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Although we have by now a precise answer, I'd like to keep the summer mood of the question and play a little more with it by an elementary arithmetic approach.

The solution I wish to sell is good for any $n$, and (just in case your favorite classic radio station entrusts you with the organization of the schedule of the year, which I think would be a finest choice) includes the issues: what piece $f(t)$ will be played at time $t$, and conversely, at what time $h(x,y)$ the piece $x$ is to be played and followed by piece $y$.

Consider the functions $(f,g):\mathbb{N}\to\mathbb N\times\mathbb N$ and $h:\mathbb N\times\mathbb{N}\to\mathbb N$ defind by $$ f(t):=\cases{ \frac{t-\lfloor\sqrt t\rfloor^2}2 &if $\; t-\lfloor\sqrt t\rfloor^2$ is even\\ \\ \lfloor\sqrt t\rfloor &if $ \;t-\lfloor\sqrt t\rfloor^2$ is odd}\qquad\qquad g(t):=f(t+1),$$ and $$h(x,y):=\cases{x^2+2x&if $\;y=0 ,$\\\\y^2+2x&if $\;0\le x <y,$\\\\ x^2+2y-1&if $\;0< y \le x $.}$$

Needless to say, $(f,g)$ and $h$ are bijective, in fact inverses of each other, which is elementary (and quite exiting) to check. Moreover (keeping your ordinal notation), they subordinate a bijection $n^2\to n\times n$ for any $n\in\mathbb N$: in other words, $n^2\ni t\mapsto (f(t),f(t+1))\in n\times n$ is injective, which exactly means "$f$ is radio-playing", and realizes the maximum cardinality $A_n$ wrto $n$ (there can not be an injective function $m\to n\times n$ for $m>n^2$, as you observed).

Although we have by now a precise answer, I'd like to keep the summer mood of the question and play a little more with it by an elementary arithmetic approach.

The solution I wish to sell is good for any $n$, and (just in case your favorite classic radio station entrusts you with the organization of the schedule of the year, which I think would be a finest choice) includes the issues: what piece $f(t)$ will be played at time $t$, and conversely, at what time $h(x,y)$ the piece $x$ is to be played and followed by piece $y$.

Consider the functions $(f,g):\mathbb{N}\to\mathbb N\times\mathbb N$ and $h:\mathbb N\times\mathbb{N}\to\mathbb N$ defind by $$ f(t):=\cases{ \frac{t-\lfloor\sqrt t\rfloor^2}2 &if $\; t-\lfloor\sqrt t\rfloor^2$ is even\\ \\ \lfloor\sqrt t\rfloor &if $ \;t-\lfloor\sqrt t\rfloor^2$ is odd}\qquad\qquad g(t):=f(t+1),$$ and $$h(x,y):=\cases{x^2+2x&if $\;y=0 ,$\\\\y^2+2x&if $\;0\le x <y,$\\\\ x^2+2y-1&if $\;0< y \le x $.}$$

Needless to say, $(f,g)$ and $h$ are bijective, in fact inverses of each other, which is elementary (and quite exciting) to check. Moreover (keeping your ordinal notation), they subordinate a bijection $n^2\to n\times n$ for any $n\in\mathbb N$: in other words, $n^2\ni t\mapsto (f(t),f(t+1))\in n\times n$ is injective, which exactly means "$f:n^2+1\to n$ is radio-playing", and realizes the maximum cardinality $A_n=n^2+1$ wrto $n$ (there can not be an injective function $m\to n\times n$ for $m>n^2$, as you observed).

Although we have by now a precise answer, I'd like to keep the summer mood of the question and play a little more with it by an elementary arithmetic approach.

The solution I wish to sell is good for any $n$, and (just in case your favorite classic radio station entrusts you with the organization of the schedule of the year, which I think would be a finest choice) includes the issues:

• what piece is to play at time $t$,
• when the piece $x$ is to be followed by piece $y$,
• when the piece $y$ will follow piece $x$.

Consider the functions $(f,g):\mathbb{N}\to\mathbb N\times\mathbb N$ and $h:\mathbb N\times\mathbb{N}\to\mathbb N$ defind by $$ f(t):=\cases{ \frac{t-\lfloor\sqrt t\rfloor^2}2 &if $\; t-\lfloor\sqrt t\rfloor^2$ is even\\ \\ \lfloor\sqrt t\rfloor &if $ \;t-\lfloor\sqrt t\rfloor^2$ is odd}\qquad\qquad g(t):=f(t+1),$$ and $$h(x,y):=\cases{x^2+2x&if $\;y=0 ,$\\\\y^2+2x&if $\;0\le x <y,$\\\\ x^2+2y-1&if $\;0< y \le x $.}$$

Needless to say, $(f,g)$ and $h$ are bijective, in fact inverses of each other, which is elementary (and quite exiting) to check. Moreover (keeping your ordinal notation), they subordinate a bijection $n^2\to n\times n$ for any $n\in\mathbb N$: in other words, $n^2\ni t\mapsto (f(t),f(t+1))\in n\times n$ is injective, which exactly means "$f$ is radio-playing", and realizes the maximum cardinality $A_n$ wrto $n$ (there can not be an injective function $m\to n\times n$ for $m>n^2$, as you observed).

Although we have by now a precise answer, I'd like to keep the summer mood of the question and play a little more with it by an elementary arithmetic approach.

The solution I wish to sell is good for any $n$, and (just in case your favorite classic radio station entrusts you with the organization of the schedule of the year, which I think would be a finest choice) includes the issues: what piece $f(t)$ will be played at time $t$, and conversely, at what time $h(x,y)$ the piece $x$ is to be played and followed by piece $y$.

Consider the functions $(f,g):\mathbb{N}\to\mathbb N\times\mathbb N$ and $h:\mathbb N\times\mathbb{N}\to\mathbb N$ defind by $$ f(t):=\cases{ \frac{t-\lfloor\sqrt t\rfloor^2}2 &if $\; t-\lfloor\sqrt t\rfloor^2$ is even\\ \\ \lfloor\sqrt t\rfloor &if $ \;t-\lfloor\sqrt t\rfloor^2$ is odd}\qquad\qquad g(t):=f(t+1),$$ and $$h(x,y):=\cases{x^2+2x&if $\;y=0 ,$\\\\y^2+2x&if $\;0\le x <y,$\\\\ x^2+2y-1&if $\;0< y \le x $.}$$

Needless to say, $(f,g)$ and $h$ are bijective, in fact inverses of each other, which is elementary (and quite exiting) to check. Moreover (keeping your ordinal notation), they subordinate a bijection $n^2\to n\times n$ for any $n\in\mathbb N$: in other words, $n^2\ni t\mapsto (f(t),f(t+1))\in n\times n$ is injective, which exactly means "$f$ is radio-playing", and realizes the maximum cardinality $A_n$ wrto $n$ (there can not be an injective function $m\to n\times n$ for $m>n^2$, as you observed).

Although we have by now a precise answer, I'd like to keep the summer mood of the question and play a little more with it by an elementary arithmetic approach.

The solution I wish to sell is good for any $n$, and (just in case your favorite classic radio station entrusts you with the organization of the schedule of the year, which I think would be a finest choice) includes the issues:

• what piece is to play at time $t$,
• when the piece $x$ is to be followed by piece $y$,
• when the piece $y$ will follow piece $x$.

Consider the functions $(f,g):\mathbb{N}\to\mathbb N\times\mathbb N$ and $h:\mathbb N\times\mathbb{N}\to\mathbb N$ defind by $$ f(t):=\cases{ \frac{t-\lfloor\sqrt t\rfloor^2}2 &if $\; t-\lfloor\sqrt t\rfloor^2$ is even\\ \\ \lfloor\sqrt t\rfloor &if $ \;t-\lfloor\sqrt t\rfloor^2$ is odd}\qquad\qquad g(t):=f(t+1),$$ and $$h(x,y):=\cases{x^2+2x&if $\;y=0 ,$\\\\y^2+2x&if $\;0\le x <y,$\\\\ x^2+2y-1&if $\;0< y \le x $.}$$

Needless to say, $(f,g)$ and $h$ are bijective, in fact inverses of each other, which is elementary (and quite exiting) to check. Moreover (keeping your ordinal notation), they subordinate a bijection $n^2\to n\times n$ for any $n\in\mathbb N$: in other words, $n^2\ni t\mapsto (f(t),f(t+1))\in n\times n$ is injective, which exactly means "radio-playing", and realizes the maximum cardinality $A_n$ wrto $n$ (there can not be an injective function $m\to n\times n$ for $m>n^2$, as you observed).

Although we have by now a precise answer, I'd like to keep the summer mood of the question and play a little more with it by an elementary arithmetic approach.

The solution I wish to sell is good for any $n$, and (just in case your favorite classic radio station entrusts you with the organization of the schedule of the year, which I think would be a finest choice) includes the issues:

• what piece is to play at time $t$,
• when the piece $x$ is to be followed by piece $y$,
• when the piece $y$ will follow piece $x$.

Consider the functions $(f,g):\mathbb{N}\to\mathbb N\times\mathbb N$ and $h:\mathbb N\times\mathbb{N}\to\mathbb N$ defind by $$ f(t):=\cases{ \frac{t-\lfloor\sqrt t\rfloor^2}2 &if $\; t-\lfloor\sqrt t\rfloor^2$ is even\\ \\ \lfloor\sqrt t\rfloor &if $ \;t-\lfloor\sqrt t\rfloor^2$ is odd}\qquad\qquad g(t):=f(t+1),$$ and $$h(x,y):=\cases{x^2+2x&if $\;y=0 ,$\\\\y^2+2x&if $\;0\le x <y,$\\\\ x^2+2y-1&if $\;0< y \le x $.}$$

Needless to say, $(f,g)$ and $h$ are bijective, in fact inverses of each other, which is elementary (and quite exiting) to check. Moreover (keeping your ordinal notation), they subordinate a bijection $n^2\to n\times n$ for any $n\in\mathbb N$: in other words, $n^2\ni t\mapsto (f(t),f(t+1))\in n\times n$ is injective, which exactly means "$f$ is radio-playing", and realizes the maximum cardinality $A_n$ wrto $n$ (there can not be an injective function $m\to n\times n$ for $m>n^2$, as you observed).