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The values (), …, of the partition function (1, 2, 3, 5, 7, 11, 15, and 22) can be determined by counting the Young diagrams for the partitions of the numbers from 1 to 8. In number theory, the partition function p(n) represents the number of possible partitions of a non-negative integer n.
Such a partition is called a partition with distinct parts. If we count the partitions of 8 with distinct parts, we also obtain 6: 8; 7 + 1; 6 + 2; 5 + 3; 5 + 2 + 1; 4 + 3 + 1; This is a general property. For each positive number, the number of partitions with odd parts equals the number of partitions with distinct parts, denoted by q(n).
Their numbers can be arranged into a triangle, the triangle of partition numbers, in which the th row gives the partition numbers (), (), …, (): [1] k n
The Bell numbers themselves, on the left and right sides of the triangle, count the number of ways of partitioning a finite set into subsets, or equivalently the number of equivalence relations on the set. Sun & Wu (2011) provide the following combinatorial interpretation of each value in the triangle.
Greedy number partitioning – loops over the numbers, and puts each number in the set whose current sum is smallest. If the numbers are not sorted, then the runtime is O( n ) and the approximation ratio is at most 3/2 ("approximation ratio" means the larger sum in the algorithm output, divided by the larger sum in an optimal partition).
However, the coefficient of x 12 is −1 because there are seven ways to partition 12 into an even number of distinct parts, but there are eight ways to partition 12 into an odd number of distinct parts, and 7 − 8 = −1. This interpretation leads to a proof of the identity by canceling pairs of matched terms (involution method). [1]
In plain words, e.g., the first congruence means that If a number is 4 more than a multiple of 5, i.e. it is in the sequence 4, 9, 14, 19, 24, 29, . . . then the number of its partitions is a multiple of 5. Later other congruences of this type were discovered, for numbers and for Tau-functions.
If there is a remainder in solving a partition problem, the parts will end up with unequal sizes. For example, if 52 cards are dealt out to 5 players, then 3 of the players will receive 10 cards each, and 2 of the players will receive 11 cards each, since 52 5 = 10 + 2 5 {\textstyle {\frac {52}{5}}=10+{\frac {2}{5}}} .