A323346 Square array read by ascending antidiagonals: T(p,q) is the number of bases e such that e^2 = 1 (including e = 1) in Clifford algebra Cl(p,q)(R).

1, 2, 1, 3, 3, 1, 4, 6, 4, 2, 6, 10, 10, 6, 6, 12, 16, 20, 16, 12, 16, 28, 28, 36, 36, 28, 28, 36, 64, 56, 64, 72, 64, 56, 64, 72, 136, 120, 120, 136, 136, 120, 120, 136, 136, 272, 256, 240, 256, 272, 256, 240, 256, 272, 256, 528, 528, 496, 496, 528, 528, 496, 496, 528, 528, 496

Offset: 0

Author

Jianing Song, Jan 12 2019

Keywords

• nonn
• tabl

Comments

See A323100 for an introduction of Clifford algebras.

Examples

			Table begins
p\q|  0   1   2    3    4    5  ...
---+-------------------------------
0  |  1,  1,  1,   2,   6,  16, ...
1  |  2,  3,  4,   6,  12,  28, ...
2  |  3,  6, 10,  16,  28,  56, ...
3  |  4, 10, 20,  36,  64, 120, ...
4  |  6, 16, 36,  72, 136, 256, ...
5  | 12, 28, 64, 136, 272, 528, ...
...
See A323100 for an example that shows T(1,3) = 6.
		

Links

• Jianing Song, Antidiagonals n = 0..99, flattened
• Wikipedia, Clifford algebras

Crossrefs

Cf. A038503(n+1) (first row), A038504(n+1) (first column), A007582 (main diagonal).

A323100 is the complement sequence.

Programs

Maple

s := sqrt(2): h := n -> [ 0, -s, -2, -s, 0, s, 2,  s][1 + modp(n+1, 8)]:
T := proc(n, k) option remember;
if n = 0 then return 2^k*(1 - 1/2) - 2^((k - 3)/2)*h(k + 2) fi;
if k = 0 then return 2^n*(1 - 1/2) - 2^((n - 3)/2)*h(n) fi;
T(n, k-1) + T(n-1, k) end:
for n from 0 to 9 do seq(T(n, k), k=0..9) od; # Peter Luschny, Jan 12 2019

Mathematica

T[n_, k_] := 2^(n + k) - Sum[Binomial[n, i] Binomial[k, j] Mod[Binomial[i - j, 2], 2], {i, 0, n}, {j, 0, k}];
Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] (* Jean-François Alcover, Jun 19 2019 *)

PARI

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

Formula

T(p,q) = Sum_{i=0..p} Sum_{j=0..q} binomial(p, i)*binomial(q, j)*(1 - (binomial(i - j, 2) mod 2)).

T(p,q) = 2^(p+q) - A323100(p,q).