%I #40 Aug 31 2015 15:34:59

%S 1,1,1,1,1,1,1,3,3,1,1,2,2,2,1,1,5,10,10,5,1,1,3,15,5,15,3,1,1,7,7,35,

%T 35,7,7,1,1,4,28,28,14,28,28,4,1,1,9,36,84,126,126,84,36,9,1,1,5,15,

%U 30,210,42,210,30,15,5,1

%N Triangle (by rows): T(n,k) = A007318(n,k) / A003989(n+1,k+1).

%C 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

%H N. Alexeev, J. Andersen, R. Penner, P. Zograf, <a href="http://arxiv.org/abs/1307.0967">Enumeration of chord diagrams on many intervals and their non-orientable analogs</a>, arXiv:1307.0967 [math.CO], 2013-2014 [From Tom Copeland, Jun 18 2015]

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

%F T(n,k) = binomial(n,k)/A003989(n+1,k+1), 0<=k<=n. - _R. J. Mathar_, Sep 04 2013

%F 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

%e The triangle T(n,k) begins:

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

%e 0: 1

%e 1: 1 1

%e 2: 1 1 1

%e 3: 1 3 3 1

%e 4: 1 2 2 2 1

%e 5: 1 5 10 10 5 1

%e 6: 1 3 15 5 15 3 1

%e 7: 1 7 7 35 35 7 7 1

%e 8: 1 4 28 28 14 28 28 4 1

%e 9: 1 9 36 84 126 126 84 36 9 1

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

%e ... reformatted. - _Wolfdieter Lang_, Aug 24 2015

%t 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_ *)

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

%K nonn,tabl,easy

%O 0,8

%A _Jason Richardson-White_, Jun 15 2009

%E Name changed, and _R. J. Mathar_'s formula corrected, by _Wolfdieter Lang_, Aug 24 2015