Square array A(n,k): least positive m such that n|m and k|(m+1)

[Ali Sada]

Hi everyone,

Hope all is well. Is the array below suitable for the OEIS?

Best,

Ali 

Square array A(n,k), n,k≥1, where A(n,k) is the least positive integer m such that n divides m and k divides m+1; if no such m exists, then A(n,k)=0.

[M F Hasler]

On Tue, Jun 23, 2026 at 12:15 AM Ali Sada <ali....@gmail.com> wrote:

Hi everyone,

Hope all is well. Is the array below suitable for the OEIS?

Square array A(n,k), n,k≥1, where A(n,k) is the least positive integer m such that n divides m and k divides m+1; if no such m exists, then A(n,k)=0.

Looks interesting enough to me.

Maybe XREF  A127465, clearly not the same but it has remarkable overlap:

1, 1,2, 2,0,3, 3,2,3,4, 4,0,0,0,5, ... = your array by falling antidiagonals

1, 1,2, 2,0,3, 2,2,0,4 ,4,0,0,0,5, ... =  triangle A127465 read by rows

I think that's not a pure coincidence.

At least the "border values" can probably be easily proved to be equal.

- Maximilian

 

[Gareth McCaughan]

I think it's "mostly" coincidence, though I admit the possibility that I'm missing things.

The k'th entry of the n'th falling antidiag of A.S.'s array (n>=1; 1<=k<=n) is his A(k,n+1-k) and is therefore the least positive m such that k|n and n+1-k|m+1.

The k'th entry of the n'th row in A127465 is what that entry calls T(n,k) = k phi(n/k) where phi is the Euler totient function extended to be zero at all non-integers.

Border values are k=1 and k=n.

k=1: we want A(1,n) = T(n,1). LHS is smallest m with 1|m and n|m+1, which is just m=n-1 except for n=1. RHS is phi(n). These agree when n is prime, and also as it happens when n=1, but not otherwise; e.g., the start of the 4th antidiag of A.S.'s array is 3 whereas the start of the 4th row of A127465 is 2.

k=n: we want A(n,1) = T(n,n). LHS is smallest m with n|m and 1|m+1, which is just m=n. RHS is n phi(1) = n. So these do agree.

It looks as if zero entries in A are necessarily zero in T but not vice versa. That is: if there's no m with k|m and n+1-k|m+1 then it's not true that k|n. That is: if k|n then there is m such that k|m and n+1-k|m+1. That's true because we can take m=n.

So: one border agrees, one border agrees at prime-or-1 indices but not elsewhere, zeros in one are a subset of zeros in the other. But I don't see all that much similarity between these two arrays beyond that, nor do I see any particular reason why there should be much similarity.

Am I missing something?

-- 
g

[Ali Sada]

Thank you very much Maximilian. I really appreciate your response. The proposed array is https://oeis.org/draft/A396382.

Best,

Ali 

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