A118685 Signs of entries in the multiplication table for hypercomplex numbers with Cayley-Dickson construction (by antidiagonals).

0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0

Offset

0,1

Comments

The signs in the second line of the table give the Thue-Morse sequence (A010060).

Links

Table of n, a(n) for n=0..104.

Joerg Arndt, Matters Computational (The Fxtbook), section 39.14.1 "The Cayley-Dickson construction", pp.815-818

Joerg Arndt, Demo program

John C. Baez, The Octonions, Bull. Amer. Math. Soc., 39 (2002), 145-205.

Example

Multiplication table for the octonions:
Let e0,e1,...e7 be the units.
The third entry in the second row is 3+, meaning that e1*e2==+e3.
The product is anti-commutative unless one factor is e0.
0 1 2 3 4 5 6 7
0: 0+ 1+ 2+ 3+ 4+ 5+ 6+ 7+
1: 1+ 0- 3- 2+ 5- 4+ 7+ 6-
2: 2+ 3+ 0- 1- 6- 7- 4+ 5+
3: 3+ 2- 1+ 0- 7- 6+ 5- 4+
4: 4+ 5+ 6+ 7+ 0- 1- 2- 3-
5: 5+ 4- 7+ 6- 1+ 0- 3+ 2-
6: 6+ 7- 4- 5+ 2+ 3- 0- 1+
7: 7+ 6+ 5- 4- 3+ 2+ 1- 0-
For the multiplication er*ec = +-ep we have p = r XOR c
The sign is given in the following array:
0 1 2 3 4 5 6 7
0: + + + + + + + +
1: + - - + - + + -
2: + + - - - - + +
3: + - + - - + - +
4: + + + + - - - -
5: + - + - + - + -
6: + - - + + - - +
7: + + - - + + - -
Now replace all + by 0 and all - by 1.
Read by antidiagonals (rising order) to obtain the sequence.
Cayley-Dickson construction:
Multiplication rule is
(a,b)*(A,b) = (a*A - B*conj(b), conj(a)*B + A*b)
where conj(a,b) := (conj(a), -b) and conj(x):=x for x real
[ Transposed rule/table is obtained if rule is changed to
(a,b)*(A,b) = (a*A - conj(B)*b, b*conj(A) + B*a) ]

Prog

(C++)
void cp2(ulong a, ulong b, ulong &u, ulong &v) { u=a; v=b; }
int CD_sign(ulong r, ulong c, ulong n) // (returns +1 or -1)
{
  int s = +1;
  while ( true )
  {
      if ( (r==0) || (c==0) )  return s;
      if ( c==r )  return -s;
      if ( c>r )   { std::swap(r, c); s=-s; }
      n >>= 1;
      if ( c>=n )  cp2(c-n, r-n, r, c);
      else if ( r>=n )  cp2(c, r-n, r, c);
  }
}

Crossrefs

Cf. A096809, A010060.

Sequence in context: A188187 A341996 A341999 * A244063 A080343 A011664

Adjacent sequences: A118682 A118683 A118684 * A118686 A118687 A118688

Keyword

nonn,tabl

Author

Joerg Arndt, May 20 2006

Status

approved