Báo cáo toán học: "The Number of Solutions of X 2 = 0 in Triangular Matrices Over GF" pdf

... The Number of Solutions of X 2 =0 in Triangular Matrices Over GF(q) Shalosh B. EKHAD 1 and Doron ZEILBERGER 1 Abstract: We prove an explicit formula for the number of n × n upper triangular matrices, ... matrix, is given by the polynomial C n (q), where, C 2n (q )=  j  2n n − 3j  −  2n n − 3j − 1  · q n 2 −3j 2 −j , C 2n+1 (q )=  j  2n +1 n − 3...

Ngày tải lên: 07/08/2014, 06:20

... 468 30 8 400 0 20 1 60 10 26 37 1 329 51 835 800 997 9 20 1814 40 11 1 125 21 08 61 391 524 0 14757 1 20 127 00 800 12 195 20 5 905 105 49168 103 6 929 60 26 4 600 000 13 123 795 1 80 923 68 423 9184 32 2 623 622 400 14 409 50 20 545 9 20 ... 11197 300 32 15189491 5 20 15 57 12 153375 60 20 1 403 0656 571784 3 20 80 16 71937 60 25 16883516 148486353165 17 1 922...

Ngày tải lên: 07/08/2014, 07:21

... journal of combinatorics 13 ( 20 06 ), #R65 12 q p N 1 N 2 N 3 Total 13 3 0 0 0 8 29 7 4 0 0 1 20 53 13 0 4 0 384 149 37 16 16 0 4896 173 43 20 20 0 6888 26 9 67 36 24 0 166 32 293 73 28 28 0 184 32 317 ... 166 32 293 73 28 28 0 184 32 317 79 12 32 0 1 90 32 389 97 48 36 0 345 60 509 127 40 40 0 519 12 557 139 56 32 0 623 76 653 163 52 7...

Ngày tải lên: 07/08/2014, 13:21

... M´exico, Circuito Exterior, C.U. Coyoac´an 04 5 10, M´exico D.F. merino@matem.unam.mx Submitted: Nov 21 , 20 07 ; Accepted: Jul 11, 20 08 ; Published: Jul 21 , 20 08 Mathematics Subject Classifications: 05 A19 Abstract If ... that  n 0 J n +2 (−1)( 2) n t n n! = 1 (cos(t) + sin(t)) 2 . (5) Taking x = 2 and y = −1 in Equation (3), we get the following identiti...

Ngày tải lên: 07/08/2014, 15:22

... become f [00 ] w = 1 + f [00 ] w (zf [00 ] w−1 + z 3 /2 f [01 ] w−1 + z 3 /2 f [ 10] w−1 + z 2 f [11] w−1 ) (16) f [01 ] w = f [00 ] w (z 1 /2 f [00 ] w−1 + zf [01 ] w−1 ) (17) f [ 10] w = f [00 ] w (z 1 /2 f [00 ] w−1 + ... zf [ 10] w−1 ) (18) f [11] w = f [00 ] w−1 + f [01 ] w (z 1 /2 f [00 ] w−1 + zf [01 ] w−1 ). (19) We clean up our notation now. Le...

Ngày tải lên: 07/08/2014, 15:22

... B 0 + B 2 G 0 = exp(C 0 ) C 1 = 0 G 1 = 0 C 2 = −R G 2 = exp(C 0 )C 2 C 3 = 0 G 3 = 0 C 4 = − 1 2  R + R 2 2B 4 − R  G 4 = exp(C 0 )  C 4 + C 2 2 2  C 5 = B 5  1 − 2B 4 R  − 5 2 G 5 = exp(C 0 )C 5 the ... 15374 406 6t 6 − 415 828 997t 5 − 522 964992t 4 −3 4 20 73344t 3 − 121 237 504 t 2 − 22 151168t − 1...

Ngày tải lên: 07/08/2014, 21:20

... Then  i d i = n. Draw a punctured polygon with n the electronic journal of combinatorics 16 ( 20 09 ), #R49 14 n d(n) 3 4 4 6 5 26 6 80 7 24 6 n d(n) 8 8 10 9 27 04 10 925 2 11 3 20 66 12 1 127 20 Table 1: ... ICRA, Torun ( 20 07 ). [V] Vatne D. F. The mutation class of D n quivers, arXiv :08 10. 4789v1 ( 20 08 ). the electronic journal of combinatorics 16 ( 20 09...

Ngày tải lên: 07/08/2014, 21:21

... 01 001 100 0 100 11 101 100 0 100 1 10 0 100 1 100 0 100 11 101 100 101 1 100 01 001 100 0 100 11 101 100 101 1 101 01 001 100 0 100 11 101 100 11 100 10 0 100 11 101 100 0 100 1 101 000 1 100 01 001 1 101 100 0 100 1 101 000 1 101 28 01 001 100 0 100 11 101 100 0 100 1 100 01 001 100 0 100 11 101 100 0 100 1 101 01 001 100 0 100 11 101 100 101 1 100 1 01 001 1 101 100 0 100 1 101...

Ngày tải lên: 08/08/2014, 12:23

... Jun 28 , 20 10; Accepted: Jan 27 , 20 11; Published: Feb 14, 20 11 Mathematics Subject Classification: 05 C 30; 05 A16, 05 C05, 05 C69, 92E 10 Abstract For graphs F and G an F -matching in G is a subgraph of ... with integer coefficients g(r) =  d i=1 c i g(r − i). Assume that g (0) = 0 and c d = 1. Then for every positive integer m > 0 there exists an index r...

Ngày tải lên: 08/08/2014, 12:23

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Ngày tải lên: 05/08/2014, 10:20