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S Nov 11, 2023 at 9:04 history bounty ended CommunityBot
S Nov 11, 2023 at 9:04 history notice removed CommunityBot
Nov 10, 2023 at 10:24 comment added Alex Ravsky To ensure the needed multiplicities we have to start count these numbers from at least $\max\{a_1^1,\dots,a_n^1\}$.
Nov 10, 2023 at 10:13 answer added Alex Ravsky timeline score: 1
Nov 3, 2023 at 19:19 answer added Jean-Armand Moroni timeline score: 1
Nov 3, 2023 at 12:01 history edited user242318 CC BY-SA 4.0
Provided necessary condition
Nov 3, 2023 at 12:00 comment added user242318 The necessity of $\sum_i 1/N_i \leq 1$ is easy to see from counting the numbers used in the sequences $a_i$ up to $N_1 \cdot \dots \cdot N_n$.
Nov 3, 2023 at 11:09 answer added Command Master timeline score: 4
Nov 3, 2023 at 9:20 comment added Alex Ravsky @Jean-ArmandMoroni I expect I can easily prove that this condition is necessary. On the other hand, it is not sufficient. For instance, it is easy to show that for $(N_1,N_2,N_3)=(2,3,7)$ there are no required sequences.
S Nov 3, 2023 at 7:13 history bounty started user242318
S Nov 3, 2023 at 7:13 history notice added user242318 Canonical answer required
Oct 10, 2023 at 13:33 comment added Jean-Armand Moroni I would bet that the correct condition is: $\sum_i \frac 1 {N_i} \le 1$, but no proof in mind.
Oct 10, 2023 at 13:18 history edited user242318 CC BY-SA 4.0
Unclear formulation
Oct 10, 2023 at 13:16 comment added user242318 @Jean-ArmandMoroni : I mean the for each sequence $a_i=(a_i^1, a_i^2, \dots)$ we have $a_i^{j+1} - a_i^j \leq N_i$.
Oct 10, 2023 at 13:11 comment added Jean-Armand Moroni Could you please explain what you mean by "with consecutive differences bounded by $N_1, N_2, \dots, N_n$"?
Oct 10, 2023 at 12:45 history asked user242318 CC BY-SA 4.0