A Request Regarding A393845 and A394793

[Ali Sada]

Hi everyone,

Hope all is well. I have just proposed A393845 (a(n) is the smallest positive multiple of n whose decimal representation contains every distinct digit of n-1.) and A394793 (a(n) is the smallest positive multiple of n whose decimal representation contains every distinct digit of n+1.) Naturally, half of the terms are even numbers, and most of the other half are also even. I would really appreciate it if someone could calculate the percentage of odd numbers in each sequence for large n and compare it to the percentage of composite numbers < n. 

Best,

Ali   

[M. F. Hasler]

It is somewhat ambiguous to say that most composite numbers are even.

It is also true that exactly 50% of all composite numbers are even, 

and exactly 50% of all composite numbers are odd.

Among the numbers up to  N, only approximately  N/log N  numbers are prime, 

the others are composite. So, up to N we have

~ N/2 (- 1) even composite numbers,

~ N/log N  primes

~ N/2 - N/log N  odd composite numbers.

So it is true that the difference of  "even composites up to N" - "odd composites up to N"  

is equal to the number of primes up to N,  ~ N/log N , which is unbounded.

But at the same time, this fraction of  1 / log N  of all numbers up to  N  tends to zero,

so exactly 0% of all numbers are primes, and therefore even and odd composite numbers

both account for exactly 50% of all (composite) numbers.

Indeed, almost all numbers are pandigital, and have all digits even more than once.

Therefore, for large N, you will almost always have a(N) = N.

(Again, because N as well as N+-1 will have all digits 0 - 9.)

Therefore the percentage of odd terms of both sequences will eventually approach 50%.

(Actually, the fraction of odd terms of the sequence grows above 35% already around 2 x 10^9,

where we are still far from having "almost all" numbers pandigital.)

- Maximilian

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- Maximilian

[Jon Awbrey]

Theorem. In the long run, numbers (positive integers) just get even.

In other words, on average numbers are even.

Hint. Weighted average over degrees of even‑ness.

Regards,

Jon

On 4/2/2026 10:53 PM, M. F. Hasler wrote:
> It is somewhat ambiguous to say that most composite numbers are even.
> It is also true that exactly 50% of all composite numbers are even,
> and exactly 50% of all composite numbers are odd.

<...>

[David desJardins]

On Thu, Apr 2, 2026 at 8:36 PM Jon Awbrey <jaw...@att.net> wrote:

Theorem. In the long run, numbers (positive integers) just get even.
In other words, on average numbers are even.

Sure. Because once you have two odd numbers, then their average is even. QED. 

[Ali Sada]

Thank you Maximilian. I really appreciate it. Maybe I should have proposed "non-trivial" and see what happens. For the first hundreds of terms, it seemed that the n-1 version was different from the n+1 version. I'm not sure if that continues further. 

Going back to "non-trivial", I saw that my late friend David Sycamore came up with this sequence  A380885: "a(n) is the smallest multiple m*n (m > 1) of n which contains every decimal digit of n, including repetitions". Would it be adequate to add another version excluding the trivial terms (adding zeros or repeating the same digits.) It would be great if the editors allow me to introduce this sequence and add David as co-author since it's his idea. 

Best,

Ali 

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