Is that right ! IF : Σ_(k = 1) ^n (⌊(n/k)⌋−⌊((n−1)/k)⌋) = 2 so n is a prime number .


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Question Number 182934 by moh777 last updated on 17/Dec/22

$I s t h a t r i g h t!$ $I F :$ $\underset{k = 1}{\sum^{n}} (⌊ \frac{n}{k} ⌋ - ⌊ \frac{n - 1}{k} ⌋) = 2$ $s o n i s a p r i m e n u m b e r .$
	Answered by dre23 last updated on 17/Dec/22
	$⇔∀n≥2 Σ_(k=2) ^n ([(n/k)]−[((n−1)/k)])=1 n est premiers n=2 worcks ∀n≥3 ⇔Σ_(k=2) ^(n−1) ([(n/k)]−[((n−1)/k)])=0 f(n,k)=[(n/k)]−[((n−1)/k)] if n is not prime ∃s∈N−{1,n} such n=sp (n/s)=p and((n−1)/s)=p−(1/p), [(n/k)]=p,[((n−1)/k)]=p−1 Σ_(k=2) ^(n−1) ([(n/k)]−[((n−1)/k)])=Σ_(k=2,k≠k) ^(n−1) f(n,k)+p−(p−1)≥p−(p−1)=1 ⇔Σ_(k=2) ^(n−1) f(n,k)≥1>0⇒not true so Σ_(k=1) ^n f(n,k)=2⇒n is prime true$
	$\Leftrightarrow \forall n ⩾ 2$ $\underset{k = 2}{\sum^{n}} ([\frac{n}{k}] - [\frac{n - 1}{k}]) = 1 n e s t p r e m i e r s$ $n = 2 w o r c k s \forall n ⩾ 3$ $\Leftrightarrow \underset{k = 2}{\sum^{n - 1}} ([\frac{n}{k}] - [\frac{n - 1}{k}]) = 0$ $f (n, k) = [\frac{n}{k}] - [\frac{n - 1}{k}]$ $i f n i s n o t p r i m e \exists s \in N - {1, n} s u c h n = s p$ $\frac{n}{s} = p a n d \frac{n - 1}{s} = p - \frac{1}{p},$ $[\frac{n}{k}] = p, [\frac{n - 1}{k}] = p - 1$ $\underset{k = 2}{\sum^{n - 1}} ([\frac{n}{k}] - [\frac{n - 1}{k}]) = \underset{k = 2, k \neq k}{\sum^{n - 1}} f (n, k) + p - (p - 1) ⩾ p - (p - 1) = 1$ $\Leftrightarrow \underset{k = 2}{\sum^{n - 1}} f (n, k) ⩾ 1 > 0 \Rightarrow n o t t r u e$ $s o \underset{k = 1}{\sum^{n}} f (n, k) = 2 \Rightarrow n i s p r i m e t r u e$
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