A091044 One half of odd-numbered entries of even-numbered rows of Pascal's triangle A007318.

1, 2, 2, 3, 10, 3, 4, 28, 28, 4, 5, 60, 126, 60, 5, 6, 110, 396, 396, 110, 6, 7, 182, 1001, 1716, 1001, 182, 7, 8, 280, 2184, 5720, 5720, 2184, 280, 8, 9, 408, 4284, 15912, 24310, 15912, 4284, 408, 9, 10, 570, 7752, 38760, 83980, 83980, 38760, 7752, 570, 10, 11

Offset

1,2

Comments

The odd-numbered columns of this triangle can be reduced: see triangle A091043.

The odd-numbered rows coincide with the ones of the reduced triangle A091043.

binomial(2*n,2*m+1) is even for n >= m + 1 >= 1, hence every T(n,m) is a positive integer.

The GCD (greatest common divisor) of the entries of each odd-numbered row n=2*k+1, k>=0, is 1.

The GCD of the entries of the even-numbered row n=2*k, k>=1, is A006519(n) (highest power of 2 in n=2*k).

Links

Indranil Ghosh, Rows 1..125, flattened

Matthew Blair, Rigoberto Flórez, and Antara Mukherjee, Honeycombs in the Pascal triangle and beyond, arXiv:2203.13205 [math.HO], 2022. See p. 4.

Kevin Buchin, Man-Kwun Chiu, Stefan Felsner, Günter Rote, and André Schulz, The Number of Convex Polyominoes with Given Height and Width, arXiv:1903.01095 [math.CO], 2019.

Hernan de Alba, W. Carballosa, J. Leaños, and L. M. Rivera, Independence and matching numbers of some token graphs, arXiv preprint arXiv:1606.06370 [math.CO], 2016.

Wolfdieter Lang, First 9 rows.

Formula

T(n, m)= binomial(2*n, 2*m+1)/2, n >= m + 1 >= 1, else 0.

Put a(n) = n!*(n+1/2)!/(1/2)!. T(n+1,k) = (n+1)*a(n)/(a(k)*a(n-k)).

T(n-1,k-1)*T(n,k+1)*T(n+1,k) = T(n-1,k)*T(n,k-1)*T(n+1,k+1). Cf. A111910. - Peter Bala, Oct 13 2011

From Peter Bala, Jul 29 2013: (Start)

O.g.f.: 1/(1 - 2*t*(x + 1) + t^2*(x - 1)^2)= 1 + (2 + 2*x)*t + (3 + 10*x + 3*x^2)*t^2 + ....

The n-th row polynomial R(n,x) = 1/(4*sqrt(x))*( (1 + sqrt(x))^(2*n) - (sqrt(x) - 1)^(2*n) ) and has n-1 real zeros given by the formula -cot^2(k*Pi/(2*n)) for k = 1,2,...,n-1. Cf A091042.

The row polynomial R(n,x) satisfies (x - 1)^n*R(n,x/(x - 1)) = U(n,2*x - 1), the n-th row polynomial of A053124.

Row sums A000302. Sum {k = 0..n-1} 2^k*T(n,k) = A001109(n). (End)

From Werner Schulte, Jan 13 2017: (Start)

(1) T(n,m) = T(n-1,m) + T(n-1,m-1)*(2*n-1-m)/m for 0 < m < n-1 with T(n,0) = n and T(n,n) = 0;

(2) T(n,m) = 2*T(n-1,m) + 2*T(n-1,m-1) - T(n-2,m) + 2*T(n-2,m-1) - T(n-2,m-2) for 0 < m < n-1 with T(n,0) = T(n,n-1) = n and T(n,m) = 0 if m < 0 or m >= n;

(3) The row polynomials p(n,x) = Sum_{m=0..n-1} T(n,m)*x^m satisfy the recurrence equation p(n+2,x) = (2+2*x)*p(n+1,x) - (x-1)^2*p(n,x) for n >= 1 with initial values p(1,x) = 1 and p(2,x) = 2+2*x.

(End)

G.f.: x*y /(1 - 2*(x+y) + (x-y)^2) with the entries regarded as an infinite square array A(i, j) read by antidiagonals. - Michael Somos, Oct 15 2017

Example

Triangle begins:
  [1];
  [2,2];
  [3,10,3];
  [4,28,28,4];
  [5,60,126,60,5];
  [6,110,396,396,110,6];
  ...
n = 6 = 2*3: gcd(6,110,396) = 2 = A006519(6);
n = 5: gcd(5,60,126) = 1 = A006519(5).

Mathematica

Flatten[Table[Binomial[2n, 2m+1]/2, {n, 1, 11}, {m, 0, n-1}]] (* Indranil Ghosh, Feb 22 2017 *)

Prog

(PARI) {A(i, j) = binomial(2*i + 2*j - 2, 2*i - 1) / 2}; /* Michael Somos, Oct 15 2017 */

Crossrefs

Cf. A000302 (row sums), A001109, A006519, A007318, A053124, A091042, A091043, A111910.

Sequence in context: A019234 A032172 A032103 * A079661 A220644 A338372

Adjacent sequences: A091041 A091042 A091043 * A091045 A091046 A091047

Keyword

nonn,easy,tabl

Author

Wolfdieter Lang, Jan 23 2004

Status

approved