The number-of-divisors function $d$, and the sum-of-divisors function $\sigma$, are defined by $$ d(n) = \sum_{d \mid n} 1, $$ $$ \sigma(n) = \sum_{d \mid n} d, $$ respectively. Now let $N$ be a square-free positive integer and consider the difference $$ d(N) - \dfrac{\sigma(N)}{N}. $$ Is there any kind of smooth function giving an upper bound for this?
2026-04-08 18:09:23.1775671763
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Upper bound for the difference between number-of-divisors and sum-of-divisors functions
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The number of divisors is usually so small that it doesn't make a difference, so the records are usually the highly abundant numbers, A002093 in the OEIS.
As an example, the $10^4$-th highly abundant number is $N=7442466942548913946301030400=2^{13}\cdot3^4\cdot5^2\cdot7^2\cdot11\cdot13\cdot17\cdot19\cdot23\cdot29\cdot31\cdot37\cdot41\cdot43\cdot47\cdot53\cdot59$ which has 5160960 divisors summing to 53306653535048357847760896000.
If you restrict to squarefree numbers only, then Will's answer is spot on: you can't do better than primorials.
for primorial $$N = \prod_{p \leq x} p,$$ we find $$ d(N) = 2^{\pi(x)} \approx 2^{x/\log x} $$ which is quite large, while $$ \sigma(N)/N = \prod_{p \leq x} 1 + \frac{1}{p} \approx \frac{6 e^\gamma \log x}{ \pi^2}, $$ much smaller. If the difference is not always an upper bound it will do until the real thing comes along.
Try it on computer. Note that i have written in terms of $x$ rather than writing some $p_n;$ that tends to create or erase logs.
Below is the result of the first 100 primes, the second column is the prime, the third column is the cumulative product of $1 + (1/p).$