Školjka

%V0 Let $(a_n)^{\infty}_{n=1}$ be a sequence of integers with $a_{n} < a_{n+1}, \quad \forall n \geq 1.$ For all quadruple $(i,j,k,l)$ of indices such that $1 \leq i < j \leq k < l$ and $i + l = j + k$ we have the inequality $a_{i} + a_{l} > a_{j} + a_{k}.$ Determine the least possible value of $a_{2008}$.

%V0 neka je $n, n \in \mathbb{N}$, te $x_1, x_2, \dots , x_n, y_1, y_2, \dots, y_n \in \mathbb{C}$ postoji li $k, k \in \mathbb{N}$, takav da se iz jednakosti $$x_1 + x_2 + \dots + x_n = y_1 + y_2 + \dots + y_n$$ $$x_1^2 + x_2^2 + \dots + x_n^2 = y_1^2 + y_2^2 + \dots + y_n^2$$ $$ \vdots $$ $$x_1^k + x_2^k + \dots + x_n^k = y_1^k + y_2^k + \dots + y_n^k$$ moze zakljuciti da su "$x$-evi" do na permutaciju jednaki "$y$-ima". ako postoji, koji je najmanji takav $k$. dodatno, postoji li $l, l \in \mathbb{N}$ takav da se za proizvoljnih $l$ brojeva $a_1, a_2, \dots , a_l \in \mathbb{Z} \setminus \{0\}$, takvih da $\forall i \neq j$ vrijedi $M(a_i, a_j) = 1$ te odgovarajucih jednakosti $$x_1^{a_1} + x_2^{a_1} + \dots + x_n^{a_1} = y_1^{a_1} + y_2^{a_1} + \dots + y_n^{a_1}$$ $$x_1^{a_2} + x_2^{a_2} + \dots + x_n^{a_2} = y_1^{a_2} + y_2^{a_2} + \dots + y_n^{a_2}$$ $$ \vdots $$ $$x_1^{a_l} + x_2^{a_l} + \dots + x_n^{a_l} = y_1^{a_l} + y_2^{a_l} + \dots + y_n^{a_l}$$ moze zakljuciti da su "$x$-evi" do na permutaciju jednaki "$y$-ima". ako postoji, koji je najmanji takav $l$. takoder, uz pretpostavku $x_1, x_2, \dots , x_n, y_1, y_2, \dots, y_n \in \mathbb{R}^+$, smijemo li uzeti $a_1, a_2, \dots , a_l \in \mathbb{R} \setminus \{0\}$?

%V0 Let $f$ and $g$ be two integer-valued functions defined on the set of all integers such that (a) $f(m + f(f(n))) = -f(f(m+ 1) - n$ for all integers $m$ and $n;$ (b) $g$ is a polynomial function with integer coefficients and g(n) = $g(f(n))$ $\forall n \in \mathbb{Z}.$

%V0 A function $f$ defined on the positive integers (and taking positive integers values) is given by: $\begin{matrix} f(1) = 1, f(3) = 3 \\ f(2n) = f(n) \\ f(4n + 1) = 2f(2n + 1) - f(n) \\ f(4n + 3) = 3f(2n + 1) - 2f(n)\text{,} \end{matrix}$ for all positive integers $n.$ Determine with proof the number of positive integers $\leq 1988$ for which $f(n) = n.$

%V0 Let $f(n)$ be a function defined on the set of all positive integers and having its values in the same set. Suppose that $f(f(n) + f(m)) = m + n$ for all positive integers $n,m.$ Find the possible value for $f(1988).$

%V0 $(FRA 5)$ Let $\alpha(n)$ be the number of pairs $(x, y)$ of integers such that $x+y = n, 0 \le y \le x$, and let $\beta(n)$ be the number of triples $(x, y, z)$ such that$x + y + z = n$ and $0 \le z \le y \le x.$ Find a simple relation between $\alpha(n)$ and the integer part of the number $\frac{n+2}{2}$ and the relation among $\beta(n), \beta(n -3)$ and $\alpha(n).$ Then evaluate $\beta(n)$ as a function of the residue of $n$ modulo $6$. What can be said about $\beta(n)$ and $1+\frac{n(n+6)}{12}$? And what about $\frac{(n+3)^2}{6}$? Find the number of triples $(x, y, z)$ with the property $x+ y+ z \le n, 0 \le z \le y \le x$ as a function of the residue of $n$ modulo $6.$What can be said about the relation between this number and the number $\frac{(n+6)(2n^2+9n+12)}{72}$?