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%I #23 Apr 08 2024 09:13:15
%S 0,1,0,-4,2,0,-12,12,3,0,32,-48,-24,4,0,80,-160,-120,40,5,0,-192,480,
%T 480,-240,-60,6,0,-448,1344,1680,-1120,-420,84,7,0,1024,-3584,-5376,
%U 4480,2240,-672,-112,8,0,2304,-9216,-16128,16128,10080,-4032,-1008,144,9,0
%N Triangle L, read by rows, equal to the matrix log of A118435, with the property that L^2 consists of a single diagonal (two rows down from the main diagonal).
%C L = log(A118435) = log(H*[C^-1]*H], where C=Pascal's triangle and H=A118433 where H^2 = I (identity matrix).
%F For even exponents of L, L^(2m) is a single diagonal:
%F if n == k+2m, then [L^(2m)](n,k) = n!/k!*2^(n-k-2m)/(n-k-2m)!; else if n != k+2m: [L^(2m)](n,k) = 0.
%F For odd exponents of L:
%F if n >= k+2m+1, then [L^(2m+1)](n,k) = n!/k!*2^(n-k-2m-1)/(n-k-2m-1)!*(-1)^(m+[(n+1)/2]-[k/2]+n-k); else if n < k+2m+1: [L^(2m)](n,k) = 0.
%F Unsigned row sums equals A027471(n+1) = n*3^(n-1).
%e The matrix log, L = log(H*[C^-1]*H], begins:
%e 0;
%e 1, 0;
%e -4, 2, 0;
%e -12, 12, 3, 0;
%e 32, -48, -24, 4, 0;
%e 80, -160, -120, 40, 5, 0;
%e -192, 480, 480, -240, -60, 6, 0;
%e -448, 1344, 1680, -1120, -420, 84, 7, 0;
%e 1024, -3584, -5376, 4480, 2240, -672, -112, 8, 0;
%e 2304, -9216, -16128, 16128, 10080, -4032, -1008, 144, 9, 0;
%e ...
%e The matrix square, L^2, is a single diagonal:
%e 0;
%e 0, 0;
%e 2, 0, 0;
%e 0, 6, 0, 0;
%e 0, 0, 12, 0, 0;
%e 0, 0, 0, 20, 0, 0;
%e 0, 0, 0, 0, 30, 0, 0;
%e ...
%e From _Peter Luschny_, Apr 23 2020: (Start)
%e In unsigned form and without the main diagonal, as computed by the Maple script:
%e [0], [0]
%e [1], [1]
%e [2], [4, 2]
%e [3], [12, 12, 3]
%e [4], [32, 48, 24, 4]
%e [5], [80, 160, 120, 40, 5]
%e [6], [192, 480, 480, 240, 60, 6]
%e [7], [448, 1344, 1680, 1120, 420, 84, 7] (End)
%p # Generalized Worpitzky transform of the harmonic numbers.
%p CL := p -> PolynomialTools:-CoefficientList(expand(p), x):
%p H := n -> add(1/k, k=1..n):
%p Trow := proc(n) local k,v; if n=0 then return [0] fi;
%p add(add((-1)^(n-v)*binomial(k,v)*H(k)*(-x+v-1)^n, v=0..k), k=0..n); CL(%) end:
%p for n from 0 to 7 do Trow(n) od; # _Peter Luschny_, Apr 23 2020
%t nmax = 12;
%t h[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2] - Quotient[k, 2]+n-k);
%t H = Table[h[n, k], {n, 0, nmax}, {k, 0, nmax}];
%t Cn = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, nmax}];
%t L = MatrixLog[H.Inverse[Cn].H ];
%t Table[L[[n+1, k+1]], {n, 0, nmax}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Apr 08 2024 *)
%o (PARI) /* From definition of L as matrix log of H*C^-1*H: */
%o {L(n,k)=local(H=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1)*(-1)^(r\2-(c-1)\2+r-c))),C=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1))),N=(H*C^-1*H)); Log=sum(p=1,n+1,-(N^0-N)^p/p);Log[n+1,k+1]}
%o for(n=0, 10, for(k=0, n, print1(L(n, k), ", ")); print(""))
%o (PARI) /* The matrix power L^m is given by: */
%o {L(n,k,m)=if(m%2==0,if(n==k+m,n!/k!*2^(n-k-m)/(n-k-m)!), if(n>=k+m,n!/k!*2^(n-k-m)/(n-k-m)!*(-1)^(m\2+(n+1)\2-k\2+n-k)))}
%o for(n=0, 10, for(k=0, n, print1(L(n, k,1), ", ")); print(""))
%Y Cf. A118435 (exp(L)), A118442 (column 0), A118443 (row sums), A027471 (unsigned row sums); A118433 (self-inverse triangle), A001815 (column 1?), A001789 (third of column 2?).
%K sign,tabl
%O 0,4
%A _Paul D. Hanna_, Apr 28 2006
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