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%I #46 Jan 06 2021 12:20:07
%S 1,1,1,1,8,1,1,27,27,1,1,64,216,64,1,1,125,1000,1000,125,1,1,216,3375,
%T 8000,3375,216,1,1,343,9261,42875,42875,9261,343,1,1,512,21952,175616,
%U 343000,175616,21952,512,1,1,729,46656,592704,2000376,2000376,592704,46656,729,1
%N Triangle of cubed binomial coefficients, T(n,k) = C(n,k)^3, read by rows.
%C Diagonal of rational function R(x,y,z,t) = 1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z) with respect to x, y, z, i.e., T(n,k) = [(xyz)^n*t^k] R(x,y,z,t). - _Gheorghe Coserea_, Jul 01 2018
%H Indranil Ghosh, <a href="/A181543/b181543.txt">Rows 0..120 of triangle, flattened</a>
%H Jeffrey S. Geronimo, Hugo J. Woerdeman, and Chung Y. Wong, <a href="https://arxiv.org/abs/2101.00525">The autoregressive filter problem for multivariable degree one symmetric polynomials</a>, arXiv:2101.00525 [math.CA], 2021.
%F Row sums equal A000172, the Franel numbers.
%F Central terms are A002897(n) = C(2n,n)^3.
%F Antidiagonal sums equal A181545;
%F The g.f. of the antidiagonal sums is Sum_{n>=0} (3n)!/(n!)^3 * x^(3n)/(1-x-x^2)^(3n+1).
%F G.f. for column k: [Sum_{j=0..2k} A181544(k,j)*x^j]/(1-x)^(3k+1), where the row sums of A181544 equals De Bruijn's s(3,n) = (3n)!/(n!)^3.
%F G.f.: A(x,y) = Sum_{n>=0} (3n)!/n!^3 * x^(2n)*y^n/(1-x-x*y)^(3n+1). - _Paul D. Hanna_, Nov 04 2010
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 8, 1;
%e 1, 27, 27, 1;
%e 1, 64, 216, 64, 1;
%e 1, 125, 1000, 1000, 125, 1;
%e 1, 216, 3375, 8000, 3375, 216, 1;
%e 1, 343, 9261, 42875, 42875, 9261, 343, 1;
%e 1, 512, 21952, 175616, 343000, 175616, 21952, 512, 1;
%e 1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1;
%e ...
%p T:= (n, k)-> binomial(n, k)^3:
%p seq(seq(T(n, k), k=0..n), n=0..10); # _Alois P. Heinz_, Jan 06 2021
%t Flatten[Table[Binomial[n,k]^3,{n,0,10},{k,0,n}]] (* _Harvey P. Dale_, May 23 2011 *)
%o (PARI) T(n,k)=binomial(n,k)^3
%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print())
%o (PARI) T(n,k)=polcoeff(polcoeff(sum(m=0,n,(3*m)!/m!^3*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(3*m+1)),n,x),k,y)
%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ _Paul D. Hanna_, Nov 04 2010
%o (PARI)
%o diag(expr, N=22, var=variables(expr)) = {
%o my(a = vector(N));
%o for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
%o for (n = 1, N, a[n] = expr;
%o for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
%o return(a);
%o };
%o x='x; y='y; z='z; t='t;
%o concat(apply(Vec, diag(1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z), 10, [x, y, z]))) \\ _Gheorghe Coserea_, Jul 01 2018
%Y Cf. A000172 (row sums), A181545 (antidiagonal sums), A002897, A181544, A248658.
%Y Variants: A008459, A007318.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Oct 30 2010
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