# [seqfan] Re: Very nice new sequence A329126 [1, 6, 42, 60, 139810, 126, ...]

Don Reble djr at nk.ca
Tue Nov 9 07:34:52 CET 2021

```Seqfans:

> %S A329126 1,110,101010,111100,100010001000100010,1111110
> %N a(n) is the lexicographically earliest string of digits which
>    yields a multiple of n when read in any numeric base.

Does the author really mean "lexicographically earliest"?
There are many candidates for a(2): 110, 1010, 10010, 100010, ...,
Each is divisible by 2 in any base. Lexicographically,
each is earlier than the previous, and there is no earliest.

I think he means simply "the least positive integer which yields
a multiple...".

---

Here are a(n) formulas for some kinds of n-values.
In the following, B is 2 or 10 (for A329000 vs. A329126).
Let L(n) be Carmichael's lambda function (A002322)
L-(n) = L(n) / gcd[L(n),n]           (A174824)
L+(n) = lcm[L(n),n]                  (A268336)
W-(n) = [B ^ L-(n)] - 1
W+(n) = [B ^ L+(n)] - 1
Z(n)  =  B ^ A051903(n)

If n is in A124240 or n is prime,
a(n) = Z(n) * W+(n) / W-(n)
If n = 2*q (doubled odd prime),
a(n) = Z(n) * (B+1) * W+(q) / W-(q)
If n = 3*r (r prime, greater than 6),
a(n) = Z(n) * [(B^6-1) / (B^2-1)] * W+(r) / W-(r)

A051903(n) is the maximum exponent in the prime factorization of n.
A124240 has values n for which L(n) divides n.

a(3*5) is irregular because L(3*5) < 3*L(3).
(Yeah, there's more to it, but this is a brief message.)
For primes r>6, L(3*r) >= 3*L(3); they're all regular.

--
Don Reble  djr at nk.ca

```