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%I #30 Jun 19 2019 09:31:13
%S 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,
%T 36,64,56,64,72,64,56,64,72,136,120,120,136,136,120,120,136,136,272,
%U 256,240,256,272,256,240,256,272,256,528,528,496,496,528,528,496,496,528,528,496
%N 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).
%C See A323100 for a introduction of Clifford algebras.
%H Jianing Song, <a href="/A323346/b323346.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)*(1 - (binomial(i - j, 2) mod 2)).
%F T(p,q) = 2^(p+q) - A323100(p,q).
%e Table begins
%e p\q| 0 1 2 3 4 5 ...
%e ---+-------------------------------
%e 0 | 1, 1, 1, 2, 6, 16, ...
%e 1 | 2, 3, 4, 6, 12, 28, ...
%e 2 | 3, 6, 10, 16, 28, 56, ...
%e 3 | 4, 10, 20, 36, 64, 120, ...
%e 4 | 6, 16, 36, 72, 136, 256, ...
%e 5 | 12, 28, 64, 136, 272, 528, ...
%e ...
%e See A323100 for an example that shows T(1,3) = 6.
%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 - 1/2) - 2^((k - 3)/2)*h(k + 2) fi;
%p if k = 0 then return 2^n*(1 - 1/2) - 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_] := 2^(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. A038503((n+1) (first row), A038504(n+1) (first column), A007582 (main diagonal).
%Y A323100 is the complement sequence.
%K nonn,tabl
%O 0,2
%A _Jianing Song_, Jan 12 2019
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