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Let $f(n)$ be the number of positive integers $x\le n$ such that $\lambda(k)=x$ has no solution, where $\lambda(k)$ denotes the Carmichael-function.

Does $$\lim_{n\rightarrow \infty} \frac{f(n)}{n}$$ exist , and if yes, is it $1$ or some smaller value ?

The last few lines in a numerical analysis were :

579000000  0.77069874093264248704663212435233160622
580000000  0.77070391551724137931034482758620689655
581000000  0.77070938382099827882960413080895008606
582000000  0.77071472164948453608247422680412371134
583000000  0.77071994511149228130360205831903945112
584000000  0.77072559931506849315068493150684931507
585000000  0.77073061196581196581196581196581196581

This indicates a slow increase, but when I tried small ranges with larger values , the frequency seemed to still increase (above $0.8$). Any ideas ?

Peter
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1 Answers1

5

Let $V_{\lambda}(x)=\#\{\lambda(n)\leq x : n\geq 1\}$.

By a result of Florian Luca and Carl Pomerance: https://math.dartmouth.edu/~carlp/rangeoflambda13.pdf

The number of images of lambda function has an upper bound of the form $$ V_{\lambda}(x)\leq \frac x{(\log x)^{\eta+o(1)}} $$ where $\eta=1-(1+\log\log 2)/\log 2=0.0860713..$

Therefore, almost all of $m\leq x$ is non-image of $\lambda$ function. Hence, we have by $f(x)=\lfloor x \rfloor - V_{\lambda}(x)$, $$ \lim_{x\rightarrow\infty} \frac{f(x)}x=1. $$

Later by Kevin Ford, Florian Luca, and Carl Pomerance, https://faculty.math.illinois.edu/~ford/wwwpapers/lambda_range.pdf it is known that the earlier exponent in the upper bound was in fact the correct one. That is, $$ V_{\lambda}(x)= \frac x{(\log x)^{\eta+o(1)}}. $$

K.defaoite
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Sungjin Kim
  • 20,850