论文标题
递归丰富和递归完美的数字
Recursively abundant and recursively perfect numbers
论文作者
论文摘要
除数函数$σ（n）$总和$ n $的除数。当$σ（n） - n $时，我们将$ n $称为$ n $，当$σ（n） - n = n $时，我们将其称为$ n $。我最近引入了递归除数函数$ a（n）$，即除数函数的递归类似物。它衡量了一个数字高度分解成各个部分的程度，从而使各个部分都可以分为小节。正如除数函数激发了丰富而完美的数字一样，递归除数功能激发了它们的递归类似物，我在这里介绍了它们。如果$ a（n）> n $递归递归或丰富，则递归完美或原始，如果$ a（n）= n $。丰富和完美的数字及其递归对应物之间存在惊人的相似之处。两个足够数字的乘积是足够的，而且数字丰富或奇怪的数字。存在奇怪的数字，但很少见，我猜想这样的数字不被第一个$ k $ primes排除 - 对于大量数字来说，这是正确的。除了1个以外的原始数量，有很多原始数量，但是它们不可能奇怪。原始数量是对某些二磷剂方程的两个和奇数质量解决方案的能力的产物，让人想起两个和梅森恩素的能力的完美数量。这类数字之间的相似之处暗示了除数函数与其递归类似物之间的更深层次的联系，值得进一步研究。
The divisor function $σ(n)$ sums the divisors of $n$. We call $n$ abundant when $σ(n) - n > n$ and perfect when $σ(n) - n = n$. I recently introduced the recursive divisor function $a(n)$, the recursive analog of the divisor function. It measures the extent to which a number is highly divisible into parts, such that the parts are highly divisible into subparts, so on. Just as the divisor function motivates the abundant and perfect numbers, the recursive divisor function motivates their recursive analogs, which I introduce here. A number is recursively abundant, or ample, if $a(n) > n$ and recursively perfect, or pristine, if $a(n) = n$. There are striking parallels between abundant and perfect numbers and their recursive counterparts. The product of two ample numbers is ample, and ample numbers are either abundant or odd perfect numbers. Odd ample numbers exist but are rare, and I conjecture that there are such numbers not divisible by the first $k$ primes -- which is known to be true for the abundant numbers. There are infinitely many pristine numbers, but that they cannot be odd, apart from 1. Pristine numbers are the product of a power of two and odd prime solutions to certain Diophantine equations, reminiscent of how perfect numbers are the product of a power of two and a Mersenne prime. The parallels between these kinds of numbers hint at deeper links between the divisor function and its recursive analog, worthy of further investigation.
