A118441 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).

0, 1, 0, -4, 2, 0, -12, 12, 3, 0, 32, -48, -24, 4, 0, 80, -160, -120, 40, 5, 0, -192, 480, 480, -240, -60, 6, 0, -448, 1344, 1680, -1120, -420, 84, 7, 0, 1024, -3584, -5376, 4480, 2240, -672, -112, 8, 0, 2304, -9216, -16128, 16128, 10080, -4032, -1008, 144, 9, 0

Offset

0,4

Comments

L = log(A118435) = log(H*[C^-1]*H], where C=Pascal's triangle and H=A118433 where H^2 = I (identity matrix).

Links

Table of n, a(n) for n=0..54.

Formula

For even exponents of L, L^(2m) is a single diagonal:

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.

For odd exponents of L:

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.

Unsigned row sums equals A027471(n+1) = n*3^(n-1).

Example

The matrix log, L = log(H*[C^-1]*H], begins:
     0;
     1,     0;
    -4,     2,      0;
   -12,    12,      3,     0;
    32,   -48,    -24,     4,     0;
    80,  -160,   -120,    40,     5,     0;
  -192,   480,    480,  -240,   -60,     6,     0;
  -448,  1344,   1680, -1120,  -420,    84,     7,   0;
  1024, -3584,  -5376,  4480,  2240,  -672,  -112,   8,  0;
  2304, -9216, -16128, 16128, 10080, -4032, -1008, 144,  9,  0;
  ...
The matrix square, L^2, is a single diagonal:
  0;
  0, 0;
  2, 0,  0;
  0, 6,  0,  0;
  0, 0, 12,  0,  0;
  0, 0,  0, 20,  0,  0;
  0, 0,  0,  0, 30,  0,  0;
  ...
From Peter Luschny, Apr 23 2020: (Start)
In unsigned form and without the main diagonal, as computed by the Maple script:
  [0], [0]
  [1], [1]
  [2], [4,   2]
  [3], [12,  12,   3]
  [4], [32,  48,   24,   4]
  [5], [80,  160,  120,  40,   5]
  [6], [192, 480,  480,  240,  60,  6]
  [7], [448, 1344, 1680, 1120, 420, 84, 7] (End)

Maple

# Generalized Worpitzky transform of the harmonic numbers.
CL := p -> PolynomialTools:-CoefficientList(expand(p), x):
H := n -> add(1/k, k=1..n):
Trow := proc(n) local k, v; if n=0 then return [0] fi;
add(add((-1)^(n-v)*binomial(k, v)*H(k)*(-x+v-1)^n, v=0..k), k=0..n); CL(%) end:
for n from 0 to 7 do Trow(n) od; # Peter Luschny, Apr 23 2020

Mathematica

nmax = 12;
h[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2] - Quotient[k, 2]+n-k);
H = Table[h[n, k], {n, 0, nmax}, {k, 0, nmax}];
Cn = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, nmax}];
L = MatrixLog[H.Inverse[Cn].H ];
Table[L[[n+1, k+1]], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 08 2024 *)

Prog

(PARI) /* From definition of L as matrix log of H*C^-1*H: */
{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]}
for(n=0, 10, for(k=0, n, print1(L(n, k), ", ")); print(""))
(PARI) /* The matrix power L^m is given by: */
{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)))}
for(n=0, 10, for(k=0, n, print1(L(n, k, 1), ", ")); print(""))

Crossrefs

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?).

Sequence in context: A299769 A091435 A330472 * A244131 A206428 A357012

Adjacent sequences: A118438 A118439 A118440 * A118442 A118443 A118444

Keyword

sign,tabl,changed

Author

Paul D. Hanna, Apr 28 2006

Status

approved