%I #14 Sep 30 2024 18:41:20
%S 1,1,2,3,0,-1,5,0,-2,-1,8,0,-5,-2,0,0,0,1,13,0,-10,-5,1,0,0,2,0,1,21,
%T 0,-20,-10,3,2,0,5,0,2,0,0,0,0,0,-1,34,0,-38,-20,9,6,1,10,0,3,0,0,0,0,
%U 0,-2,0,0,-1,55,0,-71,-38,22,18,2,20,0,4,-2,-2,0,0
%N Irregular table T(n, k), n >= 0, k >= 0, read by rows with row polynomials P(n-1, n-1) for n > 0 with T(0, 0) = 1 where P(n, k) = (-1)^(n-k)*z^f(n-k+1)*P(n, k-1) + P(k-1, k-1) for 0 < k <= n with P(n, k) = 0 for k > n, P(n, 0) = 1 for n >= 0 and where f(2n+1) = n, f(2n) = 2n for n >= 0.
%C Row n length is A131355(n).
%C Conjecture: antidiagonal sums equal A000041.
%F Conjectures: (Start)
%F T(n, 0) = A000045(n+1) for n >= 0.
%F T(n, 1) = 0 for n >= 0.
%F Sum_{j=0..2*(floor(k/2)+1)} A084610(floor(k/2)+1, j)*T(n+j, k) = 0 for n >= b(k), k >= 0 where b(k) is some nonnegative integer sequence (with a single exception at k = 1).
%F G.f. for k-th column is Q_k(x)/(1-x-x^2)^(floor(k/2)+1) for k >= 0 where Q_k(x) is some family of polynomials (with a single exception at k = 1). (End)
%e Irregular table begins:
%e 1;
%e 1;
%e 2;
%e 3, 0, -1;
%e 5, 0, -2, -1;
%e 8, 0, -5, -2, 0, 0, 0, 1;
%e 13, 0, -10, -5, 1, 0, 0, 2, 0, 1;
%e 21, 0, -20, -10, 3, 2, 0, 5, 0, 2, 0, 0, 0, 0, 0, -1;
%e 34, 0, -38, -20, 9, 6, 1, 10, 0, 3, 0, 0, 0, 0, 0, -2, 0, 0, -1;
%o (PARI)
%o f(n) = if(n%2, (n-1)/2, n)
%o rows_upto(n) = my(v1); v1 = vector(n+1, i, 1); for(i=2, n, for(j=i+1, n+1, v1[j] = v1[i] + (-1)^(j-i+1)*z^f(j-i)*v1[j])); v1 = vector(n+1, i, Vecrev(v1[i]))
%Y Cf. A000041, A000045, A084610, A131355.
%K sign,tabf
%O 0,3
%A _Mikhail Kurkov_, Sep 29 2024