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Pennsylvania State University Math Club Meeting Monday, January 23, 2006 6:00pm 106 McAllister 6:00  Meeting begins Recap from last meeting: Mike had a brainstorm to step up the math lectures Sellers will be presenting arithmetic properties of hyper mary partitions Elections for a new secretary (Spring semester)  Sandi and Jenn Bowling Lauren  $10 for unlimited pizza and bowling Snowtubing Monday  2 for 1 Tuesday  $10.50/person Thursday $9.80/person Friday (710) $10/person James Sellers  has won 2 awards for faculty of the year, one at cedarville university (scholar) and distinguished undergraduate teaching from penn state math 54 journal articles, 90+ presentations, referee for 16 different journals 6:09 – Meeting ends, Lecture begins Research done two years ago with Kevin Cortright(?)  published two papers together, one as he graduated, one after he left Husband of one wife, and the father of five kids Director of undergraduate mathematics at PSU Partitions  We'll talk about a certain family of partitions  hyperbinary Definition with an example: n = 5 All the ways positive integers can be summed to get to 5 – 7 ways 5 4+1 3+2 3+1+1 2+2+1 2+1+1+1 1+1+1+1+1 Thus, p(5) = 7 p(n) can be calculated easily n p(n) 1 1 2 2 3 3 4 5 5 7 6 11 7 15 8 22 9 30 10 42 And so on… First studied by Leonhard Euler 17071783 Leonhard Euler  arguably greatest mathematician of the century Wrote about 500 mathematical papers Made contributions to number theory, geometry, calculus, and graph theory Also had 13 kids  only 5 of them survived adolescence Contributing to the study of partitions  the study of pentagonal theorem p(n) has a cool reoccurrence: p(n) = p(n1) + p(n2)  p(n5)  p(n7) + p(n12) + p(n15) 1, 2, 5, 7, 12, and 15 are all numbers of the form (k(3k+/1))/2) where k is a positive integer It goes , , +, +, and so on This does stop sooner or later (k(3k+/1))/2) are the pentagonal numbers Euler: The Master of Us All by William Dunham 2030 page history on Euler, and then 89 topical chapters Srinivasa Ramanujan  18871920  phenomenal mathematician Came up with a formula for p(n) and recognized some amazing results with p(n) Realized that every fourth value in a group of (5 p(4), p(9), etc.) is a multiple of 5 5p(5n+4)  mathematical formula stating the above p(7n + 5) is p(5), p(12), etc… all multiples of 7 7p(7n+5) Could there be a pattern? 11p(11n+6) This began a study of what is now known as modular forms This is the impetus of creating partitions for other numbers And now for the main topic… The function of the night! Let h2(n) = number of hyperbinary partitions of n = numbers of partitions of n into powers of 2 wherein each part can appear at most twice Partitions of 5 that are hyperbinary… 5 X 4+1 3+1+1 X 2+2+1 2+1+1+1 X 1+1+1+1+1 X So… h2(5) = 2 Bruce Resnick  studied various partition functions where he only wanted to use powers of 2 In the MAA monthly in 2000, Recounting the Rationals by Wilf and Caulkin Proved a result about the positive rational numbers, that they are a countable set When they did their work, they never thought of it as partitions  Kevin and James picked it up and studied it as its generating function: n h2(n) 1 1 2 2 3 1 4 3 5 2 6 3 7 1 8 4 9 3 10 5 And so on… h2(2n+1) = h2(2n)  h2(2n1) There seems to be symmetry around 2's Number of values around 2 seems closely related to powers of 2 h2(2n+1) = h2(n) h2(2^n  1) = 1 h2(2n) = h2(n) + h2(n1) Are there any cool results for an arithmetic progression? h2((2^(n1))(2^(m)1) = m(n1)+1 Theorem: hs(3n+2) is always even. 2h2(3n+2) Corollary: the ONLY even values of h2 occur at 3n+2 Proof: by contradiction Assume, to contradict (*), that N is the smallest positive integer such that h2(3N+2) is odd. Case 1: Let N=2J+1 Then h2(3N+2) =h2(3(2J+1)+2) =h2(6J+5)  this is odd =h2((6J+4)/2) =h2(3J+2) This is our contradiction, because J is smaller than N Case 2: N=4J Then h2(2N+2) =h2(3(4J)+2) =h2(12J+2) =h2(6J+1)+h2(6J)  first is odd, second is even =h2(3J)+h2(3J)+h2(3J1) =2h2(3J)+h2(3J1)  the second must be odd, this second part is…: h2(3(J1+2), and J1 is smaller than N Case 3: Let N = 4J+2 Then h2(3N+2) =h2(12J+8) =h2(6J+4) + h2(6J+3) =h2(3J+2)+h2(3J+1)+h2(3J+1) =h2(3J+2)+2h2(3J+1)  h2(3J+2) has to be odd: h2(3J+2), and J is smaller than N Kevin and James defined a new function hm(n) = numbers of partitions of n into powers of m used at most m times. Theorem: For m≥3, hm(m^(j)n + m^(j1)k) = j*hm(n) Pick an m≥3, pick a j≥1, and pick a m1≥k≥2 7:19 – Lecture ends 