Non-Associativity of the Double Minus Operation MADISON MICKEY UNIVERSITY OF NEBRASKA AT KEARNEY PRESENTED: 3/14/2017 JOINT WORK WITH J. HUANG AND J. XU Double Minus Operation ο΅ Define binary operation: π Σ© b = βa β b ο΅ Let πΆΣ©,π,π be the number of distinct results with exactly r plus signs from π₯0 Σ© π₯1 Σ© β¦ Σ© π₯π ο΅ Let πΆΣ©,π = 0 β€π β€π+1 πΆΣ©,π,π for π β₯ 0 which is the total number of distinct results from the previous expression. ο΅ πΆΣ©,π measures the non-associativity of the operation Σ© Example (n = 3) β β βπ₯0 β π₯1 β π₯2 β π₯3 = βπ₯0 β π₯1 + π₯2 β π₯3 β βπ₯0 β π₯1 β βπ₯2 β π₯3 = π₯0 + π₯1 + π₯2 + π₯3 β βπ₯0 β βπ₯1 β π₯2 β π₯3 = π₯0 β π₯1 β π₯2 β π₯3 βπ₯0 β β βπ₯1 β π₯2 β π₯3 = βπ₯0 β π₯1 β π₯2 + π₯3 βπ₯0 β βπ₯1 β βπ₯2 β π₯3 = βπ₯0 + π₯1 β π₯2 β π₯3 CΣ©,3,0 = 0 CΣ©,3,1 = 4 = 4 1 CΣ©,3,2 = 0 CΣ©,3,3 = 0 CΣ©,3,4 = 1 = CΣ©,3 4 4 4 4 = + =5 1 4 --+++++ +-----+ -+-- Theorem (Huang, M., and Xu) For π β₯ 1 and 0 β€ π β€ π + 1: πΆΣ©,π,π = π+1 π π+1 β1 π 0 ππ π + π β‘ 1 πππ 3 πππ π β 2π β 2; ππ π + π β‘ 1 πππ 3 πππ π = 2π β 2; ππ π + π β’ 1 πππ 3 . Theorem (Huang, M., and Xu) For π β₯ 1 we have 2π+1 β1 πΆΣ©,π = 3 2π+1 β2 3 , ππ π ππ πππ , ππ π ππ ππ£ππ. Proof by Example ο΅ ο΅ 6 0 + 6 6 + 6 1 =? + 6 4 6 2 =? 6 5 =? π₯5 + 6 6 + Using the binomial expansion: 6 0 6 1 π₯1 + 6 2 π₯2 + 6 2 6 3 + ο΅ (1 + π₯)6 = ο΅ π₯ = 1: ο΅ π₯ = β1: ο΅ Add together and divide by 2: ο΅ ο΅ 6 3 6 0 6 1 + 6 0 + β + 6 1 + + 6 2 β 6 3 6 4 + 6 3 6 4 π₯3 + + 6 4 6 5 β 6 0 + + 6 5 π₯4 + π₯6 6 6 + 6 2 6 5 + 6 6 6 4 + 6 6 Result based on ±1 = 1 and 1 + (β1) = 0 (Second Roots of Unity) Used Third Roots of Unity to calculate binomial coefficient totals ο΅ 13 = 1, π3 = 1, (π2 )3 = 1 and 1 + π + π2 = 0 Examining Non-Associativity ο΅ Related to Sequence A000975 in the OEIS (On-line Encyclopedia of Integer Sequences) ο΅ This sequence is characterized by a recurrence relation. ο΅ This sequence also gives binary operations with alternating ones and zeros. ο΅ A(1) = 1 = 12 = 1 × 20 A(2) = 2 = 102 = 1 × 21 + 0 × 20 A(3) = 5 = 1012 = 1 × 22 + 0 × 21 + 1 × 20 A(4) = 10 = 10102 = 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20 A(5) = 21 = 101012 = 1 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 1 × 20 This sequence occurs in other interesting places. THANK YOU!
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