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Suppose a certain apportionment method gives two agents , some , seats respectively, and then these two agents form a coalition, and the method is re-activated. An apportionment method always encourages coalitions if a coalition of two parties receives at least a i + a j {\displaystyle a_{i}+a_{j}} seats (in other words, it is split-proof - a ...
In other words: every part of a fair allocation is fair too. This immediately follows from the min-max inequality. Moreover: Every apportionment method that is uniform, symmetric and balanced must be a rank-index method. [1]: Thm.8.3 Every apportionment method that is uniform, house-monotone and balanced must be a rank-index method. [2]
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).
Two sequences that differ in the order of their terms define different compositions of their sum, while they are considered to define the same integer partition of that number. Every integer has finitely many distinct compositions. Negative numbers do not have any compositions, but 0 has one composition, the empty sequence.
The frame rule states that the only two allocations that a party can receive should be either the lower or upper frame. [1] If at any time an allocation gives a party a greater or lesser number of seats than the upper or lower frame, that allocation (and by extension, the method used to allocate it) is said to be in violation of the quota rule.
The Alabama paradox is when an increase in the total number of seats leads to a decrease in the number of seats allocated to a certain party. In the example below, when the number of seats to be allocated is increased from 25 to 26, parties D and E end up with fewer seats, despite their entitlements increasing. With 25 seats, the results are:
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Two of the problems are trivial (the number of equivalence classes is 0 or 1), five problems have an answer in terms of a multiplicative formula of n and x, and the remaining five problems have an answer in terms of combinatorial functions (Stirling numbers and the partition function for a given number of parts).