This is the sequence of $n$ so that $\frac{n}{\lambda(n)}$ is greatest up until $n$, where $\lambda$ is the Carmichael function. The analogous quantity $\frac{n}{\varphi(n)}$ is maximized when $n$ is a primorial, but as you can see in the link, the version with the Carmichael function is more complicated.

If we fix a list of primes, and write $n=2^{a_0} p_1^{a_1}\dots p_k^{a_k}$ where only $a_0$ can be $0$, we can find what the exponents should be. There are four different cases for $a_0=0,1,2$ and $a_0>2$. I'll do the last one and the others are analogous.


and now we choose the $a_i$ to be equal to or smaller than one (2 for $a_0$) plus the largest exponent of $p_i$ which occurs in any of the other parentheses, because if we chose $a_i$ to be greater than that it would have the effect of multiplying both $n$ and $\lambda(n)$ which would cancel out.

So this means that for any list of primes we know that it appears on the sequence a finite number of times and we can find what the exponents are. For example if we want the primes that divide $n$ to be $2,3,5,7$ we have that the possibilities of $n$ are $105,210,315,420,630,840,1260,2520$ and of those only $1260$ appears on the sequence. One thought I had is that we would want $n$ to be divisible by some prime $p$ so that $p-1$ has a prime raised to a large power in its factorization, but for example n which is divisible by $2,17$ does not appear on the sequence. Larger Fermat primes don't work either. Some of the terms in the sequence are primorials like for the totient function, but some don't appear, for example 30 and 210 don't.

So, is it possible to determine what the $n$ which appear in the sequence are, or what their prime factorizations look like?


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