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%I #32 Jun 03 2018 02:06:15
%S 1,1,1,1,1,1,1,1,2,0,1,1,3,4,0,1,1,4,11,9,0,1,1,5,21,46,21,0,1,1,6,34,
%T 127,207,51,0,1,1,7,50,268,833,979,127,0,1,1,8,69,485,2299,5763,4797,
%U 323,0,1,1,9,91,794,5130,20838,41401,24138,835,0
%N Array of coefficients A(n,k) of the formal power series P(n,x) read by upwards antidiagonals, where P(n,x) = Sum_{k>=0} A(n,k)*x^k = 1+x*P(n,x)^(1*n)+x^2*P(n,x)^(2*n) for n >= 0.
%C The terms define the array A(n,k):
%C n\k: 0 1 2 3 4 5 6 7 8 9 10 ...
%C 0: 1 1 1 0 0 0 0 0 0 0 0 ...
%C 1: 1 1 2 4 9 21 51 127 323 835 ...
%C 2: 1 1 3 11 46 207 979 4797 24138 123998 ...
%C 3: 1 1 4 21 127 833 5763 41401 305877 2309385 ...
%C 4: 1 1 5 34 268 2299 20838 ...
%C 5: 1 1 6 50 485 5130 ...
%C 6: 1 1 7 69 794 ...
%C 7: 1 1 8 91 ...
%C 8: 1 1 9 116 ...
%C 9: 1 1 10 144 ...
%C 10: 1 ...
%C etc.
%C For row 1 see A001006, for row 2 see A006605, and for row 3 see A255673.
%C Be careful if you use the formulas for n < 0 (DIV0, signed values)!
%C Nevertheless, it might be interesting ...
%C Conjecture: The A(n,k), here n > 0, are the number of lattice paths, if
%C (a) length of path is k*n for the k-th term of row n,
%C (b) allowed steps are (1,-1), (1,-1+n) and (1,-1+2*n) for terms of row n,
%C (c) you start at (0,0), end at (k*n,0), and
%C (d) never cross the x-axis.
%C This is proved for row 1 (A001006) and row 2 (A006605).
%C Conjecture: The coefficients B(m,n,k) of the P(n,x)^m (see the formula below), m > 0 and n > 0, are the number of lattice paths, if
%C (a) length of path is k*n+m-1 (k-th coefficient of P(n,x)^m),
%C (b) allowed steps are (1,-1), (1,-1+n), and (1,-1+2*n),
%C (c) you start at (0,m-1), end at (k*n+m-1,0), and
%C (d) never cross the x-axis.
%C This is proved for B(1,1,k) (A001006), and B(1,2,k) (A006605). - _Werner Schulte_, Aug 30 2015
%F A(n,k) = 1/(n*k+1)*Sum_{j=0..k} (-1)^j*binomial(n*k+1, j)*binomial(2*n*k+2-2*j, k-j) (conjectured).
%F The g.f. P(n,x) of row n of the array A(n,k) satisfy:
%F P(n,x) = (1 + x*P(n,x)^n)^2/(1 + x*P(n,x)^(n-1)), n > 0.
%F P(n,x) = P(n-1,x*P(n,x)), n > 0.
%F P(n,x) = P(n-2,x*P(n,x)^2), n > 1.
%F etc.
%F P(n,x) = P(0,x*P(n,x)^n), n >= 0.
%F The coefficients B(m,n,k) of the P(n,x)^m are:
%F B(m,n,k) = m/(n*k + m)*(Sum_{j=0..k} (-1)^j*binomial(n*k+m, j)* binomial(2*n*k + 2*m - 2*j, k - j)), if m > 0, and n > 0 (conjectured).
%F A(n,0) = A(n,1) = 1, n >= 0.
%F A(n,2) = n+1, n >= 0.
%F A(n,3) = n*(3*n + 5)/2, n >= 0.
%F A(n,4) = n*(8*n^2 + 18*n + 1)/3, n >= 0.
%F A(n,5) = n*(125*n^3 + 350*n^2 + 55*n - 26)/24, n >= 0.
%F P(n,x) = exp(Sum_{k>=1} 1/(n*k)*(Sum{j=0..k} (-1)^j*binomial(n*k,j)* binomial(2*n*k-2*j,k-j))) for n > 0 (conjectured). - _Werner Schulte_, Sep 20 2015
%F P(n,x/(1+x+x^2)^n) = 1+x+x^2 for n >= 0. - _Werner Schulte_, Oct 20 2015
%e The terms of the array A(n,k) read by upwards antidiagonals define the triangle T(n,m) = A(n-m,m) for 0 <= m <= n, i.e.
%e 1;
%e 1, 1;
%e 1, 1, 1;
%e 1, 1, 2, 0;
%e 1, 1, 3, 4, 0;
%e 1, 1, 4, 11, 9, 0;
%e 1, 1, 5, 21, 46, 21, 0;
%e etc.
%Y Cf. A001006, A006605, A255673.
%K nonn,tabl,easy
%O 0,9
%A _Werner Schulte_, Aug 18 2015