k==0 or k==n): return 1 if (n==0) else 2^(n-1) # A011782
    else: return 2*T(n-1, k-1) + 2*T(n-1, k) - (2 - 0^(n-2))*T(n-2, k-1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 02 2022

T(n, n-1) = A001792(n-1).
T(2*n, n) = A052141(n).
Sum_{k=0..n} T(n, k) = A003480(n).
G.f.: U(z, w) = Sum_{n >= 0, k >= 0} U(n, k)*z^n*w^k = Sum{n >= 0, k >= 0} T(n, k)*z^(n-k)*w^k = (1-z)*(1-w)/(1 - 2*w - 2*z + 2*z*w).
Maple code gives another explicit formula for U(n, k).
From Jon Stadler (jstadler(AT)capital.edu), Apr 30 2003: (Start)
U(n,k) is the number of ways of writing the vector (n,k) as an ordered sum of vectors, equivalently, the number of paths from (0,0) to (n,k) in which steps may be taken from (i,j) to (p,q) provided (p,q) is to the right or above (i,j).
2*U(n,k) = Sum_{i <= n, j <= k} U(i,j).
U(n,k) = 2*U(n-1,k) + Sum_{i < k} U(n,i).
U(n,k) = Sum_{j=0..n+k} C(n,j-k+1)*C(k,j-n+1)*2^j. (End)
T(n, k) = 2*(T(n-1, k-1) + T(n-1, k)) - (2 - 0^(n-2))*T(n-2, k-1) for n > 1 and 1 < k < n; T(n, 0) = T(n, n) = 2*T(n-1, 0) for n > 0; and T(0, 0) = 1. - Reinhard Zumkeller, Dec 03 2004
From Emeric Deutsch, Oct 12 2010: (Start)
Sum_{k=0..n} k*T(n,k) = A181292(n).
T(n,k) = Sum_{j=0..min(k, n-k)} (-1)^j*2^(n-j-1)*binomial(k, j)*binomial(n-j, k) for (n,k) != (0,0).
G.f.: G(t,z) = (1-z)*(1-t*z)/(1 - 2*z - 2*t*z + 2*t*z^2). (End)
U(n,k) = 0 if k < 0; else U(k,n) if k > n; else 1 if n <= 1; else 3 if n = 2 and k = 1; else 2*U(n,k-1) + 2*U(n-1,k) - 2*U(n-1,k-1). - David W. Wilson; corrected in the case k > n by Robert Israel, Jun 15 2011 [Corrected by Petros Hadjicostas, Jul 16 2020]
U(n,k) = binomial(n,k) * 2^(n-1) * hypergeom([-k,-k], [n+1-k], 2) if n >= k >= 0 with (n,k) <> (0,0). - Robert Israel, Jun 15 2011 [Corrected by Petros Hadjicostas, Jul 16 2020]
U(n,k) = Sum_{0 <= i+j <= n+k-1} (-1)^j*C(i+j+1, j)*C(n+i, n)*C(k+i, k). - Masato Maruoka, Dec 10 2019
T(n, k) = 2^(n - 1)*binomial(n, k)*hypergeom([-k, k - n], [-n], 1/2) = A059474(n, k)/2 for n >= 1. - Peter Luschny, Nov 26 2021
From G. C. Greubel, Sep 02 2022: (Start)
T(n, n-k) = T(n, k).
T(n, 0) = T(n, n) = A011782(n).
T(n, n-2) =  2*A049611(n-1), n >= 2.
T(n, n-3) =  4*A049612(n-2), n >= 3.
T(n, n-4) =  8*A055589(n-3), n >= 4.
T(n, n-5) = 16*A055852(n-4), n >= 5.
T(n, n-6) = 32*A055853(n-5), n >= 6.
Sum_{k=0..floor(n/2)} T(n, k) = A181306(n). (End)

cp's OEIS Frontend

A059576 Summatory Pascal triangle T(n,k) (0 <= k <= n) read by rows. Top entry is 1. Each entry is the sum of the parallelogram above it.

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