[seqfan] Re: Xmas-challenge

[Andrew Weimholt]

here one of length 72 - probably not the max, so no attempt at a proof

52, 26, 78, 39, 13, 65,  5, 55, 11, 99,

33, 66, 22, 44, 88,  8, 64, 32, 16, 48,

96, 24, 72, 36, 18, 90, 30, 60, 20, 80,

40, 10, 50, 25, 75, 15, 45,  9, 81, 27,

54,  6, 12, 84, 28, 56, 14, 98, 49,  7,

63, 21, 42,  3, 57, 19, 38, 76,  4, 68,

34, 17, 51,  1, 69, 23, 46, 92,  2, 62,

31, 93,

On Sat, Dec 26, 2020 at 7:46 AM Peter Luschny <peter.luschny at gmail.com>
wrote:

> The lengths from respondents I know so far are 42, 55, and 63.
>
> If you found the max length prove that no longer one is possible!
>
> Then swap 99 for n and start building the sequence:
>
> a(n) is the length of the longest sequence of distinct numbers
> between 1 and n such that if k immediately follows m, then
> either k divides m or m divides k.
>
> However, if this sequence is already in the OEIS, don't reveal it (this
> year).
>
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>