[Math.NT] 3 Dec 2016 Fractal Generalized Pascal Matrices

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10000 ... 11000 ...   12100 ... P = 13310 ... .   14641 ...   ......  ......      In Sec. 2. we consider the set of generalized Pascal matrices as a group under Hadamard multiplication and introduce a special system of matrices, which implies the concept of zero generalized Pascal matrices; as an example, we consider the matrix of the q-binomial coeﬃcients for q = −1. In Sec. 3. we construct a fractal generalized Pascal matrices whose Hadamard product is the Pascal matrix. In Sec. 4. we consider the fractal zero generalized Pascal matrices that were previously considered in [7] – [11] (generalized Sierpinski matrices and others).

2 Special system of generalized Pascal matrices

− Elements of the matrix Pc(x), – denote them Pc(x) n,m = cmcn m/cn =(n, m) for gener- ality which will be discussed later, – satisfy the identities (n, 0)=1, (n, m)=(n, n − m) , (1)

(n + q, q)(n + p, m + p)(m + p,p)=(n + p,p)(n + q, m + q)(m + q, q) , (2) q, p =0, 1, 2, . . . It means that each matrix Pc(x) can be associated with the algebra of formal power series whose elements are multiplied by the rule

a (x) ◦ b (x)= g (x) , gn = (n, m) ambm−n, m=0 X that is, if (A)n,m = an−m (n, m), (B)n,m = bn−m (n, m), (G)n,m = gn−m (n, m), then AB = BA = G:

n −1 gn =(n + p,p) (n + p, m + p)(m + p,p) an−mbm = m=0 X n −1 =(n + q, q) (n + q, m + q)(m + q, q) an−mbm. m=0 X The set of generalized Pascal matrix is a group under Hadamard multiplication (we denote this operation ×):

∞ n Pc(x) × Pg(x) = Pc(x)×g(x), c (x) × g (x)= cngnx . n=0 X Introduce the special system of matrices

q−1 − xq 1 P = P (ϕ)= P , c (ϕ,q,x)= xn 1 − , q> 1. ϕ,q q c(ϕ,q,x) ϕ n=0 ! X

2 Then 1 1 c = , 0 ≤ i < q; c − = , 0 < i ≤ q, qn+i ϕn qn i ϕn−1 n n cqm+jcq(n−m)+i−j ϕ ϕ = m n−m =1, i ≥ j; = m n−m−1 = ϕ, i < j, cqn+i ϕ ϕ ϕ ϕ or

(ϕ,qP )n,m =1, n (modq) ≥ m (modq); = ϕ, n (modq) < m (modq) .

For example, ϕ,2P , ϕ,3P : 100000000 ... 100000000 ... 110000000 ... 110000000 ...     1 ϕ 1000000 ... 111000000 ... 111100000 ... 1 ϕ ϕ 100000 ...     1 ϕ 1 ϕ 10000 ... 1 1 ϕ 110000 ...     111111000 ... , 111111000 ... .     1 ϕ 1 ϕ 1 ϕ 1 0 0 ... 1 ϕ ϕ 1 ϕ ϕ 1 0 0 ...     111111110 ... 1 1 ϕ 1 1 ϕ 1 1 0 ...     1 ϕ 1 ϕ 1 ϕ 1 ϕ 1 ... 111111111 ...     ......  ......  ......  ......          Elements of the matrix ϕ,qP × Pc(x) satisfy the identities (1), (2) for any values ϕ, so it makes sense to consider also the case ϕ =0 since it corresponds to a certain algebra of formal power series (which, obviously, contains zero divisors; for example, in the algebra 2 associated with the matrix 0,2P × Pc(x) the product of the series of the form xa (x ) is zero). It is clear that in this case the series c (ϕ,q,x) is not deﬁned. Matrix 0,qP × Pc(x) and Hadamard product of such matrices will be called zero generalized Pascal matrix. Remark. Zero generalized Pascal matrix appears when considering the set of generalized Pascal matrices Pg(q,x): − x n 1 P = [xn] = qm, q ∈ R. g(q,x) n,1 (1 − x) (1 − qx) m=0 X − Here g (0, x)=(1 − x) 1, g (1, x)= ex. In other cases (the q-umbral calculus [2]), except q = −1, ∞ (q − 1)n n g (q, x)= xn, (qn − 1)! = (qm − 1), q0 − 1 !=1. (qn − 1)! n=0 m=1 X Y −1 Matrices Pg(q,x), Pg(q,x) also can be deﬁned as follows:

n n−1 n m −1 −1 m [↑, n]Pg(q,x) = x (1 − q x) , [n, →]Pg(q,x) = (x − q ). m=0 m=0 Y Y −1 When q = −1 we get the matrices Pg(−1,x), Pg(−1,x): 1000000 ... 1 0 0 0 000 ... 1100000 ... −11 0 0 000 ...     1010000 ... −10 1 0 000 ... 1111000 ...  1 −1 −11 000 ...     1020100 ... ,  1 0 −20 100 ... ,     1122110 ... −1 1 2 −2 −1 1 0 ...     1030301 ... −10 3 0 −3 0 1 ...     ......   ......  ......   ......          3 where the series g (−1, x) is not deﬁned. Since

1−j 2m+j 2n+i (1 + x) x n Pg(−1,x) = x = , i ≥ j; =0, i

q−1 n xq 0,qP × Pc(x), c (x)= x e , n=0 ! X n n − 1 P = , i ≥ j; = n , i

n [x ] ϕ,qb (x)=1, n (modq) =0;6 = ϕ, n (modq)=0, so that

Pc(x) = 2P (b2) × 3P (b3) × 4P (b4/b2 ) × 5P (b5) × 6P (b6/b2b3 ) × 7P (b7) ×

×8P (b8/b4 ) × 9P (b9/b3 ) × 10P (b10/b2b5 ) × 11P (b11) × 12P (b12b2/b4b6 ) × ... and so on. Let eq is a basis vector of an infinite-dimensional vector space. Mapping of the set of generalized Pascal matrices in an infinite-dimensional vector space such that ϕ,qP → eq log |ϕ| is a group homomorphism whose kernel consists of all involutions in the group of generalized Pascal matrices, i.e. from matrices whose non-zero elements equal to ±1. Thus, the set of generalized Pascal matrices whose elements are non-negative numbers is an infinite-dimensional vector space. Zero generalized Pascal matrices can be viewed as points at infinity of space.

3 Fractal generalized Pascal matrices

Matrices, which will be discussed (precisely, isomorphic to them), are considered in [3] (p-index Pascal triangle), [4], [5, p. 80-88]. These matrices are introduced explicitly in [6] in connection with the generalization of the theorems on the divisibility of binomial coeﬃcients. We consider them from point of view based on the system of matrices ϕ,qP . We start with the matrix

[2]P = 2,2P × 2,22 P × 2,23 P × ... × 2,2k P × ...

4 1000000000000000 ... 1100000000000000 ...   1210000000000000 ... 1111000000000000 ...   1424100000000000 ...   1122110000000000 ...   1214121000000000 ...   1111111100000000 ...   1848284810000000 ... [2]P =   . 1144224411000000 ...   1218242812100000 ...   1111222211110000 ...   1424184814241000 ...   1122114411221100 ...   1214121812141210 ...   1111111111111111 ...   ......  ......      The ﬁrst column of the matrix [2]P , – denote it b (x), – is the Hadamard product of the series 2,2k b (x) , k =1, 2 ,... ,

n k k [x ] 2,2k b (x)=1, n mod2 =0;6 =2, n mod2 =0. It is the generating function of the distribution of divisors 2kin the series of natural numbers:

b (x)= x +2x2 + x3 +4x4 + x5 +2x6 + x7 +8x8 + x9 +2x10 + x11 +4x12+

+x13 +2x14 + x15 + 16x16 + x17 +2x18 + x19 +4x20 + x21 +2x22 + x23 + ... Theorem. k b2kn+i = bi, 0

Proof. It’s obvious that b2n =2bn, b2n+1 =1. Then

− ∞ k 1 n 2n n 2n x 2 x k 2 x b (x)= +2b x2 = +2kb x2 = , 1 − x2 1 − x2n+1 1 − x2n+1 n=0 n=0 X X or ∞ ∞ n 2n(2m+1) b (x)= (n)a (x), (n)a (x)= 2 x , n=0 m=0 X Xn (n)a2n(2m+1) = (n)a2kp+2n(2m+1) =2 , k > n, k and (n)am =0 in other cases. Hence, (n)a2kp+i = (n)ai, 0

b2kn+i!= b2kn!bi!= b2knb2kn−1!bi!= b2k bnb2k(n−1)+2k−1!bi!=

= b2k bnb2k(n−1)!b2k−1!bi!= bnb2k(n−1)!b2k+i!, k where b2kn =2 bn = b2k bn. For denominator:

b2km+j!b2k(n−m)+i−j!= b2km!bj!b2k(n−m−1)+2k+i−j!= b2km!bj!b2k(n−m−1)!b2k+i−j!. Thus, 2kn + i n − 1 2k + i n 2k + i = b = b , 2km + j n m j m+1 m +1 j 2 2 2 2 2 0 ≤ i< 2k, i < j. For example, 000000000000 ... 000000000000 ...   000000000000 ... 000000000000 ...   042400000000 ...   002200000000 ...   000400000000 ... [2]P − [2]P × 0,4P =   . 000000000000 ...   084808480000 ...   004400440000 ...   000800080000 ...   000000000000 ...   ......  ......      6 Identities 2n +1 2n +1 2n n = = = , 2m +1 2m 2m m 2 2 2 2 2n n − 1 n =2b =2b 2m +1 n m m+1 m +1 2 2 2 means that rows and columns of the matrix [2]P , denote them un (x) and gn (x), form the recurrent sequences:

2 2 2 u2n (x)= un x +2bnxun−1 x , u2n+1 (x)=(1+ x) un x ;

2 2 2 g2n (x)=(1+ x) gn x , g2n+1 (x)= xgn x +2bn+1gn+1 x . Turning to generalize, consider the matrix

[3]P = 3,3P × 3,32 P × 3,33 P × ... × 3,3k P × ...

100000000000000000 ... 110000000000000000 ...   111000000000000000 ... 133100000000000000 ...   113110000000000000 ...   111111000000000000 ...   133133100000000000 ...   113113110000000000 ...   111111111000000000 ...   199399399100000000 ... [3]P =   . 119339339110000000 ...   111333333111000000 ...   133199399133100000 ...   113119339113110000 ...   111111333111111000 ...   133133199133133110 ...   113113119113113111 ...   111111111111111111 ...   ......  ......    We will use the same notation as in the previous example. The ﬁrst column of the matrix k [3]P is the generating function of the distribution of divisors 3 in the series of natural numbers:

b (x)= x + x2 +3x3 + x4 + x5 +3x6 + x7 + x8 +9x9 + x10 + x11 +3x12+

+x13 + x14 +3x15 + x16 + x17 +9x18 + x19 + x20 +3x21 + x22 + x23 + ..., k k b3kn =3 bn, b3n+1 = b3n+2 =1, b3kn+i = bi, 0

b3n+i!= b3n!, 0 ≤ i< 3; b3n−i!= b3(n−1)!, 0 < i ≤ 3,

k− 3 1 n n 2 b3n!=3 bn!, b3kn!=3 bn!, ∞ n n (1 + x) x 1+ x3 3nx3 b (x)= +3b x3 = , 1 − x3 1 − x3n+1 n=0 X c c = c = c = n , 3n 3n+1 3n+2 3n

7 ∞ n n x3 x3 x2(3 ) c (x)= 1+ x + x2 c = 1+ + . 3 3(3n−1)/2 33n−1 n=0 Y If b ! n n = , b !b − ! m m n m 3 then 3kn + i n i = , 0 ≤ i< 3k, i ≥ j; 3km + j m j 3 3 3 3kn + i n − 1 3k + i n 3k + i = b = b , 3km + j n m j m+1 m +1 j 3 3 3 3 3 0 ≤ i< 3k, i < j. From 3n + i n n − 1 n = , i ≥ j; =3b =3b , i < j, 3m + j m n m m+1 m +1 3 3 3 3 we see that if [n, →] [3]P = un (x) , [↑, n] [3]P = gn (x) , then 3 3 u3n (x)= un x +3bn (1 + x) xun−1 x , 3 2 3 u3n+1 (x)=(1+ x) un x +3bnx un−1 x , 2 3 u3n+2 (x)= 1+x + x un x ; 2 3 g3n (x)= 1+ x + x gn x , 3 3 g3n+1 (x)=(1+ x) xgn x +3 bn+1gn+1 x , 2 3 3 g3n+2 (x)= x gn x +3bn+1 (1 + x) gn+1 x . Introduce the notation m n wm (x)= x , w−1 (x)=0. n=0 X For the matrix [q]P = q,qP × q,q2 P × q,q3 P × ... × q,qk P × ..., q =2, 3, 4, . . . , we have (in the same notation as in the previous examples):

k k bqkn = q bn, bqn+i =1, 0

k− q 1 n q−1 bqkn!= q bn!, ∞ qn n qn w − (x) x wq−2 x q x b (x)= q 2 + qb (xq)= ; 1 − xq 1 − xqn+1 n=0 X c c = n , 0 ≤ i < q, qn+i qn ∞ q qn−1 x qn c (x)= w − (x) c = w − x /q q−1 ; q 1 q q 1 n=0 Y qkn + i n i = , 0 ≤ i < qk, i ≥ j, qkm + j m j q q q 8 qkn + i n − 1 qk + i n qk + i = b = b , i

According to the deﬁnition of the matrix [q]P ,

P = [2]P × [3]P × [5]P × ... × [p]P × ..., where p is a member of the sequence of prime numbers. Respectively, the exponential series is the Hadamard product of the series

∞ n n p −1 p p−1 wp−1 x /p . n=0 Y Generalization of the matrix [q]P is the matrix

[ϕ,q]P = ϕ,qP × ϕ,q2 P × ϕ,q3 P × ... × ϕ,qk P × ...,

[q,q]P = [q]P . If ϕ =06 , identities for the matrix [ϕ,q]P can be obtained from the identities k n k n for the matrix [q]P by replacing bqkn = q bn, cqn+i = cn/q on bqkn = ϕ bn, cqn+i = cn/ϕ . If q is ﬁxed, matrices [ϕ,q]P form the group:

[ϕ,q]P × [β,q]P = [ϕβ,q]P,

− where [1,q]P = 1,qP = P(1−x) 1 is the identity element of the group of generalized Pascal matrices. If ϕ =0, we get zero generalized Pascal matrix

[0,q]P = 0,qP × 0,q2 P × 0,q3 P × ... × 0,qk P × ...

For example (Pascal triangle modulo 2),

1000000000000000 ... 1100000000000000 ...   1010000000000000 ... 1111000000000000 ...   1000100000000000 ...   1100110000000000 ...   1010101000000000 ...   1111111100000000 ...   1000000010000000 ... [0,2]P =   . 1100000011000000 ...   1010000010100000 ...   1111000011110000 ...   1000100010001000 ...   1100110011001100 ...   1010101010101010 ...   1111111111111111 ...   ......  ......      k k k k [0,q]P n,m =1, n modq ≥ m modq ; =0, n modq < m modq , k =1, 2,... 9 4 Fractal zero generalized Pascal matrices

Denote n P = . [0,q] n,m m 0,q Theorem. qkn + i n i = , 0 ≤ i, j < qk. (3) qkm + j m j 0,q 0,q 0,q Proof. k k n By deﬁnition, if n modq < m modq for some value of k, then m 0,q = 0. We represent the numbers n, m in the form ∞ ∞ i i n = niq , m = miq , 0 ≤ ni, mi < q. i=0 i=0 X X Then k−1 k−1 k i k i n modq = niq , m modq = miq . i=0 i=0 X X k k n If n modq < m modq , then ni < mi at least for one i. Since m 0,q =1, if mi ≥ ni, then true the identity ∞ n n = i . (4) m mi 0,q i=0 0,q Y It remains to note that if ∞ i k k n = niq =q s + j, 0 ≤ j < q , i=0 X then k−1 ∞ i i j = niq , s = ni+kq . i=0 i=0 X X In the works [7], [8], [9] matrices [0,q]P are called generalized Sierpinski matrices. The property (3) is represented as

Sq = Sq,k ⊗ Sq,k ⊗ Sq,k ⊗ ..., k =1, 2,...,

k where Sq=[0,q]P , Sq,k is the matrix consisting of q ﬁrst rows of the matrix Sq, ⊗ denotes the Kronecker multiplication. Matrices Sq have a generalizations, one of which can be represented as

Sq (a (x)) = Sq,k (a (x)) ⊗ Sq,k (a (x)) ⊗ Sq,k (a (x)) ⊗ ..., where Sq,1 (a (x)) is the matrix consisting of q ﬁrst rows of the matrix A:

n [↑, n] A = x a (x) , (A)n,m = an−m,

k Sq,k (a (x)) is the matrix consisting of q ﬁrst rows of the matrix Sq (a (x)). Then

Sq (a (x)) Sq (b (x)) = Sq (a (x) b (x)) .

The results of these works have been developed in [10], [11]. Introduced matrices

(q) (q) (q) (q) T = Tk ⊗ Tk ⊗ Tk ⊗ ..., k =1, 2,...,

10 (q) (q) where T1 is the matrix consisting of q ﬁrst rows of the Pascal matrix, Tk is the matrix consisting of qk ﬁrst rows of the matrix T (q). For example,

100000000 ... 110000000 ...   121000000 ... 100100000 ...   110110000 ... (3)   T = 121121000 ... .   100200100 ...   110220110 ...   121242121 ...   ......  ......      Denote n T (q) = . n,m m q Then ∞ ∞ ∞ n ni i i = , n = niq , m = miq , 0 ≤ ni, mi < q, m mi q i=0 i=0 i=0 Y X X n ∞ n n i i xm = 1+ xq . m m=0 q i=0 X Y These results become more transparent if we use algebra of the matrices (a (x) |q):

a0b0 0000000 ... a b a b 000000 ...  0 1 0 0  a1b0 0 a0b0 0 0 0 0 0 ... a b a b a b a b 0 0 0 0 ...  1 1 1 0 0 1 0 0  a b 0 0 0 a b 0 0 0 ... (a (x) , b (x) |2, 1) =  2 0 0 0  , a b a b 0 0 a b a b 0 0 ...  2 1 2 0 0 1 0 0  a b 0 a b 0 a b 0 a b 0 ...  3 0 2 0 1 0 0 0  a b a b a b a b a b a b a b a b ...  3 1 3 0 2 1 2 0 1 1 1 0 0 1 0 0   ......   ......     

11 a0b0 0000000 ... a b a b 000000 ...  0 1 0 0  a0b2 0 a0b0 0 0 0 0 0 ... a b a b a b a b 0 0 0 0 ...  0 3 0 2 0 1 0 0  a b 0 0 0 a b 0 0 0 ... (a (x) , b (x) |2, 2) =  1 0 0 0  . a b a b 0 0 a b a b 0 0 ...  1 1 1 0 0 1 0 0  a b 0 a b 0 a b 0 a b 0 ...  1 2 1 0 0 2 0 0  a b a b a b a b a b a b a b a b ...  1 3 1 2 1 1 1 0 0 3 0 2 0 1 0 0   ......   ......    Since   qk−1 k k xq n+i b xn a xq = a b , 0 ≤ i < qk,  n  n i n=0 h i X   then qkn + i, qkm + j -th element of the matrix (a (x) , b (x) |q,k) is equal to

k q n + i n i k a − b − = a − b − , 0 ≤ i, j < q . n m i j qkm + j n m m i j j 0,q 0,q 0,q Thus, (a (x) , b (x) |q,k) is the block matrix, (n, m)-th block of which is the matrix

n a − (b (x) |q) k , n m m q 0,q k where (b (x) |q)qk is the matrix consisting of q ﬁrst rows of the matrix (b (x) |q). Hence,

(a (x) , b (x) |q,k)(c (x) ,d (x) |q,k)=(a (x) ◦ c (x) , b (x) ◦ d (x) |q,k) .

If (a (x) |q)=(a (x) , a (x) |q, 1) , as in the case of the matrix [0,q]P , then (a (x) |q)=(a (x) , a (x) |q,k) for all k =1, 2,... :

q−1 qk−1 ∞ qk−1 k mk a (x)= a xn a (xq)= a xn a xq = a xnq , n  n   n  n=0 ! n=0 m=0 n=0 X X Y X     k a0 =1, aqkn+i = anai, 0 ≤ i < q ,

((a (x) |q))qkn+i,qkm+j = ((a (x) |q))n,m((a (x) |q))i,j. (5) Hence ∞ ∞ i an = ani , n = niq = n0 + q (n1 + q (n2 + ...)) , 0 ≤ ni < q. i=0 i=0 Y X In view of the identity (4),

∞ ∞ ∞ n i i a − = a − , n = n q , m = m q , n m m ni mi i i 0,q i=0 i=0 i=0 Y X X n − if ni ≥ mi for all i, and an m m 0,q =0 in other case. I.e. ∞

((a (x) |q))n,m = ((a (x) |q))ni,mi . (6) i=0 Y 12 From (5), (6) we see that rows of the matrix (a (x) |q), denote them un (x), form the recurrent sequences: n m q un (x)= an−mx , 0 ≤ n < q; uqn+i (x)= un (x ) ui (x) , 0 ≤ i < q; m=0 X or ∞ ∞ qi i un (x)= uni x , n = niq , 0 ≤ ni < q. i=0 i=0 Y X If (a (x) |q)(b (x) |q)=(a (x) ◦ b (x) |q) , where q−1 q−1 n q n q a (x)= anx a (x ) , b (x)= bnx b (x ) , n=0 ! n=0 ! X X then ∞ ∞

n ni i [x ] a (x) ◦ b (x)= cni , cni = [x ] a (x) b (x) , n = niq , 0 ≤ ni < q. i=0 i=0 Y X For example, 1 000000000000000 ... 2 100000000000000 ...   2 010000000000000 ...  4 221000000000000 ...    2 000100000000000 ...    4 200210000000000 ...    4 020201000000000 ...    8 442422100000000 ... 2   2 1  2 000000010000000 ... [0,2]P = |2 =   , 1 − x  4 200000021000000 ...    4 020000020100000 ...    8 442000042210000 ...    4 000200020001000 ...    8 400420042002100 ...    8 040402040202010 ...   16884844284424221 ...    ......   ......     ∞ ∞  − − n 1 1 ni i [x ](1 − x) ◦ (1 − x) = 2 , n = ni2 , 0 ≤ ni < 2, i=0 i=0 Y ∞ X ni 2 2i [n, →] [0,2]P = 2+ x . i=0 Y Let q−1 n q c (x)= cnx c (x ) . n=0 ! X Then cqkn+icqk(n−m)+i−j cncicn−mci−j Pc(x) qkn+i,qkm+j = = = Pc(x) n,m Pc(x) i,j, cqkm+j cmcj k 0 ≤ i, j < q , i ≥ j. Hence, matrices Pc(x) × [0,q]P and (c (x) |q) have the same fractal properties. Applied to the matrix T (q): (q) −1 T = Pc(x) × [0,q]P, cn =(n!) , 0 ≤ n < q.

13 References

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