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Problem 2089

National Olympiad second round; IMO P1/P4
Combinatorics Difficulty 7.5 Prove it Iranian Mathematical Olympiad · Iran

Let f1:NNf_1 : \mathbb{N} \to \mathbb{N} be a function. Function f2f_2 is constructed from f1f_1 such that for each kNk \in \mathbb{N}, f2(k)f_2(k) is the number of times kk has appeared in the range of f1f_1, i.e. f2(k)f_2(k) is the number of natural numbers nn for which f1(n)=kf_1(n) = k.
Suppose that the range of f1f_1 is such that f2f_2 can be defined over all natural numbers, so no number appears in the range of f1f_1 an infinite or zero number of times. Similarly, function f3f_3 can be constructed from function f2f_2 and consequently for each natural number ii, function fi+1f_{i+1} can be constructed from function fif_i.

a) Prove that for each natural number TT, there exists a function f1f_1 such that for each natural number ii, fif_i is definable and also the sequence f1,f2,f3,f_1, f_2, f_3, \dots is periodic with period TT, i.e. TT is the smallest number such that fT+1=f1f_{T+1} = f_1.

b) Does there exist a function f1f_1 such that for each natural number ii, fif_i is definable and these functions are mutually unequal?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

a) Consider a table with T+1T+1 rows and an infinite number of columns (towards right) such that the first column is filled by numbers 1,2,,T1, 2, \dots, T and 11. An algorithm is introduced for filling the cells of the table, then fi(j)f_i(j) is defined as the value of the intersection cell of the ithi^{th} row and the jthj^{th} column of the table.

1...
2...
3...
......
T1T-1...
TT...
1...

Label the cells of the table, except those of the first row, by a1,a2,a3,a_1, a_2, a_3, \dots, like the following table.
a1a_1a3a_3a6a_6a10a_{10}...
a2a_2a5a_5a9a_9...
a4a_4a8a_8...
a7a_7...
......

In each step of the algorithm, some cells of the table are filled in a way that after the jthj^{th} step aja_j is filled. Suppose that aja_j is at the intersection of the kthk^{th} row and the lthl^{th} column.

i. If aja_j was empty before the jthj^{th} step, the smallest natural number that has not yet appeared in the table is written in aja_j at that step.

ii. The number in aja_j should be equal to the number of repetitions of ll in the (k1)th(k-1)^{th} row of the table ()(*). For this reason, enough ll's should be put in the first empty cells of the (k1)th(k-1)^{th} row so that the number of ll's becomes equal to the number written in aja_j. Using this algorithm, if l>Tl > T, before this step no ll exists in the (k1)th(k-1)^{th} row; and if lTl \le T, at most one such ll exists (the one in the first column). The reason is that every number that has been written in the (k1)th(k-1)^{th} row before this step, except the first number, is written when the algorithm was used for the first l1l-1 cells of the row, and in those steps only numbers less than ll were written. Since the number in aja_j is positive, it was not less than the number of ll's in the (k1)th(k-1)^{th} row (before this step), so a sufficient number of ll's (maybe zero) can always be added to satisfy ()(*).

iii. If aja_j is in the second row, after the previous step the numbers in the first row should be copied into the cells of the last row.

Using this algorithm, all of the cells of the table are filled with natural numbers, and if fi(j)f_i(j) is defined to be the number in the jthj^{th} cell of the ithi^{th} row, fif_i's satisfy all of the problem's conditions.

a1a_1a3a_3a6a_6a10a_{10}...
a2a_2a5a_5a9a_9a14a_{14}...
a4a_4a8a_8a13a_{13}...
a7a_7a12a_{12}...
a11a_{11}...
\vdots\vdots\vdots\vdots\vdots

b) Yes! To construct such functions an approach similar to the previous part is used, utilizing a table which is infinite from right and below. Like part (a), jj is written in the jthj^{th} cell of the first column and the cells of the table, except those of the first row, are labeled by a1,a2,a_1, a_2, \dots like the following table.
1...
2...
3...
4...
5...
6...
\vdots\vdots\vdots\vdots\vdots

The table can be filled using the algorithm of part (a) (since no bottom row exists in this table, part iii of the algorithm is not necessary). Like before, it can be showed that aja_j is filled at the jthj^{th} step and if aja_j is the intersection of the kthk^{th} row and the lthl^{th} column, the number in aja_j is not less than the number of ll's in the (k1)th(k-1)^{th} row.

Again, if fi(j)f_i(j) is defined as the number in the intersection of the ithi^{th} row and the jthj^{th} column, fif_i's satisfy all of the problem's conditions; and since the numbers in the first column are all distinct, functions fif_i are mutually unequal.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.