Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\},$$ then we define the power series $F(x) = \sum_{n=0}^{\infty} P(n) x^n$. Prove the equality $F(x) = \prod_{i=1}^{k} \frac{1}{1 - x^{a_i}}$ for all $x \in ]-1, 1[$.
Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\},$$ then we define the power series $F(x) = \sum_{n=0}^{\infty} P(n) x^n$. Deduce that $P$ is a quasi-polynomial function.
Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\},$$ then we define the power series $F(x) = \sum_{n=0}^{\infty} P(n) x^n$. Calculate the leading coefficient of $P$.
Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\}.$$ We assume $k = 2$. We assume in this question that $(a_1, a_2) = (2, 3)$. Construct a function $\phi : \mathbb{Z} \rightarrow \mathbb{Z}$ of period 6 such that $P(n) = \frac{n + \phi(n)}{6}$ for all $n \in \mathbb{N}$.
Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\}.$$ We assume $k = 2$. We set $a = a_1, b = a_2, \omega_a = \exp(2\mathrm{i}\pi/a), \omega_b = \exp(2\mathrm{i}\pi/b)$. From a partial fraction decomposition of the fraction $\frac{1}{(1 - x^a)(1 - x^b)}$, show the formula $$P(n) = \frac{1}{2a} + \frac{1}{2b} + \frac{n}{ab} + \frac{1}{a} \sum_{j=1}^{a-1} \frac{\omega_a^{-jn}}{1 - \omega_a^{jb}} + \frac{1}{b} \sum_{k=1}^{b-1} \frac{\omega_b^{-kn}}{1 - \omega_b^{ka}}$$ for all integer $n \geq 0$.
Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\}.$$ We assume $k = 2$. Prove that $$P(n) = \frac{n}{ab} - \left\{\frac{b^* n}{a}\right\} - \left\{\frac{a^* n}{b}\right\} + 1$$ for all integer $n \geq 0$, where $a^*$ and $b^*$ are integers satisfying $a^* a = 1$ modulo $b$ and $b^* b = 1$ modulo $a$ respectively. Hint. One may use the formula $(*)$ for $b = 1$.
Let $n \in \mathbb { N } ^ { * }$. Justify that $\zeta ( 2 ) = \sum _ { k = 1 } ^ { + \infty } \frac { 1 } { k ^ { 2 } }$ is well-defined, then show that we can write $$\sum _ { k = 1 } ^ { n } \frac { 1 } { k ^ { 2 } } = \frac { p _ { n } } { q _ { n } }$$ with $p _ { n } \in \mathbb { N } ^ { * }$ and $q _ { n } = d _ { n } ^ { 2 }$.
113. We classify the natural numbers in groups such that each group contains consecutive numbers, i.e., $\ldots, \{4,5,6\}, \{2,3\}, \{1\}, \ldots$ What is the sum of the numbers in the group containing 350? $$4125 \quad (1) \qquad 4050 \quad (2) \qquad 4015 \quad (3) \qquad 3980 \quad (4)$$
For any $n \geq 5$, the value of $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n - 1}$ lies between (A) 0 and $\frac{n}{2}$ (B) $\frac{n}{2}$ and $n$ (C) $n$ and $2n$ (D) none of the above
What is the limit of $\sum _ { k = 1 } ^ { n } \frac { e ^ { - k / n } } { n }$ as $n$ tends to $\infty$ ? (A) The limit does not exist. (B) $\infty$ (C) $1 - e ^ { - 1 }$ (D) $e ^ { - 0.5 }$
Let $a _ { n }$ be the number of subsets of $\{ 1,2 , \ldots , n \}$ that do not contain any two consecutive numbers. Then (A) $a _ { n } = a _ { n - 1 } + a _ { n - 2 }$ (B) $a _ { n } = 2 a _ { n - 1 }$ (C) $a _ { n } = a _ { n - 1 } - a _ { n - 2 }$ (D) $a _ { n } = a _ { n - 1 } + 2 a _ { n - 2 }$.
Consider the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5 , \ldots$ obtained by writing one 1 , two 2's, three 3's and so on. What is the $2020 ^ { \text {th} }$ term in the sequence? (A) 62 (B) 63 (C) 64 (D) 65
Let $a _ { 1 } = b _ { 1 } = 1$ and $a _ { n } = a _ { n - 1 } + ( n - 1 ) , b _ { n } = b _ { n - 1 } + \mathrm { a } _ { n - 1 } , \forall n \geq 2$. If $\mathrm { S } = \sum _ { \mathrm { n } = 1 } ^ { 10 } \left( \frac { b _ { n } } { 2 ^ { n } } \right)$ and $\mathrm { T } = \sum _ { n = 1 } ^ { 8 } \frac { \mathrm { n } } { 2 ^ { n - 1 } }$ then $2 ^ { 7 } ( 2 S - T )$ is equal to $\_\_\_\_$