cp's OEIS Frontend

A258820 Reversed rows of A178252 presented as diagonals of an irregular triangle.

%I A258820 #55 Apr 11 2016 10:21:35
%S A258820 1,1,1,1,1,1,1,3,1,1,2,1,1,5,2,1,1,3,10,1,1,7,5,5,1,1,4,7,5,1,1,9,28,
%T A258820 35,3,1,1,5,12,14,7,1,1,11,15,21,14,7,1,1,6,55,30,126,28,1,1,13,22,
%U A258820 165,42,21,4,1
%N A258820 Reversed rows of A178252 presented as diagonals of an irregular triangle.
%C A258820 The diagonals of T are the reversed rows of A178252. E.g., the fifth diagonal of T is (1,2,2,1,1) from the example below, which is the fifth reversed row of A178252.
%C A258820 Factoring out the greatest common divisor (gcd) of the coefficients of the sub-polynomials in the indeterminate q of the polynomials in s presented on p. 12 of the Alexeev et al. link and then evaluating the sub-polynomials at q=1 gives the first nine rows of T given in the example below. E.g., for k=6 (the seventh row), 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 (1,2+3,1+1,1)=(1,5,2,1).
%C A258820 The row length sequence of this irregular triangle is A008619(n) = 1 + floor(n/2). - _Wolfdieter Lang_, Aug 25 2015
%H A258820 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.
%F A258820 T(n,k) =  A178252(n-k,n-2k) = A055151(n,k) / A161642(n,k) = A007318(n,2k) * A000108(k) / A161642(n,k) = n! / [(n-2k)! k! (k+1)! A161642(n,k)] =  A003989(n-k+1,k+1) * (n-k)! / [ (n-2k)! (k+1)! ], where A003989(j,k) = gcd(j,k).
%e A258820 The irregular triangle T(n,k) starts
%e A258820 n\k  0 1  2  3 4 5 ...
%e A258820 0:   1
%e A258820 1:   1
%e A258820 2:   1 1
%e A258820 3:   1 1
%e A258820 4:   1 3  1
%e A258820 5:   1 2  1
%e A258820 6:   1 5  2  1
%e A258820 7:   1 3 10  1
%e A258820 8:   1 7  5  5 1
%e A258820 9:   1 4  7  5 1
%e A258820 10:  1 9 28 35 3 1
%e A258820 ... reformatted. - _Wolfdieter Lang_, Aug 25 2015
%t A258820 max = 15; coes = Table[ PadRight[ CoefficientList[ BernoulliB[n, x], x], max], {n, 0, max-1}]; inv = Inverse[coes] // Numerator; t[n_, k_] := inv[[n, k]]; t[n_, k_] /; k == n+1 = 1; Table[t[n-k+1, k], {n, 2, max+1}, {k, 2, Floor[n/2]+1}] // Flatten (* _Jean-François Alcover_, Jul 22 2015 *)
%Y A258820 Cf. A050169, A161642, A055151, A088617, A178252, A107711, A165661, A003989, A008619.
%K A258820 nonn,tabf,easy
%O A258820 0,8
%A A258820 _Tom Copeland_, Jun 18 2015