A262082 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) + x^3*P(n,x)^(3*n) for n >= 0.
1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 5, 0, 1, 1, 4, 12, 13, 0, 1, 1, 5, 22, 54, 36, 0, 1, 1, 6, 35, 139, 262, 104, 0, 1, 1, 7, 51, 284, 953, 1337, 309, 0, 1, 1, 8, 70, 505, 2509, 6894, 7072, 939, 0, 1, 1, 9, 92, 818, 5455, 23426, 51796, 38426, 2905, 0, 1, 1, 10
Offset: 0
Examples
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., n\m 0 1 2 3 4 5 6 7 8 9 ... 0: 1 1: 1 1 2: 1 1 1 3: 1 1 2 1 4: 1 1 3 5 0 5: 1 1 4 12 13 0 6: 1 1 5 22 54 36 0 7: 1 1 6 35 139 262 104 0 8: 1 1 7 51 284 953 1337 309 0 9: 1 1 8 70 505 2509 6894 7072 939 0 etc. [reformatted by _Wolfdieter Lang_, Oct 15 2015]
Formula
A(n,k) = 1/(n*k+1) * Sum_{j=0..k} (-2)^j*binomial(n*k+1,j)* binomial(3*n*k+3-2*j,k-j) for n >= 0, and k >= 0. (conjectured)
A(n,0) = A(n,1) = 1, n >= 0;
A(n,2) = n+1, n >= 0;
A(n,3) = (3*n^2+5*n+2)/2, n >= 0;
A(n,4) = (8*n^3+18*n^2+13*n)/3, n >= 0;
A(n,5) = (125*n^4+350*n^3+355*n^2+34*n)/24, n >= 0.
The g.f. P(n,x) of row n of the array A(n,k) satisfy:
P(n,x) = P(n-1,x*P(n,x)), n > 0;
P(n,x) = P(n-2,x*P(n,x)^2), n > 1;
etc.
P(n,x) = P(0,x*P(n,x)^n), n >= 0.
The coefficients B(m,n,k) of the P(n,x)^m are:
B(m,n,k) = m/(n*k+m) * Sum_{j=0..k} (-2)^j*binomial(n*k+m,j)* binomial(3*n*k+3*m-2*j,k-j) for m > 0, n > 0, and k >= 0. (conjectured)
P(n,x) = exp(Sum_{k>=1} 1/(n*k)*(Sum_{j=0..k} (-2)^j*binomial(n*k,j)* binomial(3*n*k-2*j,k-j))) for n > 0 (conjectured); (see for n=1: A036765, for n=2: A186241, and for n=3: A200731).
P(n,x/(1+x+x^2+x^3)^n) = 1+x+x^2+x^3 for n >= 0. - Werner Schulte, Nov 20 2015
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