0,8

Taking each row polynomial listed on p. 12 of the Alexeev et al. link and listing the GCD of each sub-polynomial in the indeterminate q gives the left half of this entry's symmetric/palindromic triangle. E.g., for k=6, q*s^6 + (6*q + 9*q^2) s^4 + (15*q + 15*q^2) s^2 + 5 = q*s^6 + 3*(2*q + 3*q^2)*s^4 + 15*(q + q^2)*s^2 + 5 generates (1,3,15,5). See also A055151. - Tom Copeland, Jun 18 2015

N. Alexeev, J. Andersen, R. Penner, P. Zograf, Enumeration of chord diagrams on many intervals and their non-orientable analogs, arXiv:1307.0967 [math.CO], 2013-2015.

T(2n,n) = A000108(n).

T(n,k) = binomial(n,k)/A003989(n+1,k+1), 0<=k<=n. - R. J. Mathar, Sep 04 2013

For first half (k <= floor(n/2)) of each palindromic row, T(n,k) = A055151(n,k) / A258820(n,k) = A007318(n,2k) * A000108(k) / A258820(n,k) = n! / [(n-2k)! k! (k+1)! A258820(n,k)]. - Tom Copeland, Jun 18 2015

The triangle T(n,k) begins:

n\k 0 1 2 3 4 5 6 7 8 9 10 ...

0: 1

1: 1 1

2: 1 1 1

3: 1 3 3 1

4: 1 2 2 2 1

5: 1 5 10 10 5 1

6: 1 3 15 5 15 3 1

7: 1 7 7 35 35 7 7 1

8: 1 4 28 28 14 28 28 4 1

9: 1 9 36 84 126 126 84 36 9 1

10: 1 5 15 30 210 42 210 30 15 5 1

... reformatted. - Wolfdieter Lang, Aug 24 2015

T[n_, k_] := Binomial[n, k]/GCD[n-k+1, k+1]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 06 2015, after R. J. Mathar *)

Cf. A000108, A003989, A055151, A258820, A007318.

Sequence in context: A255916 A367825 A107333 * A152141 A098505 A178395

Adjacent sequences: A161639 A161640 A161641 * A161643 A161644 A161645

nonn,tabl,easy

Jason Richardson-White, Jun 15 2009

Name changed, and R. J. Mathar's formula corrected, by Wolfdieter Lang, Aug 24 2015

approved