An integer partition, is a way of writing n as a sum of positive integers. Two sums that differ only in the order of their summands are considered the same partition. [quote]For example, 4 can be partitioned in five distinct ways: 4 3 + 1 2 + 2 2 + 1 + 1 1 + 1 + 1 + 1[/quote] The number of partitions of n is given by the partition function $p\left ( n \right )$. So $p\left ( 4 \right ) = 5$ . Determine all the positive integers so that $p\left ( n \right )+p\left ( n+4 \right )=p\left ( n+2 \right )+p\left ( n+3 \right )$.

For each positive integer $ n$, let $ f(n)$ denote the number of ways of representing $ n$ as a sum of powers of 2 with nonnegative integer exponents. Representations which differ only in the ordering of their summands are considered to be the same. For instance, $ f(4) \equal{} 4$, because the number 4 can be represented in the following four ways: 4; 2+2; 2+1+1; 1+1+1+1. Prove that, for any integer $ n \geq 3$ we have $ 2^{\frac {n^2}{4}} < f(2^n) < 2^{\frac {n^2}2}$.

An integer partition, is a way of writing n as a sum of positive integers. Two sums that differ only in the order of their summands are considered the same partition. [quote]For example, 4 can be partitioned in five distinct ways: 4 3 + 1 2 + 2 2 + 1 + 1 1 + 1 + 1 + 1[/quote] The number of partitions of n is given by the partition function $p\left ( n \right )$. So $p\left ( 4 \right ) = 5$ . Determine all the positive integers so that $p\left ( n \right )+p\left ( n+4 \right )=p\left ( n+2 \right )+p\left ( n+3 \right )$.

For each positive integer $ n$, let $ f(n)$ denote the number of ways of representing $ n$ as a sum of powers of 2 with nonnegative integer exponents. Representations which differ only in the ordering of their summands are considered to be the same. For instance, $ f(4) \equal{} 4$, because the number 4 can be represented in the following four ways: 4; 2+2; 2+1+1; 1+1+1+1. Prove that, for any integer $ n \geq 3$ we have $ 2^{\frac {n^2}{4}} < f(2^n) < 2^{\frac {n^2}2}$.

Let $ A_n $ be the set of partitions of the sequence $ 1,2,..., n $ into several subsequences such that every two neighbouring terms of each subsequence have different parity,and $ B_n $ the set of partitions of the sequence $ 1,2,..., n $ into several subsequences such that all the terms of each subsequence have the same parity ( for example,the partition $ {(1,4,5,8),(2,3),(6,9),(7)} $ is an element of $ A_9 $,and the partition $ {(1,3,5),(2,4),(6)} $ is an element of $ B_6 $ ). Prove that for every positive integer $ n $ the sets $ A_n $ and $ B_{n+1} $ contain the same number of elements.

For a positive integer $n$, a [i]sum-friendly odd partition[/i] of $n$ is a sequence $(a_1, a_2, \ldots, a_k)$ of odd positive integers with $a_1 \le a_2 \le \cdots \le a_k$ and $a_1 + a_2 + \cdots + a_k = n$ such that for all positive integers $m \le n$, $m$ can be [b]uniquely[/b] written as a subsum $m = a_{i_1} + a_{i_2} + \cdots + a_{i_r}$. (Two subsums $a_{i_1} + a_{i_2} + \cdots + a_{i_r}$ and $a_{j_1} + a_{j_2} + \cdots + a_{j_s}$ with $i_1 < i_2 < \cdots < i_r$ and $j_1 < j_2 < \cdots < j_s$ are considered the same if $r = s$ and $a_{i_l} = a_{j_l}$ for $1 \le l \le r$.) For example, $(1, 1, 3, 3)$ is a sum-friendly odd partition of $8$. Find the number of sum-friendly odd partitions of $9999$.

For a positive integer $n$, a [i]sum-friendly odd partition[/i] of $n$ is a sequence $(a_1, a_2, \ldots, a_k)$ of odd positive integers with $a_1 \le a_2 \le \cdots \le a_k$ and $a_1 + a_2 + \cdots + a_k = n$ such that for all positive integers $m \le n$, $m$ can be [b]uniquely[/b] written as a subsum $m = a_{i_1} + a_{i_2} + \cdots + a_{i_r}$. (Two subsums $a_{i_1} + a_{i_2} + \cdots + a_{i_r}$ and $a_{j_1} + a_{j_2} + \cdots + a_{j_s}$ with $i_1 < i_2 < \cdots < i_r$ and $j_1 < j_2 < \cdots < j_s$ are considered the same if $r = s$ and $a_{i_l} = a_{j_l}$ for $1 \le l \le r$.) For example, $(1, 1, 3, 3)$ is a sum-friendly odd partition of $8$. Find the number of sum-friendly odd partitions of $9999$.