... is repeated Define φ((λ, µ)) = (( 1 , 1 , λ2 , ), (µ2 , µ3, )), if 1 < 1 or 1 = 1 = 2s + 1, ((λ2 , λ3 , ), ( 1 , 1 , µ2 , )), if 1 > 1 or 1 = 1 = 2s It is straightforward to verify ... of (1. 2) may be written as 1 (az, q; qa; q, b) (1 − a) (1 .4) By the Heine’s transformation (III .1) in Gasper and Rahman [7, p 2 41 ], (1 .4) is equal to (q, abz; q)∞ 1 (a, b; abz; q, q), (1 − ... odd parts Theorem 2 .1 was established by Chapman [5] in his proof of the q-identity ∞ −q 2n 1 + q 2n n − q 2n n =1 n 1 j =1 ∞ − q 2j 1 − q 2j 1 = − q 2j − q 2j j =1 ∞ ( 1) d d =1 qd , − qd which is...
... of 10 counted by B (10 ), C (10 ) and D (10 ) Those counted by B (10 ) are {1, 1, 1, 1, 1, 1, 1, 1, 1, 1} , {4, 1, 1, 1, 1, 1, 1} , {4, 4, 1, 1} , {6, 1, 1, 1, 1} , {6, 4} , {9, 1} , those counted by C (10 ) ... sum ′ a1 ≥0 q ma1 m +1 k+a1 ) k =1 (1 + bq a2 ≥a1 q ma2 m +1 m+k+a2 ) k =1 (1 + bq ′ ··· an 1 ≥an−2 q man 1 m +1 (n−2)m+k+an 1 ) k =1 (1 + bq ′ an ≥an 1 q man (3.6) m +1 (1 + bq (n 1) m+k+an ) k =1 Next, ... combinatorics 18 (2 011 ), #P60 Proof We rewrite the left side of (3 .1) as the nested sum q ma1 m +1 k+a1 ) k =1 (1 + bq a2 ≥a1 a1 ≥0 q ma2 m +1 (m +1) +k+a2 ) k =1 (1 + bq q man 1 m +1 (n−2)(m +1) +k+an 1 ) k =1 (1...
... ReheapDown(child, lastPosition) return 10 Insert new element into min-heap Insert 14 : 14 14 14 The new element is put to the last position, and ReheapUp is called for 11 that position InsertHeap ... success 13 End InsertHeap Delete minimum element from min-heap 31 31 31 31 The element in the last position is put to the position of the root, and ReheapDown is called for that position 14 Delete ... complete binary tree • If the height is h, the number of nodes n is between 2h -1 and (2h -1) • Complete tree: n = 2h -1 when last level is full • Nearly complete: All nodes in the last level are...
... = K[x, y, z, t, T1 , T2 , T3 , T4 ]/J , where the ideal J is generated by z T1 + wT2 + xT3 , y T1 + xT2 + wT4 , xw2 T1 + 2 z T2 + y T3 , x2 wT1 + y T2 + z T4 , xwT1 + T2 + T3 T4 From this it ... Theorem 2 .1 The Rees ring R(I) is equal to K[x, T ]/(L1 , L2 , L3 , L4 , Q), i.e J = (L1 , L2 , L3 , L4 , Q) Denote by A the ideal (L1 , L2 , L3 , L4 , Q) Let us first remark that xs L1 − xt L2 ... Lsat = 111 −3 1 More precisely, the ideal ILsat is the definition ideal of the curve (s4 , s3 t, st3 , t4 ), and IL = (xz − y w, y z − x2 w2 , z − xw3 , y − x3 w) Applying Theorem 2 .1, we...