... by 2n, and then replace l by l + n, and finally use the binomial theorem Similarly for A2 n +1 (q) References [K] A. A Kirillov, Onthenumber of solutions to the equation X = in triangular matrices ... odd and r = (n − 1) /2, in which case it is a monomial in q times (1 w) (1+ w) , wr +1 and applying CTw kills it all the same, thanks to the symmetry of the Chu-Pascal triangle Summing the expression ... qEKHAD The input files inZ0, inZ1, inZ2 as well as the corresponding output files, outZ0, outZ1, outZ2 can be obtained by anonymous ftp to ftp.math.temple.edu, directory pub/ekhad/sasha The package...
... history, and I was hooked (Today, after the wild bull market of the late 19 90s and the brutal bear market that began in early 20 00, The Intelligent Investor reads more prophetically than ever.) Graham ... first draft of which was finished in January 19 71 At that time the DJIA was in a strong recovery from its 19 70 low of 6 32 and was advancing toward a 19 71 high of 9 51, with attendant general optimism ... its managers into raising the dividend, and came away with $11 0 per share three years later.) Despite a harrowing loss of nearly 70% during the Great Crash of 19 29 19 32, Graham survived and thrived...
... 19 92 the electronic journal of combinatorics 15 (20 08), #N28 [3] Donaghey, R.: Alternating permutations and binary increasing trees J Combinatorial Theory Ser A, 18 , 14 1 14 8, 19 75 [4] Foata, ... Thus, thenumber of 012 increasing trees on n vertices corresponds two different evaluations of the Tutte polynomial of a complete graph, that is Tn 1 (2, 1) and Tn +1 (1, 1) A similar situation ... r´arrangements et nombres d’Euler C R Acad Sci Paris e Sr A- B, 27 5, A 114 7 A 115 0, 19 72 [5] Foata, D and Strehl, V.: Rearrangements of the symmetric group and enumerative properties of the tangent...
... too far into the cusp Let a1 ≤ a2 ≤ · · · ≤ an 1 be the successive minima (in the sense of √ Minkowski) of OL Then one has automatically a1 a2 an 1 DL ; however, on account of the assumption ... de Jong and N M Katz, personal communication [10 ] A Granville, ABC allows us to count squarefrees, Internat Math Research Notices 19 (19 98), 9 91 10 09 [11 ] A Kable and A Yukie, Onthenumber of ... that K has no proper subfield, one further has for ≤ j ≤ n − that a1 aj aj +1 (indeed, were this not so, the lattice spanned by a1 , a2 , , aj would be stable under multiplication by a1 , and...
... Medical Journal of Australia 19 99, 17 0:59- 62 Page of (page number not for citation purposes) Journal of Orthopaedic Surgery and Research 20 06, 1: 3 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 ... (3 10 ) (3 10 ) Intra-capsular fracture 29 21 Hemiartroplasty 28 18 80 – 10 0 % 36 35 Osteosynthesis with 60 – 79 % 13 10 < 60 % 11 Mean 84 82 SD (16 .5) (23 .1) 0 .1 Two parallel nails Extra capsular ... implementation of a clinical pathway programme in an acute care general hospital In Singapore International Journal for Quality in Health Care 20 00, 5 ( 12 ):403- 4 12 Choong PFM, Langford AK, Dowsey MM, Santamaoa...
... equal M − 1, among the first 2 + integers 2 i + qi − 1, i = 1, 2, , 2 + 1, only 2 can be less or equal 2 Hence there must be an i1 ≤ 2 + with 2 i1 + qi1 − ≥ 2 + Continuing in the same manner ... (x + 1) n n −n, + x − n, + x ; 1 + ; 1 F1 −n − x 1 n−x n+x F1 To the F1 -series we apply the quadratic transformation (see [11 , (3 .2) ]) F1 a 4z , +a a, b a ; z = (1 + z) F1 2 ; 1+ a b + a − ... n) Also here, we write x + m for y everywhere A combination of (3 .11 ) and (3 .25 ) yields (m+n) /2 − m /2 P1 (x; m, n) = 2n 2 i =1 (2x + m /2 + 2i) (x + m /2 + 1) 2n 2 n 1 × i =1 m /2 (x + m + 1) 2n−2...
... gcd (A) , d (A) = d (A − A) , Λ (A) = maxA − minA, Λ (A) = Λ (A) /d (A) Furthermore, let κ (A) = Λ (A) − , |A| − θ (A) = max (A) , Λ (A) T (A) = ( |A| − 2) ( κ (A) + − κ (A) ) + and hA = {a1 + · · · + ah : a1 ... the electronic journal of combinatorics (20 00) , #R30 o odd numbers is (2, 1) -sum-free we have SF n ≥ (n +1) /2 In fact Erd˝s and 2 ,1 n n /2 Cameron [6] conjectured SF 2 ,1 = O (2 ) This conjecture ... proved that in this case SF n +1, = (1 + o (1) )2 (n +1) /2 The case of k being much larger than was treated by Calkin and Taylor [4] They showed that for some constant ck thenumber of (k, 1) -sum-free...
... to contain all k-permutations with substantial probability, has t(k) ∼ (k /2) 2 References [1] Alon, N., and Friedgut, E (19 99) Onthenumber of permutations avoiding a given pattern J Combinatorial ... Combinatorial Theory, Ser A, to appear [2] B´na, M (19 97) Exact and asymptotic enumeration of permutations with subsequence condio tions Ph.D Thesis, M.I.T [3] B´na, M (19 99) The solution of a conjecture ... Consider the concatenation τ of τ with τ as a permutation, τ ∈ Sm+n Clearly, τ avoids σ, establishing (2) [In detail, suppose to the contrary that τ contains σ, say at the k-tuple of positions given...
... + 11 t 1 + + 19 t 1 + 7t−7 + 21 t i=58 ( 12 6t + 11 i + 893 i =21 t 14 81 (i − 1) + 11 t 1 13t 1 13t 1 15t 1 + + + + + 15 t 1 + 17 t 1 + 17 t 12 + 19 t 1 + 19 t 1 + 21 t 1 ) + 22 t+64 (i − 1) i =22 t−798 2 + 23 t +26 7 ... xi xx1 x2 xi,8 i,8 i,8 (19 t 1) /2 (19 1t+3 01) /2+ 9i xi xx1 x2 xi,9 i,9 i,9 (21 t 1) /2 ( 21 1 t+305) /2 +10 i xi xx1 x2 xi ,10 i ,10 i ,10 (t−5 71) /2 (25 1 t +23 57) /2 +11 i xi xx1 x2 xi ,11 i ,11 i ,11 From the construction, ... + 17 t +1 + 17 t +1222 t−7 42 + 19 t +1 + 19 t +1 + 21 t +1 + t−5 71+ 2 ) + i= 7t +1 (19 t + 2i + 2) 22 + + + + = t−7 42 i= 7t +1 19t+ 7t +1 i =1 7t−7 + 21 t i=58 ( 12 6t + 11 i + 894 i =21 t 14 81 i + + 11 t +1 + 13 t+1...
... C4 contains v ∗ as a vertex, we must have that every w ∈ A appears in at least one of A or A+ (it may appear in both; in particular, v ∗ appears in both A and A+ as there are no antiparallel ... Equality (1) follows from the fact that M3 contains ρ (1 − β)mn cells The equalities in (2) follow from the fact that M1 does not contain 1 entries, has ρβmn cells, and the fact that M1 ∪ M3 has ... ρ (1 − β) the electronic journal of combinatorics (20 01) , #N6 (1) (3) u1 v1 urn urn +1 u2rn un -1 -1 M1 vbm vbm +1 M5 no -1 here M3 M2 M4 no +1 here vm Figure 1: The adjacency matrix M and...
... 3, A ∪ B = V and |A ∩ B| = Let {v} = A ∩ B, and let HA = P1 vP2 be a hamiltonian path of T [A] , and HB = Q1 vQ2 a hamiltonian path of T [B] We apply Lemma twice, and obtain paths R1 and R2 such ... reversing the arc → v1 gives the transitive tournament of order n As noted by Moon [1] , there is a bijection between partitions of V \ {v1 , } and hamiltonian paths that include the arc → v1 , and there ... tonian cycles, with β = ≈ 1. 710 References [1] J W Moon, The Minimum number of spanning paths in a strong tournament, Publ Math Debrecen 19 (19 72) ,10 1 -10 4 [2] C Thomassen, Onthenumber of Hamiltonian...
... = a Æn +1 |a| =m m H (a0 , , an )q a1 +···+nan athe electronic journal of combinatorics 13 (20 06), #N8 (5) Proof of Theorem We are free to assume n ≤ m We define the function E of the variables ... Applications, Volume 4 12 , Issue 1, January 20 06, Pages 30–38 [2] R H Brualdi and H J Ryser Combinatorial matrix theory Cambridge University Press, 19 91 [3] J H van Lint and R M Wilson A course ... because π is a polynomial in q with constant term Theorem is proved References [1] J Benton, R Snow, and N Wallach A combinatorial problem associated with nonograms, Linear Algebra and its Applications,...
... Th A 53 (19 90) 18 3 -20 8 [ 12 ] Daurat, A. , Nivat, M.: Salient and reentrant points of discrete sets, Disc Appl Math 15 1 (20 05) 10 6 - 12 1the electronic journal of combinatorics 14 (20 07), #R57 16 [13 ] ... of the tree is (1) b , which is the label of the one-cell polyomino the electronic journal of combinatorics 14 (20 07), #R57 11 (1) b (1) (2) b (1) r r (2) b (1) r (1) g (2) r (1) r (1) g (2) r (1) r (1) r ... thenumber of convex polygons onthe square and honeycomb lattices, J Phys A: Math Gen 21 (19 88) 26 35 -26 42 [11 ] Conway, J.H., Lagarias, J.C.: Tiling with polyominoes and combinatorial group theory,...