cp's OEIS Frontend

This is a table read by rows of length 4. Every row is formed from the previous one by the circular Pascal triangle-like rule: a, b, c, d -> d+a, a+b, b+c, c+d. Consider a labeled binary tree such that the root has label 0 and every node labeled k has children labeled k and (k+1) mod 4; the n-th row of this sequence counts nodes on the level n+1 with labels 0, 1, 2, 3, while the n-th row of A099423 counts nodes up to level n. - Andrey Zabolotskiy, Jan 06 2023

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.

A100215(n) (ves) = ((-1)^n)*A009116(n+3) (jes) + a(n) (les) + A038503(n+1) (tes) (Sn, below, corresponds to the generating function from above). Coefficients of Sn(z)*(1-z)/(1+z) gives match to A038504 (Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 1"). Coefficients of Sn(z)/(1+z) gives match to A038505 (Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 2"). Coefficients of Sn(z)/(1-z^2) gives match to A000749 (Number of strings over Z_2 of length n with trace 1 and subtrace 1). The elements 'i, 'j, 'k, i', j', k', 'ii', 'jj', 'kk', 'ij', 'ik', 'ji', 'jk', 'ki', 'kj', e ("floretions") are members of the quaternion product factor space Q x Q /{(1,1), (-1,-1)}. "les" sums over coefficients belonging to basis vectors which squared give the unit e (excluding e itself).

This sequence is identical to its 4th differences. - Jean-François Alcover, Nov 07 2013

Row sums are A038505. Diagonal sums are A117353. Inverse is A117354.

This sequence is connected to A124072. To see this, change the sign of every negative term and consider the differences of every line. Hence for the second line, and following lines, the four terms form periodic sequences:

0 1 0 1 0

1 0 0 1 1

1 0 1 2 1

1 1 3 3 1

2 4 6 4 2

6 10 10 6 6

16 20 16 12 16

36 36 28 28 36

72 64 56 64 72

136 120 120 136 136

256 240 256 272 256.

The lines are connected as seen by the examples: (3rd line connected to 2nd, from right to left) 1+1=2, 1+0=1, 0+0=0, 0+1=1; (11th line connected to 10th) 136+136=272, 136+120=256, 120+120=240, 120+136=256.

The 4 columns are almost known (must the first line be suppressed?): A038503 (without the first 1), A000749 (without the first 0), A038505, A038504. Like the present sequence, every sequence of A124072 beginning with a negative number (-2, -11, ...) is a "twisted" sequence (see A129339 comments, A129961 and the present 4 columns). Periodic with period 2^n.

Inverse binomial transform of A129819. - R. J. Mathar, Feb 25 2009

A000749 is identical to its fourth differences, which implies that the 2nd differences equal the 5th, the 3rd differences the 6th and so on and implies that each of the sequences of these differences obeys the recurrence a(n)=4a(n-1)-6a(n-2)+4a(n-3), n > 3.

The table containing A000749 and its first differences (essentially A038505), 2nd differences (A038504) and 3rd differences (A038503) as the 4 rows is

O, 0, 0, 1, 4, 10, 20, 36, 64, ...

0, 0, 1, 3, 6, 10, 16, 28, 56, ...

0, 1, 2, 3, 4, 6, 12, 28, 64, ...

1, 1, 1, 1, 2, 6, 16, 36, 72, ...

Columns sums are 1, 2, 4, 8, 16, 32 ... = 2^n =A000079. The sequence reads this table column by column.