554 PI MU EPSILON JOURNAL In this case, ng 2 = 2. If we begin by taking the frrst vector from r 1 we have II
Partitions of n into Prime Parts Michael Last (Student) Davis .. S'enior High School The nwnber of partitions of a number n is the number of \vays that n can be divided into unordered parts that are positive integers. For example, 5 = 3 +I + l is the same as 5 = I + 3 + 1. and other partitions of 5 include 5, 4 + I , 3 + 2, 2 + 1 + I + I, 2 + 2 + L and I + I+ I + l + I. To make things more interesting, one can look at specific types of partitions, such as those where all parts are odd, or all parts are distinct. One of the main things mathematicians investigate about partitions is the attempt to define for which types of partitions the number of partitions of n are equal for both types of partitions for all n. That is ~ they ask \vhether there exists a one-to-one correspondence between partitions of n from two different types of partitions (for example. the famous 18th century mathematician Euler proved a classical theorem that the number of partitions into odd parts is equal to the number of partitions into distinct parts for all n). Generating functions are a way of representing the number of partitions. Consider the infinite series :E aix i. When each ~ is the number of partitions of i, then the series is the generating function for the number of partitions of i. This paper investigates the consequences of restricting the partitions to prime parts, such as 5 = 5, and 5 = 3 + 2, but disregarding 5 = 2 + 2 + l, because I is not a prime. Four theorems are proven that demonstrate several properties of partitions of n into prime parts. At the end of the paper is a listing of all the prime partitions of the integers 2 through 11. Theorem 1 : The sequence of the number of partitions of n into prime parts is non-decreasing for all n. Proof: For every partition of n ~ we will find a unique partition of n + I. Write the parts in non-increasing order. Each partition of n has 2 as a part or it does not. If it does not have a 2, then add I to the last number, divide that number by 2, and replace the last number by that many 2's. For example, 22 = 7 + 7 + 5 + 3 ------ 23 = 7 + 7 + 5 + 2 + 2. <strong>No</strong>te that the new partition has at least t\vo 2's and if the new partition has a 3, then it has exactly two 2's. If the partition has one or two 2's, then change the last 2 into a 3. Since there is at most one 2 ~ this is a unique partition. If the 556 ~TITIO NS ofn into PRIME PARTS, LAST 557 non has exactly three 2's. then change two of the 2's into a 5. and keep the 2. Again. as it has only one 2, thus it is a unique partition. Finally. if there four or more 2's. then change a 2 into a 3. There are at least three 2!s ~ hence partition is again unique. A problem arises, however, when you consider the case where your partitions n mclude ... 5 + 3 + .. . + 3 + 2 + 2 and ... 3 + .. . + 3 + 2 + 2 + 2, as these will p to the same thing. However. this is solved by the follo·wing trick. In the c:ase \\here you have ... 3 + ... + 3 + 2 + 2 + 2, let the number of times 3 appears be called y. Either y is even or it is odd. If y is odd, then 3 + .. . + 3 + 2 + 2 + 2 1s odd and divisible by three. thus not a prime number. Add one to this ber, and then divide it into a bunch of 2's. For example, .. . + 3 + 3 + 3 + 3 T 3 + 2 + 2 + 2 would map to ... + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2. Smce if there is anything in the . . . (where . . . denotes the list of the numbers _ _,.--.......· n. g those written explicitly) of this partition of n + I it is at least 5, this as guaranteed not to be generated by one of our previous cases. In the case where the ... is empty, we still have a unique partition because the only numbers that will map to a partition of all 2's are prime, and 3 + 3 + 3 + 3 + 3 + 2 + 2 + 2 is not prime. lfy is even, theny < 5 or y > 5. Ify > 5, then map .. . + 3 + ... + 3 + 2 + 2 + 2 to ... + 7 + 5 + 2 + .. + 2. This is guaranteed to generate a unique partition of n + I because 5 will only appear followed by one or two 2's in the absence of3's. Ify < 6, theny = 2 or y = 4. Ify = 2, then map ... + 3 + 3 + 2 + 2 + 2 to ... + II + 2, which is not generated in any of the above cases. Likewise, when y = 4, map ... + 3 + 3 + 3 + 3 + 2 + 2 + 2 to ... + 17 + 2, which is not generated by any of the above cases. <strong>No</strong>te that every new partition has at least one 2 or one 3. Theorem 2: For n > 8, the sequence of the number of partitions of n into prime parts is strictly increasing. Proof: The partitions of 8 are 5 + 3, 3 + 3 + 2, and 2 + 2 + 2 + 2. The partitions of9 are 5 + 2 + 2, 3 + 3 + 3, 3 + 2 + 2 + 2, and 7 + 2. Hence, there are more partitions of9 then of8. For n > 9, we will find a partition of n that we have not observed from n - I by Theorem I. To do this, we will find partiti
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