vicentev wrote: Consider a function \( n \mapsto f(n) \) that satisfies the following conditions: \( f(n) \) is an integer for each \( n \). \( f(0) = 1 \). \( f(n+1) > f(n) + f(n-1) + \cdots + f(0) \) for each \( n = 0, 1, 2, \dots \). Determine the smallest possible value of \( f(2023) \). Is $n \mapsto f(n)$ same as $f : \mathbb N \mapsto \mathbb N$?

so, $f(0+1) > f(0)$ or, $f(1) > f(0)$, taking the minimum value of $f(1)$ will be $f(1)=2$ if $f$ is a function such that $N \mapsto N$ plugging in $n=1$ gives, $f(2) > f(1)+f(0)$ or, $f(2) > 3$ so taking the minimum value of $f(2)$ will be $f(2)=4$. from where we can easily see a pattern for the minimum values, that for $f(n)$ we can define it as $f(n)=1+2+....+n+(n+1)$ for $n \geq 1$, where the minimum value of $f(n)$ is taken. So by this, the minimum value of $f(2023)$ should be $1+2+...+2023+2024$ which is $2047276$. Note: my approach might be wrong so feel free to correct me if I made a mistake anywhere. Also, I re.commend using induction in proving that $f(n)_{min} = 1+2+3+...+n+(n+1) \iff f:\mathbb N \mapsto \mathbb N$.

gaussiemann144 wrote: vicentev wrote: Consider a function \( n \mapsto f(n) \) that satisfies the following conditions: \( f(n) \) is an integer for each \( n \). \( f(0) = 1 \). \( f(n+1) > f(n) + f(n-1) + \cdots + f(0) \) for each \( n = 0, 1, 2, \dots \). Determine the smallest possible value of \( f(2023) \). Is $n \mapsto f(n)$ same as $f : \mathbb N \mapsto \mathbb N$? yep is the same