cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A323100 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).

Original entry on oeis.org

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, 28, 64, 72, 64, 56, 64, 72, 64, 56, 120, 136, 136, 120, 120, 136, 136, 120, 120, 240, 256, 272, 256, 240, 256, 272, 256, 240, 256, 496, 496, 528, 528, 496, 496, 528, 528, 496, 496, 528
Offset: 0

Views

Author

Jianing Song, Jan 04 2019

Keywords

Comments

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').
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.
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.

Examples

			Table begins
p\q|  0   1   2    3    4    5  ...
---+-------------------------------
0  |  0,  1,  3,   6,  10,  16, ...
1  |  0,  1,  4,  10,  20,  36, ...
2  |  1,  2,  6,  16,  36,  72, ...
3  |  4,  6, 12,  28,  64, 136, ...
4  | 10, 16, 28,  56, 120, 256, ...
5  | 20, 36, 64, 120, 240, 496, ...
...
Example for T(1,3) = 10: (Start)
1^2 = 1;
(e_1)^2 = 1;
(e_2)^2 = -1;
(e_3)^2 = -1;
(e_4)^2 = -1;
((e_1)*(e_2))^2 = -(e_1)^2*(e_2)^2 = 1;
((e_1)*(e_3))^2 = -(e_1)^2*(e_3)^2 = 1;
((e_1)*(e_4))^2 = -(e_1)^2*(e_4)^2 = 1;
((e_2)*(e_3))^2 = -(e_2)^2*(e_3)^2 = -1;
((e_2)*(e_4))^2 = -(e_2)^2*(e_4)^2 = -1;
((e_3)*(e_4))^2 = -(e_3)^2*(e_4)^2 = -1;
((e_1)*(e_2)*(e_3))^2 = -(e_1)^2*(e_2)^2*(e_3)^2 = -1;
((e_1)*(e_2)*(e_4))^2 = -(e_1)^2*(e_2)^2*(e_4)^2 = -1;
((e_1)*(e_3)*(e_4))^2 = -(e_1)^2*(e_3)^2*(e_4)^2 = -1;
((e_2)*(e_3)*(e_4))^2 = -(e_2)^2*(e_3)^2*(e_4)^2 = 1;
((e_1)*(e_2)*(e_3)*(e_4))^2 = (e_1)^2*(e_2)^2*(e_3)^2*(e_4)^2 = -1. (End)
From _Peter Luschny_, Jan 13 2019: (Start)
The first few lines of the triangle T(i-j,j) are:
[0]   0;
[1]   0,   1;
[2]   1,   1,   3;
[3]   4,   2,   4,   6;
[4]  10,   6,   6,  10,  10;
[5]  20,  16,  12,  16,  20,  16;
[6]  36,  36,  28,  28,  36,  36,  28;
[7]  64,  72,  64,  56,  64,  72,  64,  56;
[8] 120, 136, 136, 120, 120, 136, 136, 120, 120;
[9] 240, 256, 272, 256, 240, 256, 272, 256, 240, 256; (End)

Links

Crossrefs

Cf. A038505(n+1) (first row), A000749(n+1) (first column), A006516 (main diagonal),
A321959 (antidiagonal sums).
A323346 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) + 2^((k - 3)/2)*h(k + 2) fi;
if k = 0 then return 2^(n - 1) + 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_] := 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)*(binomial(i - j, 2) mod 2).
T(p,q) = 2^(p+q) - A323346(p,q).

A323225 a(n) = ((2^n*n + i*(1 - i)^n - i*(1 + i)^n))/4, where i is the imaginary unit.

Original entry on oeis.org

0, 1, 3, 7, 16, 38, 92, 220, 512, 1160, 2576, 5648, 12288, 26592, 57280, 122816, 262144, 557184, 1179904, 2490624, 5242880, 11009536, 23067648, 48233472, 100663296, 209717248, 436211712, 905973760, 1879048192, 3892305920, 8053047296, 16642981888, 34359738368
Offset: 0

Views

Author

Peter Luschny, Mar 18 2019

Keywords

Comments

Related to Clifford algebras (see A323100 and A323346).

Links

Crossrefs

Antidiagonal sums of A323346.
Cf. A001787, A009545, A321959, A323100.

Programs

  • Maple
    a := n -> ((2^n*n + I*(1 - I)^n - I*(1 + I)^n))/4:
seq(a(n), n=0..32);
  • Mathematica
    LinearRecurrence[{6, -14, 16, -8}, {0, 1, 3, 7}, 40] (* Jean-François Alcover, Mar 20 2019 *)
Table[((2^n n + I (1 - I)^n - I (1 + I)^n))/4, {n, 0, 29}] (* Alonso del Arte, Mar 27 2020 *)

Formula

a(n) = Sum_{k = 0..n} A323346(n - k, k - 1).
a(n) = (A001787(n) + A009545(n))/2.
a(n) = [x^n] (x*(3*x^2 - 3*x + 1))/((2*x - 1)^2*(2*x^2 - 2*x + 1)).
a(n) = n! [x^n] (exp(2*x)*x + exp(x)*sin(x))/2.
a(n) = (4*n*a(n-3) + (2 - 6*n)*a(n-2) + (4*n - 2)*a(n-1))/(n - 1) for n >= 3.
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.