When is $prod_{n=2}^{k} n-frac{1}{n}$ not an integer?
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1
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For which values of $k$ does the following product not evaluate to an integer: $$prod_{n=2}^{k} n-frac{1}{n}$$
I find it somewhat surprising that it not only can evaluate to an integer, but also that it does so most of the time.
The only values of $k$ that I could find so far, such that it does not evaluate to an integer are $2$ and $18$. Do these numbers have any special characteristics which would explain this?
integers products
asked Nov 19 at 18:42
g3nuine
1208
add a comment |
up vote
1
down vote
favorite
For which values of $k$ does the following product not evaluate to an integer: $$prod_{n=2}^{k} n-frac{1}{n}$$
I find it somewhat surprising that it not only can evaluate to an integer, but also that it does so most of the time.
The only values of $k$ that I could find so far, such that it does not evaluate to an integer are $2$ and $18$. Do these numbers have any special characteristics which would explain this?
integers products
asked Nov 19 at 18:42
g3nuine
1208
Only number that it doesn't evaluate to an integer is $2$.
– Jakobian
Nov 19 at 18:46
My math app outputs 3379030566911999.5 for k=18 but maybe thats due to a bug.
– g3nuine
Nov 19 at 18:52
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For which values of $k$ does the following product not evaluate to an integer: $$prod_{n=2}^{k} n-frac{1}{n}$$
I find it somewhat surprising that it not only can evaluate to an integer, but also that it does so most of the time.
The only values of $k$ that I could find so far, such that it does not evaluate to an integer are $2$ and $18$. Do these numbers have any special characteristics which would explain this?
integers products
asked Nov 19 at 18:42
g3nuine
1208
For which values of $k$ does the following product not evaluate to an integer: $$prod_{n=2}^{k} n-frac{1}{n}$$
I find it somewhat surprising that it not only can evaluate to an integer, but also that it does so most of the time.
The only values of $k$ that I could find so far, such that it does not evaluate to an integer are $2$ and $18$. Do these numbers have any special characteristics which would explain this?
integers products
integers products
asked Nov 19 at 18:42
g3nuine
1208
asked Nov 19 at 18:42
g3nuine
1208
asked Nov 19 at 18:42
g3nuine
1208
asked Nov 19 at 18:42
g3nuine
1208
asked Nov 19 at 18:42
g3nuine
1208
1208
Only number that it doesn't evaluate to an integer is $2$.
– Jakobian
Nov 19 at 18:46
My math app outputs 3379030566911999.5 for k=18 but maybe thats due to a bug.
– g3nuine
Nov 19 at 18:52
add a comment |
Only number that it doesn't evaluate to an integer is $2$.
– Jakobian
Nov 19 at 18:46
My math app outputs 3379030566911999.5 for k=18 but maybe thats due to a bug.
– g3nuine
Nov 19 at 18:52
Only number that it doesn't evaluate to an integer is $2$.
– Jakobian
Nov 19 at 18:46
Only number that it doesn't evaluate to an integer is $2$.
– Jakobian
Nov 19 at 18:46
My math app outputs 3379030566911999.5 for k=18 but maybe thats due to a bug.
– g3nuine
Nov 19 at 18:52
My math app outputs 3379030566911999.5 for k=18 but maybe thats due to a bug.
– g3nuine
Nov 19 at 18:52
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
$$prod_{n=2}^k frac{(n+1)(n-1)}{n} = frac{frac{(k+1)!}{2}cdot(k-1)!}{k!} = frac{(k+1)(k-1)!}{2} $$
The only number for which it can't be an integer is $k=2$, and that's indeed the case.
answered Nov 19 at 18:54
Jakobian
2,344720
Thanks. At least now I know that I can't count on the products evaluated by my app.
– g3nuine
Nov 19 at 18:59
add a comment |
up vote
1
down vote
It is surprising that this is often an integer. But because it is surprising there is probably a simple reason that will be become clear when we try a few examples.
$(2 - frac 12)(3 - frac 13) = 2*3 - frac 32 - frac 23 + frac 1{2*3}$.
Okay, one thing that becomes clear is that if we have $(n - frac 1n)(m - frac 1n)$ our terms are going to be $nm$ (which is clearly an integer) $-frac nm, -frac nm$ and $frac 1{mmn}$. When we add $-frac mn -frac nm$ we get $-(frac {m^2}{mn} + frac {n^2}{mn})=-frac {m^2 + n^2}{nm}$ and we get $mn - frac {m^2 + n^2-1}{mn}$.
Well, no reason that $mn|m^2 + n^2 -1$ but if $m$ and $n$ are consecutive, i..e. $(n = m+1)$ we get $$m(m+1) - frac {m^2 + (m+1)^2 - 1}{m(m+1)} = m(m+1) - frac {2m^2 + 2m}{m(m+1)} = m(m+1) - frac {2m(m+1)}{m(m+1)} = m(m+1) -2$$ and that.... is an integer.
So $(m - frac 1m)((m+1) - frac 1{m+1}$ must be an integer and as integers multiply together to integers we have for a product of even number of terms:
$prod_{n=2}^{2m+ 1} (n-frac 1n) = prod_{k=1}^m [(2k-frac 1{2k})((2k+1)-frac 1{2k+1})] = prod_{k=1}^m (2k(2k+1) -2)=Kin mathbb Z$.
But if we have a product of odd terms:
$prod_{n=2}^{2m+ 2}(n-frac 1n) =(prod_{n=2}^{2m+ 1}(n-frac 1n))((2m+2)-frac 1{2m+2})=K( 2m+2 - frac 1{2m+2})$ and as $( 2m+2 - frac 1{2m+2})$ is not an integer $K( 2m+2 - frac 1{2m+2})$ won't be an integer.....
.... unless $2m+2|K$.
Okay.... we are halfway done.
When will $2m+2|K$?
I'm embarrassed to admit how long the following took me to figure out but:
$K=prod_{k=1}^m(2k(2k+1) -2)$ and for $2m(2m+1) -2=2(m+1)(2m -1)$ so $2m+2$ will always divide $K$ and this will always be an integer. (Unless we only multiply one term $(2 - frac 12)$.)
answered Nov 19 at 20:02
fleablood
67.4k22684
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$prod_{n=2}^k frac{(n+1)(n-1)}{n} = frac{frac{(k+1)!}{2}cdot(k-1)!}{k!} = frac{(k+1)(k-1)!}{2} $$
The only number for which it can't be an integer is $k=2$, and that's indeed the case.
answered Nov 19 at 18:54
Jakobian
2,344720
Thanks. At least now I know that I can't count on the products evaluated by my app.
– g3nuine
Nov 19 at 18:59
add a comment |
up vote
3
down vote
accepted
$$prod_{n=2}^k frac{(n+1)(n-1)}{n} = frac{frac{(k+1)!}{2}cdot(k-1)!}{k!} = frac{(k+1)(k-1)!}{2} $$
The only number for which it can't be an integer is $k=2$, and that's indeed the case.
answered Nov 19 at 18:54
Jakobian
2,344720
Thanks. At least now I know that I can't count on the products evaluated by my app.
– g3nuine
Nov 19 at 18:59
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$prod_{n=2}^k frac{(n+1)(n-1)}{n} = frac{frac{(k+1)!}{2}cdot(k-1)!}{k!} = frac{(k+1)(k-1)!}{2} $$
The only number for which it can't be an integer is $k=2$, and that's indeed the case.
answered Nov 19 at 18:54
Jakobian
2,344720
$$prod_{n=2}^k frac{(n+1)(n-1)}{n} = frac{frac{(k+1)!}{2}cdot(k-1)!}{k!} = frac{(k+1)(k-1)!}{2} $$
The only number for which it can't be an integer is $k=2$, and that's indeed the case.
answered Nov 19 at 18:54
Jakobian
2,344720
answered Nov 19 at 18:54
Jakobian
2,344720
answered Nov 19 at 18:54
Jakobian
2,344720
answered Nov 19 at 18:54
Jakobian
2,344720
2,344720
Thanks. At least now I know that I can't count on the products evaluated by my app.
– g3nuine
Nov 19 at 18:59
add a comment |
Thanks. At least now I know that I can't count on the products evaluated by my app.
– g3nuine
Nov 19 at 18:59
Thanks. At least now I know that I can't count on the products evaluated by my app.
– g3nuine
Nov 19 at 18:59
Thanks. At least now I know that I can't count on the products evaluated by my app.
– g3nuine
Nov 19 at 18:59
add a comment |
up vote
1
down vote
It is surprising that this is often an integer. But because it is surprising there is probably a simple reason that will be become clear when we try a few examples.
$(2 - frac 12)(3 - frac 13) = 2*3 - frac 32 - frac 23 + frac 1{2*3}$.
Okay, one thing that becomes clear is that if we have $(n - frac 1n)(m - frac 1n)$ our terms are going to be $nm$ (which is clearly an integer) $-frac nm, -frac nm$ and $frac 1{mmn}$. When we add $-frac mn -frac nm$ we get $-(frac {m^2}{mn} + frac {n^2}{mn})=-frac {m^2 + n^2}{nm}$ and we get $mn - frac {m^2 + n^2-1}{mn}$.
Well, no reason that $mn|m^2 + n^2 -1$ but if $m$ and $n$ are consecutive, i..e. $(n = m+1)$ we get $$m(m+1) - frac {m^2 + (m+1)^2 - 1}{m(m+1)} = m(m+1) - frac {2m^2 + 2m}{m(m+1)} = m(m+1) - frac {2m(m+1)}{m(m+1)} = m(m+1) -2$$ and that.... is an integer.
So $(m - frac 1m)((m+1) - frac 1{m+1}$ must be an integer and as integers multiply together to integers we have for a product of even number of terms:
$prod_{n=2}^{2m+ 1} (n-frac 1n) = prod_{k=1}^m [(2k-frac 1{2k})((2k+1)-frac 1{2k+1})] = prod_{k=1}^m (2k(2k+1) -2)=Kin mathbb Z$.
But if we have a product of odd terms:
$prod_{n=2}^{2m+ 2}(n-frac 1n) =(prod_{n=2}^{2m+ 1}(n-frac 1n))((2m+2)-frac 1{2m+2})=K( 2m+2 - frac 1{2m+2})$ and as $( 2m+2 - frac 1{2m+2})$ is not an integer $K( 2m+2 - frac 1{2m+2})$ won't be an integer.....
.... unless $2m+2|K$.
Okay.... we are halfway done.
When will $2m+2|K$?
I'm embarrassed to admit how long the following took me to figure out but:
$K=prod_{k=1}^m(2k(2k+1) -2)$ and for $2m(2m+1) -2=2(m+1)(2m -1)$ so $2m+2$ will always divide $K$ and this will always be an integer. (Unless we only multiply one term $(2 - frac 12)$.)
answered Nov 19 at 20:02
fleablood
67.4k22684
add a comment |
up vote
1
down vote
It is surprising that this is often an integer. But because it is surprising there is probably a simple reason that will be become clear when we try a few examples.
$(2 - frac 12)(3 - frac 13) = 2*3 - frac 32 - frac 23 + frac 1{2*3}$.
Okay, one thing that becomes clear is that if we have $(n - frac 1n)(m - frac 1n)$ our terms are going to be $nm$ (which is clearly an integer) $-frac nm, -frac nm$ and $frac 1{mmn}$. When we add $-frac mn -frac nm$ we get $-(frac {m^2}{mn} + frac {n^2}{mn})=-frac {m^2 + n^2}{nm}$ and we get $mn - frac {m^2 + n^2-1}{mn}$.
Well, no reason that $mn|m^2 + n^2 -1$ but if $m$ and $n$ are consecutive, i..e. $(n = m+1)$ we get $$m(m+1) - frac {m^2 + (m+1)^2 - 1}{m(m+1)} = m(m+1) - frac {2m^2 + 2m}{m(m+1)} = m(m+1) - frac {2m(m+1)}{m(m+1)} = m(m+1) -2$$ and that.... is an integer.
So $(m - frac 1m)((m+1) - frac 1{m+1}$ must be an integer and as integers multiply together to integers we have for a product of even number of terms:
$prod_{n=2}^{2m+ 1} (n-frac 1n) = prod_{k=1}^m [(2k-frac 1{2k})((2k+1)-frac 1{2k+1})] = prod_{k=1}^m (2k(2k+1) -2)=Kin mathbb Z$.
But if we have a product of odd terms:
$prod_{n=2}^{2m+ 2}(n-frac 1n) =(prod_{n=2}^{2m+ 1}(n-frac 1n))((2m+2)-frac 1{2m+2})=K( 2m+2 - frac 1{2m+2})$ and as $( 2m+2 - frac 1{2m+2})$ is not an integer $K( 2m+2 - frac 1{2m+2})$ won't be an integer.....
.... unless $2m+2|K$.
Okay.... we are halfway done.
When will $2m+2|K$?
I'm embarrassed to admit how long the following took me to figure out but:
$K=prod_{k=1}^m(2k(2k+1) -2)$ and for $2m(2m+1) -2=2(m+1)(2m -1)$ so $2m+2$ will always divide $K$ and this will always be an integer. (Unless we only multiply one term $(2 - frac 12)$.)
answered Nov 19 at 20:02
fleablood
67.4k22684
add a comment |
up vote
1
down vote
up vote
1
down vote
It is surprising that this is often an integer. But because it is surprising there is probably a simple reason that will be become clear when we try a few examples.
$(2 - frac 12)(3 - frac 13) = 2*3 - frac 32 - frac 23 + frac 1{2*3}$.
Okay, one thing that becomes clear is that if we have $(n - frac 1n)(m - frac 1n)$ our terms are going to be $nm$ (which is clearly an integer) $-frac nm, -frac nm$ and $frac 1{mmn}$. When we add $-frac mn -frac nm$ we get $-(frac {m^2}{mn} + frac {n^2}{mn})=-frac {m^2 + n^2}{nm}$ and we get $mn - frac {m^2 + n^2-1}{mn}$.
Well, no reason that $mn|m^2 + n^2 -1$ but if $m$ and $n$ are consecutive, i..e. $(n = m+1)$ we get $$m(m+1) - frac {m^2 + (m+1)^2 - 1}{m(m+1)} = m(m+1) - frac {2m^2 + 2m}{m(m+1)} = m(m+1) - frac {2m(m+1)}{m(m+1)} = m(m+1) -2$$ and that.... is an integer.
So $(m - frac 1m)((m+1) - frac 1{m+1}$ must be an integer and as integers multiply together to integers we have for a product of even number of terms:
$prod_{n=2}^{2m+ 1} (n-frac 1n) = prod_{k=1}^m [(2k-frac 1{2k})((2k+1)-frac 1{2k+1})] = prod_{k=1}^m (2k(2k+1) -2)=Kin mathbb Z$.
But if we have a product of odd terms:
$prod_{n=2}^{2m+ 2}(n-frac 1n) =(prod_{n=2}^{2m+ 1}(n-frac 1n))((2m+2)-frac 1{2m+2})=K( 2m+2 - frac 1{2m+2})$ and as $( 2m+2 - frac 1{2m+2})$ is not an integer $K( 2m+2 - frac 1{2m+2})$ won't be an integer.....
.... unless $2m+2|K$.
Okay.... we are halfway done.
When will $2m+2|K$?
I'm embarrassed to admit how long the following took me to figure out but:
$K=prod_{k=1}^m(2k(2k+1) -2)$ and for $2m(2m+1) -2=2(m+1)(2m -1)$ so $2m+2$ will always divide $K$ and this will always be an integer. (Unless we only multiply one term $(2 - frac 12)$.)
answered Nov 19 at 20:02
fleablood
67.4k22684
It is surprising that this is often an integer. But because it is surprising there is probably a simple reason that will be become clear when we try a few examples.
$(2 - frac 12)(3 - frac 13) = 2*3 - frac 32 - frac 23 + frac 1{2*3}$.
Okay, one thing that becomes clear is that if we have $(n - frac 1n)(m - frac 1n)$ our terms are going to be $nm$ (which is clearly an integer) $-frac nm, -frac nm$ and $frac 1{mmn}$. When we add $-frac mn -frac nm$ we get $-(frac {m^2}{mn} + frac {n^2}{mn})=-frac {m^2 + n^2}{nm}$ and we get $mn - frac {m^2 + n^2-1}{mn}$.
Well, no reason that $mn|m^2 + n^2 -1$ but if $m$ and $n$ are consecutive, i..e. $(n = m+1)$ we get $$m(m+1) - frac {m^2 + (m+1)^2 - 1}{m(m+1)} = m(m+1) - frac {2m^2 + 2m}{m(m+1)} = m(m+1) - frac {2m(m+1)}{m(m+1)} = m(m+1) -2$$ and that.... is an integer.
So $(m - frac 1m)((m+1) - frac 1{m+1}$ must be an integer and as integers multiply together to integers we have for a product of even number of terms:
$prod_{n=2}^{2m+ 1} (n-frac 1n) = prod_{k=1}^m [(2k-frac 1{2k})((2k+1)-frac 1{2k+1})] = prod_{k=1}^m (2k(2k+1) -2)=Kin mathbb Z$.
But if we have a product of odd terms:
$prod_{n=2}^{2m+ 2}(n-frac 1n) =(prod_{n=2}^{2m+ 1}(n-frac 1n))((2m+2)-frac 1{2m+2})=K( 2m+2 - frac 1{2m+2})$ and as $( 2m+2 - frac 1{2m+2})$ is not an integer $K( 2m+2 - frac 1{2m+2})$ won't be an integer.....
.... unless $2m+2|K$.
Okay.... we are halfway done.
When will $2m+2|K$?
I'm embarrassed to admit how long the following took me to figure out but:
$K=prod_{k=1}^m(2k(2k+1) -2)$ and for $2m(2m+1) -2=2(m+1)(2m -1)$ so $2m+2$ will always divide $K$ and this will always be an integer. (Unless we only multiply one term $(2 - frac 12)$.)
answered Nov 19 at 20:02
fleablood
67.4k22684
answered Nov 19 at 20:02
fleablood
67.4k22684
answered Nov 19 at 20:02
fleablood
67.4k22684
answered Nov 19 at 20:02
fleablood
67.4k22684
67.4k22684
add a comment |
add a comment |
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Only number that it doesn't evaluate to an integer is $2$.
– Jakobian
Nov 19 at 18:46
My math app outputs 3379030566911999.5 for k=18 but maybe thats due to a bug.
– g3nuine
Nov 19 at 18:52