Mixing
Ratio of totient $varphi(n)/varphi(n+1)$
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Is there a general formula for the ratio of totients of two consecutive integers? It occurred to me in a different context, where I was trying to find a different way of estimating the probability of drawing a random rational number in the interval $(0,1)$. One could represent a rational number as an ordered pair of co-primes $(m,n)$ with $m<n$. Suppose we put a uniform, finitely additive measure on the set ${(m,n)inmathbb{N}_{>0}: m, n text{co-prime}, m<n}$. Then fixing $n$, there are only $varphi(n)$ many rational numbers with denominator $n$, which is what brought me to the question of $varphi(n)/varphi(n+1)$. Ideally it would be great if the formula could be a function of $n$ alone (sorry this might still be vague). Many thanks in advance for any help.
number-theory
asked Dec 31 '18 at 21:42
discretizerdiscretizer
1397
$endgroup$
add a comment |
$begingroup$
Is there a general formula for the ratio of totients of two consecutive integers? It occurred to me in a different context, where I was trying to find a different way of estimating the probability of drawing a random rational number in the interval $(0,1)$. One could represent a rational number as an ordered pair of co-primes $(m,n)$ with $m<n$. Suppose we put a uniform, finitely additive measure on the set ${(m,n)inmathbb{N}_{>0}: m, n text{co-prime}, m<n}$. Then fixing $n$, there are only $varphi(n)$ many rational numbers with denominator $n$, which is what brought me to the question of $varphi(n)/varphi(n+1)$. Ideally it would be great if the formula could be a function of $n$ alone (sorry this might still be vague). Many thanks in advance for any help.
number-theory
asked Dec 31 '18 at 21:42
discretizerdiscretizer
1397
$endgroup$
1
$begingroup$
You'll get a better response if you include context in your question. For example, how did this question occur to you? What have you tried? Also, most importantly, what sort of formula are you looking for? After all, $varphi(n)/varphi(n+1)$ is a formula.
$endgroup$
– jgon
Dec 31 '18 at 21:48
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Thanks a lot! I have edited the question to include my motivations.
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
I think the addition of motivation has significantly improved the question, and I've upvoted.
$endgroup$
– jgon
Dec 31 '18 at 22:16
add a comment |
$begingroup$
Is there a general formula for the ratio of totients of two consecutive integers? It occurred to me in a different context, where I was trying to find a different way of estimating the probability of drawing a random rational number in the interval $(0,1)$. One could represent a rational number as an ordered pair of co-primes $(m,n)$ with $m<n$. Suppose we put a uniform, finitely additive measure on the set ${(m,n)inmathbb{N}_{>0}: m, n text{co-prime}, m<n}$. Then fixing $n$, there are only $varphi(n)$ many rational numbers with denominator $n$, which is what brought me to the question of $varphi(n)/varphi(n+1)$. Ideally it would be great if the formula could be a function of $n$ alone (sorry this might still be vague). Many thanks in advance for any help.
number-theory
asked Dec 31 '18 at 21:42
discretizerdiscretizer
1397
$endgroup$
Is there a general formula for the ratio of totients of two consecutive integers? It occurred to me in a different context, where I was trying to find a different way of estimating the probability of drawing a random rational number in the interval $(0,1)$. One could represent a rational number as an ordered pair of co-primes $(m,n)$ with $m<n$. Suppose we put a uniform, finitely additive measure on the set ${(m,n)inmathbb{N}_{>0}: m, n text{co-prime}, m<n}$. Then fixing $n$, there are only $varphi(n)$ many rational numbers with denominator $n$, which is what brought me to the question of $varphi(n)/varphi(n+1)$. Ideally it would be great if the formula could be a function of $n$ alone (sorry this might still be vague). Many thanks in advance for any help.
number-theory
number-theory
asked Dec 31 '18 at 21:42
discretizerdiscretizer
1397
asked Dec 31 '18 at 21:42
discretizerdiscretizer
1397
edited Dec 31 '18 at 22:12
discretizer
asked Dec 31 '18 at 21:42
discretizerdiscretizer
1397
asked Dec 31 '18 at 21:42
discretizerdiscretizer
1397
asked Dec 31 '18 at 21:42
discretizerdiscretizer
1397
1397
1
$begingroup$
You'll get a better response if you include context in your question. For example, how did this question occur to you? What have you tried? Also, most importantly, what sort of formula are you looking for? After all, $varphi(n)/varphi(n+1)$ is a formula.
$endgroup$
– jgon
Dec 31 '18 at 21:48
$begingroup$
Thanks a lot! I have edited the question to include my motivations.
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
I think the addition of motivation has significantly improved the question, and I've upvoted.
$endgroup$
– jgon
Dec 31 '18 at 22:16
add a comment |
1
$begingroup$
You'll get a better response if you include context in your question. For example, how did this question occur to you? What have you tried? Also, most importantly, what sort of formula are you looking for? After all, $varphi(n)/varphi(n+1)$ is a formula.
$endgroup$
– jgon
Dec 31 '18 at 21:48
$begingroup$
Thanks a lot! I have edited the question to include my motivations.
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
I think the addition of motivation has significantly improved the question, and I've upvoted.
$endgroup$
– jgon
Dec 31 '18 at 22:16
1
1
$begingroup$
You'll get a better response if you include context in your question. For example, how did this question occur to you? What have you tried? Also, most importantly, what sort of formula are you looking for? After all, $varphi(n)/varphi(n+1)$ is a formula.
$endgroup$
– jgon
Dec 31 '18 at 21:48
$begingroup$
You'll get a better response if you include context in your question. For example, how did this question occur to you? What have you tried? Also, most importantly, what sort of formula are you looking for? After all, $varphi(n)/varphi(n+1)$ is a formula.
$endgroup$
– jgon
Dec 31 '18 at 21:48
$begingroup$
Thanks a lot! I have edited the question to include my motivations.
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
Thanks a lot! I have edited the question to include my motivations.
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
I think the addition of motivation has significantly improved the question, and I've upvoted.
$endgroup$
– jgon
Dec 31 '18 at 22:16
$begingroup$
I think the addition of motivation has significantly improved the question, and I've upvoted.
$endgroup$
– jgon
Dec 31 '18 at 22:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $;n=p_1^{a_1}cdotldotscdot p_m^{a_m};,;;p_i,,a_iinBbb N;,;;p_i$ primes, then we know that
$$phi(n)=nprod_{k=1}^mleft(1-frac1{p_k}right)$$
so you can then write
$$frac{phi(n)}{phi(n+1)}=frac n{n+1}prod_{pmid n,,qmid (n+1)}left(1-frac1pright)left(1-frac1qright)^{-1}$$
It doesn't really look very nice a formula, though...
answered Dec 31 '18 at 22:03
DonAntonioDonAntonio
177k1492225
$endgroup$
$begingroup$
Thanks! Yes ideally I would like a formula that does not invoke facts about prime factorization of $n$, but that might be impossible...
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
@discretizer I can nearly guarantee you it is impossible. There are a lot of open problems similar to this one.
$endgroup$
– Don Thousand
Dec 31 '18 at 22:19
$begingroup$
Happy New Year, my friend !
$endgroup$
– Claude Leibovici
Jan 1 at 6:26
$begingroup$
@ClaudeLeibovici Bonne Anneé, mon ami !
$endgroup$
– DonAntonio
Jan 1 at 9:47
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $;n=p_1^{a_1}cdotldotscdot p_m^{a_m};,;;p_i,,a_iinBbb N;,;;p_i$ primes, then we know that
$$phi(n)=nprod_{k=1}^mleft(1-frac1{p_k}right)$$
so you can then write
$$frac{phi(n)}{phi(n+1)}=frac n{n+1}prod_{pmid n,,qmid (n+1)}left(1-frac1pright)left(1-frac1qright)^{-1}$$
It doesn't really look very nice a formula, though...
answered Dec 31 '18 at 22:03
DonAntonioDonAntonio
177k1492225
$endgroup$
$begingroup$
Thanks! Yes ideally I would like a formula that does not invoke facts about prime factorization of $n$, but that might be impossible...
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
@discretizer I can nearly guarantee you it is impossible. There are a lot of open problems similar to this one.
$endgroup$
– Don Thousand
Dec 31 '18 at 22:19
$begingroup$
Happy New Year, my friend !
$endgroup$
– Claude Leibovici
Jan 1 at 6:26
$begingroup$
@ClaudeLeibovici Bonne Anneé, mon ami !
$endgroup$
– DonAntonio
Jan 1 at 9:47
add a comment |
$begingroup$
If $;n=p_1^{a_1}cdotldotscdot p_m^{a_m};,;;p_i,,a_iinBbb N;,;;p_i$ primes, then we know that
$$phi(n)=nprod_{k=1}^mleft(1-frac1{p_k}right)$$
so you can then write
$$frac{phi(n)}{phi(n+1)}=frac n{n+1}prod_{pmid n,,qmid (n+1)}left(1-frac1pright)left(1-frac1qright)^{-1}$$
It doesn't really look very nice a formula, though...
answered Dec 31 '18 at 22:03
DonAntonioDonAntonio
177k1492225
$endgroup$
$begingroup$
Thanks! Yes ideally I would like a formula that does not invoke facts about prime factorization of $n$, but that might be impossible...
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
@discretizer I can nearly guarantee you it is impossible. There are a lot of open problems similar to this one.
$endgroup$
– Don Thousand
Dec 31 '18 at 22:19
$begingroup$
Happy New Year, my friend !
$endgroup$
– Claude Leibovici
Jan 1 at 6:26
$begingroup$
@ClaudeLeibovici Bonne Anneé, mon ami !
$endgroup$
– DonAntonio
Jan 1 at 9:47
add a comment |
$begingroup$
If $;n=p_1^{a_1}cdotldotscdot p_m^{a_m};,;;p_i,,a_iinBbb N;,;;p_i$ primes, then we know that
$$phi(n)=nprod_{k=1}^mleft(1-frac1{p_k}right)$$
so you can then write
$$frac{phi(n)}{phi(n+1)}=frac n{n+1}prod_{pmid n,,qmid (n+1)}left(1-frac1pright)left(1-frac1qright)^{-1}$$
It doesn't really look very nice a formula, though...
answered Dec 31 '18 at 22:03
DonAntonioDonAntonio
177k1492225
$endgroup$
If $;n=p_1^{a_1}cdotldotscdot p_m^{a_m};,;;p_i,,a_iinBbb N;,;;p_i$ primes, then we know that
$$phi(n)=nprod_{k=1}^mleft(1-frac1{p_k}right)$$
so you can then write
$$frac{phi(n)}{phi(n+1)}=frac n{n+1}prod_{pmid n,,qmid (n+1)}left(1-frac1pright)left(1-frac1qright)^{-1}$$
It doesn't really look very nice a formula, though...
answered Dec 31 '18 at 22:03
DonAntonioDonAntonio
177k1492225
answered Dec 31 '18 at 22:03
DonAntonioDonAntonio
177k1492225
answered Dec 31 '18 at 22:03
DonAntonioDonAntonio
177k1492225
answered Dec 31 '18 at 22:03
DonAntonioDonAntonio
177k1492225
177k1492225
$begingroup$
Thanks! Yes ideally I would like a formula that does not invoke facts about prime factorization of $n$, but that might be impossible...
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
@discretizer I can nearly guarantee you it is impossible. There are a lot of open problems similar to this one.
$endgroup$
– Don Thousand
Dec 31 '18 at 22:19
$begingroup$
Happy New Year, my friend !
$endgroup$
– Claude Leibovici
Jan 1 at 6:26
$begingroup$
@ClaudeLeibovici Bonne Anneé, mon ami !
$endgroup$
– DonAntonio
Jan 1 at 9:47
add a comment |
$begingroup$
Thanks! Yes ideally I would like a formula that does not invoke facts about prime factorization of $n$, but that might be impossible...
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
@discretizer I can nearly guarantee you it is impossible. There are a lot of open problems similar to this one.
$endgroup$
– Don Thousand
Dec 31 '18 at 22:19
$begingroup$
Happy New Year, my friend !
$endgroup$
– Claude Leibovici
Jan 1 at 6:26
$begingroup$
@ClaudeLeibovici Bonne Anneé, mon ami !
$endgroup$
– DonAntonio
Jan 1 at 9:47
$begingroup$
Thanks! Yes ideally I would like a formula that does not invoke facts about prime factorization of $n$, but that might be impossible...
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
Thanks! Yes ideally I would like a formula that does not invoke facts about prime factorization of $n$, but that might be impossible...
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
@discretizer I can nearly guarantee you it is impossible. There are a lot of open problems similar to this one.
$endgroup$
– Don Thousand
Dec 31 '18 at 22:19
$begingroup$
@discretizer I can nearly guarantee you it is impossible. There are a lot of open problems similar to this one.
$endgroup$
– Don Thousand
Dec 31 '18 at 22:19
$begingroup$
Happy New Year, my friend !
$endgroup$
– Claude Leibovici
Jan 1 at 6:26
$begingroup$
Happy New Year, my friend !
$endgroup$
– Claude Leibovici
Jan 1 at 6:26
$begingroup$
@ClaudeLeibovici Bonne Anneé, mon ami !
$endgroup$
– DonAntonio
Jan 1 at 9:47
$begingroup$
@ClaudeLeibovici Bonne Anneé, mon ami !
$endgroup$
– DonAntonio
Jan 1 at 9:47
add a comment |
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$begingroup$
You'll get a better response if you include context in your question. For example, how did this question occur to you? What have you tried? Also, most importantly, what sort of formula are you looking for? After all, $varphi(n)/varphi(n+1)$ is a formula.
$endgroup$
– jgon
Dec 31 '18 at 21:48
$begingroup$
Thanks a lot! I have edited the question to include my motivations.
$endgroup$
– discretizer
Dec 31 '18 at 22:13
$begingroup$
I think the addition of motivation has significantly improved the question, and I've upvoted.
$endgroup$
– jgon
Dec 31 '18 at 22:16