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One half of odd-numbered entries of even-numbered rows of Pascal's triangle A007318.
4

%I #61 Mar 31 2022 03:13:10

%S 1,2,2,3,10,3,4,28,28,4,5,60,126,60,5,6,110,396,396,110,6,7,182,1001,

%T 1716,1001,182,7,8,280,2184,5720,5720,2184,280,8,9,408,4284,15912,

%U 24310,15912,4284,408,9,10,570,7752,38760,83980,83980,38760,7752,570,10,11

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

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

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

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

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

%C 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).

%H Indranil Ghosh, <a href="/A091044/b091044.txt">Rows 1..125, flattened</a>

%H Matthew Blair, Rigoberto Flórez, and Antara Mukherjee, <a href="https://arxiv.org/abs/2203.13205">Honeycombs in the Pascal triangle and beyond</a>, arXiv:2203.13205 [math.HO], 2022. See p. 4.

%H Kevin Buchin, Man-Kwun Chiu, Stefan Felsner, Günter Rote, and André Schulz, <a href="https://arxiv.org/abs/1903.01095">The Number of Convex Polyominoes with Given Height and Width</a>, arXiv:1903.01095 [math.CO], 2019.

%H Hernan de Alba, W. Carballosa, J. Leaños, and L. M. Rivera, <a href="https://arxiv.org/abs/1606.06370">Independence and matching numbers of some token graphs</a>, arXiv preprint arXiv:1606.06370 [math.CO], 2016.

%H Wolfdieter Lang, <a href="/A091044/a091044.txt">First 9 rows</a>.

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

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

%F 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

%F From _Peter Bala_, Jul 29 2013: (Start)

%F 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 + ....

%F 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.

%F 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.

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

%F From _Werner Schulte_, Jan 13 2017: (Start)

%F (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;

%F (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;

%F (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.

%F (End)

%F 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

%e Triangle begins:

%e [1];

%e [2,2];

%e [3,10,3];

%e [4,28,28,4];

%e [5,60,126,60,5];

%e [6,110,396,396,110,6];

%e ...

%e n = 6 = 2*3: gcd(6,110,396) = 2 = A006519(6);

%e n = 5: gcd(5,60,126) = 1 = A006519(5).

%t Flatten[Table[Binomial[2n,2m+1]/2,{n,1,11},{m,0,n-1}]] (* _Indranil Ghosh_, Feb 22 2017 *)

%o (PARI) {A(i, j) = binomial(2*i + 2*j - 2, 2*i - 1) / 2}; /* _Michael Somos_, Oct 15 2017 */

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

%K nonn,easy,tabl

%O 1,2

%A _Wolfdieter Lang_, Jan 23 2004