A323100 - OEIS

%I #46 Jun 19 2019 09:31:23

%S 0,0,1,1,1,3,4,2,4,6,10,6,6,10,10,20,16,12,16,20,16,36,36,28,28,36,36,

%T 28,64,72,64,56,64,72,64,56,120,136,136,120,120,136,136,120,120,240,

%U 256,272,256,240,256,272,256,240,256,496,496,528,528,496,496,528,528,496,496,528

%N Square array read by ascending antidiagonals: T(p,q) is the number of bases e such that e^2 = -1 in Clifford algebra Cl(p,q)(R).

%C Cl(p,q)(R) is a 2^(p+q)-dimensional algebraic structure generated by {e_1, e_2, ..., e_(p+q)}, where (e_1)^2 = (e_2)^2 = ... = (e_p)^2 = +1, (e_(p+1))^2 = (e_(p+2))^2 = ... = (e_q)^2 = -1, (e_i)*(e_j) = -(e_j)*(e_i) for any i != j (anti-commutativity), ((e_i)*(e_j))*(e_k) = (e_i)*((e_j)*(e_k)) for any i, j, k (associativity). So the 2^(p+q) basis are all elements of the form Product_{s=1..t} e_(i_s) where 1 <= i_1 < i_2 < ... < i_t <= p + q and {i_1, i_2, ..., i_t} runs through all 2^(p+q) subsets of {1, 2, ..., p + q} (due to the failure of commutativity, one should be careful when taking continued products). Examples include: the real numbers Cl(0,0)(R), the complex numbers Cl(0,1)(R), split-complex numbers Cl(1,0)(R), quaternions Cl(1,0)(R), etc. If p + q = p' + q', then Cl(p,q)(R) is equal to Cl(p',q')(R) if and only if T(p,q) = T(p',q').

%C It can be shown that (Product_{s=1..t} e_(i_s))^2 = (-1)^(t*(t-1)/2)*(Product_{s=1..t} (e_(i_s))^2). So (Product_{s=1..t} e_(i_s))^2 = -1 if and only if t == 0, 1 (mod 4) and #({i_1, i_2, ..., i_t} intersect {p + 1, p + 2, ..., p + q}) is odd, or t == 2, 3 (mod 4) and #({i_1, i_2, ..., i_t} intersect {p + 1, p + 2, ..., p + q}) is even.

%C In general, let A = (a_ij) be any n X n symmetric {-1,1}-matrix, we can define an algebraic structure generated by {e_1, e_2, ..., e_n} where (e_i)^2 = a_ii for i = 1..n, (e_i)*(e_j) = (a_ij)*(e_j)*(e_i) for any i != j, ((e_i)*(e_j))*(e_k) = (e_i)*((e_j)*(e_k)) for any i, j, k. Clifford algebras are the cases where a_ij = -1 for any i != j. It can be shown that (Product_{s=1..t} e_(i_s)) * (Product_{s'=1..u} e_(j_s')) = (Product_{s=1..t, s'=1..u, i_s>=j_s'} a_(i_s)(j_s')) * (Product_{s=1..v} e_(k_s)), where 1 <= k_1 < k_2 < ... < k_v <= n and {k_1, k_2, ..., k_v} is the symmetric difference between {i_1, i_2, ..., i_t} and {j_1, j_2, ..., j_u}. Specially, (Product_{s=1..t} e_(i_s)))^2 = Product_{1<=s'<=s<=n} a_ss'. The 2^n basis, together with their additive inverses, form a group of order 2^(n+1) under multiplication, which is abelian if and only if a_ij = 1 for any i != j (in this case, it is isomorphic to (C_2)^(n+1) if a_ii = 1 for i = 1..n, and (C_2)^(n-1) X C_4 otherwise). The structure of this group can be complicated. For example, when n = 2, it can be isomorphic to either (C_2)^3, C_2 X C_4, C_2 X D_4 or Q_8.

%H Jianing Song, <a href="/A323100/b323100.txt">Antidiagonals n = 0..99, flattened</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Hypercomplex_number#Clifford_algebras">Clifford algebras</a>

%F T(p,q) = Sum_{i=0..p} Sum_{j=0..q} binomial(p, i)*binomial(q, j)*(binomial(i - j, 2) mod 2).

%F T(p,q) = 2^(p+q) - A323346(p,q).

%e Table begins

%e p\q|  0   1   2    3    4    5  ...

%e ---+-------------------------------

%e 0  |  0,  1,  3,   6,  10,  16, ...

%e 1  |  0,  1,  4,  10,  20,  36, ...

%e 2  |  1,  2,  6,  16,  36,  72, ...

%e 3  |  4,  6, 12,  28,  64, 136, ...

%e 4  | 10, 16, 28,  56, 120, 256, ...

%e 5  | 20, 36, 64, 120, 240, 496, ...

%e ...

%e Example for T(1,3) = 10: (Start)

%e 1^2 = 1;

%e (e_1)^2 = 1;

%e (e_2)^2 = -1;

%e (e_3)^2 = -1;

%e (e_4)^2 = -1;

%e ((e_1)*(e_2))^2 = -(e_1)^2*(e_2)^2 = 1;

%e ((e_1)*(e_3))^2 = -(e_1)^2*(e_3)^2 = 1;

%e ((e_1)*(e_4))^2 = -(e_1)^2*(e_4)^2 = 1;

%e ((e_2)*(e_3))^2 = -(e_2)^2*(e_3)^2 = -1;

%e ((e_2)*(e_4))^2 = -(e_2)^2*(e_4)^2 = -1;

%e ((e_3)*(e_4))^2 = -(e_3)^2*(e_4)^2 = -1;

%e ((e_1)*(e_2)*(e_3))^2 = -(e_1)^2*(e_2)^2*(e_3)^2 = -1;

%e ((e_1)*(e_2)*(e_4))^2 = -(e_1)^2*(e_2)^2*(e_4)^2 = -1;

%e ((e_1)*(e_3)*(e_4))^2 = -(e_1)^2*(e_3)^2*(e_4)^2 = -1;

%e ((e_2)*(e_3)*(e_4))^2 = -(e_2)^2*(e_3)^2*(e_4)^2 = 1;

%e ((e_1)*(e_2)*(e_3)*(e_4))^2 = (e_1)^2*(e_2)^2*(e_3)^2*(e_4)^2 = -1. (End)

%e From _Peter Luschny_, Jan 13 2019: (Start)

%e The first few lines of the triangle T(i-j,j) are:

%e [0]   0;

%e [1]   0,   1;

%e [2]   1,   1,   3;

%e [3]   4,   2,   4,   6;

%e [4]  10,   6,   6,  10,  10;

%e [5]  20,  16,  12,  16,  20,  16;

%e [6]  36,  36,  28,  28,  36,  36,  28;

%e [7]  64,  72,  64,  56,  64,  72,  64,  56;

%e [8] 120, 136, 136, 120, 120, 136, 136, 120, 120;

%e [9] 240, 256, 272, 256, 240, 256, 272, 256, 240, 256; (End)

%p s := sqrt(2): h := n -> [ 0, -s, -2, -s, 0, s, 2,  s][1 + modp(n+1, 8)]:

%p T := proc(n, k) option remember;

%p if n = 0 then return 2^(k - 1) + 2^((k - 3)/2)*h(k + 2) fi;

%p if k = 0 then return 2^(n - 1) + 2^((n - 3)/2)*h(n) fi;

%p T(n, k-1 ) + T(n-1, k) end:

%p for n from 0 to 9 do seq(T(n, k), k=0..9) od; # _Peter Luschny_, Jan 12 2019

%t T[n_, k_] := Sum[Binomial[n, i] Binomial[k, j] Mod[Binomial[i - j, 2], 2], {i, 0, n}, {j, 0, k}];

%t Table[T[n-k, k], {n, 0, 10}, {k, 0, n}] (* _Jean-François Alcover_, Jun 19 2019 *)

%o (PARI) T(p,q) = sum(i=0, p, sum(j=0, q, binomial(p, i)*binomial(q, j)*(binomial(i-j, 2)%2)))

%Y Cf. A038505(n+1) (first row), A000749(n+1) (first column), A006516 (main diagonal),

%Y A321959 (antidiagonal sums).

%Y A323346 is the complement sequence.

%K nonn,tabl

%O 0,6

%A _Jianing Song_, Jan 04 2019