Permutations, products, and unit fractions
$begingroup$
Here's a question motivated by some related MathOverflow questions of Zhi-Wei Sun.
Show that, for any $n ge 1$, there is a permutation of ${1,2,ldots, n}$, i.e., some $pi in S_n$, such that $$sum_{k=1}^n frac{1}{k cdot pi(k)} = 1.$$
combinatorics number-theory permutations egyptian-fractions
asked Nov 25 '18 at 21:57
Brian HopkinsBrian Hopkins
508615
$endgroup$
add a comment |
$begingroup$
Here's a question motivated by some related MathOverflow questions of Zhi-Wei Sun.
Show that, for any $n ge 1$, there is a permutation of ${1,2,ldots, n}$, i.e., some $pi in S_n$, such that $$sum_{k=1}^n frac{1}{k cdot pi(k)} = 1.$$
combinatorics number-theory permutations egyptian-fractions
asked Nov 25 '18 at 21:57
Brian HopkinsBrian Hopkins
508615
$endgroup$
add a comment |
$begingroup$
Here's a question motivated by some related MathOverflow questions of Zhi-Wei Sun.
Show that, for any $n ge 1$, there is a permutation of ${1,2,ldots, n}$, i.e., some $pi in S_n$, such that $$sum_{k=1}^n frac{1}{k cdot pi(k)} = 1.$$
combinatorics number-theory permutations egyptian-fractions
asked Nov 25 '18 at 21:57
Brian HopkinsBrian Hopkins
508615
$endgroup$
Here's a question motivated by some related MathOverflow questions of Zhi-Wei Sun.
Show that, for any $n ge 1$, there is a permutation of ${1,2,ldots, n}$, i.e., some $pi in S_n$, such that $$sum_{k=1}^n frac{1}{k cdot pi(k)} = 1.$$
combinatorics number-theory permutations egyptian-fractions
combinatorics number-theory permutations egyptian-fractions
asked Nov 25 '18 at 21:57
Brian HopkinsBrian Hopkins
508615
asked Nov 25 '18 at 21:57
Brian HopkinsBrian Hopkins
508615
asked Nov 25 '18 at 21:57
Brian HopkinsBrian Hopkins
508615
asked Nov 25 '18 at 21:57
Brian HopkinsBrian Hopkins
508615
asked Nov 25 '18 at 21:57
Brian HopkinsBrian Hopkins
508615
508615
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Take $pi(k)=k+1$ except for $pi(n)=1$. Then
$$sum_{k=1}^nfrac1{kpi(k)}=frac1n+sum_{k=1}^{n-1}frac1{k(k+1)}
=frac1n+sum_{k=1}^{n-1}left(frac1k-frac1{k+1}right)=1.
$$
answered Nov 25 '18 at 22:09
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
$begingroup$
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
$endgroup$
– Brian Hopkins
Nov 26 '18 at 15:13
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $pi(k)=k+1$ except for $pi(n)=1$. Then
$$sum_{k=1}^nfrac1{kpi(k)}=frac1n+sum_{k=1}^{n-1}frac1{k(k+1)}
=frac1n+sum_{k=1}^{n-1}left(frac1k-frac1{k+1}right)=1.
$$
answered Nov 25 '18 at 22:09
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
$begingroup$
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
$endgroup$
– Brian Hopkins
Nov 26 '18 at 15:13
add a comment |
$begingroup$
Take $pi(k)=k+1$ except for $pi(n)=1$. Then
$$sum_{k=1}^nfrac1{kpi(k)}=frac1n+sum_{k=1}^{n-1}frac1{k(k+1)}
=frac1n+sum_{k=1}^{n-1}left(frac1k-frac1{k+1}right)=1.
$$
answered Nov 25 '18 at 22:09
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
$begingroup$
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
$endgroup$
– Brian Hopkins
Nov 26 '18 at 15:13
add a comment |
$begingroup$
Take $pi(k)=k+1$ except for $pi(n)=1$. Then
$$sum_{k=1}^nfrac1{kpi(k)}=frac1n+sum_{k=1}^{n-1}frac1{k(k+1)}
=frac1n+sum_{k=1}^{n-1}left(frac1k-frac1{k+1}right)=1.
$$
answered Nov 25 '18 at 22:09
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
Take $pi(k)=k+1$ except for $pi(n)=1$. Then
$$sum_{k=1}^nfrac1{kpi(k)}=frac1n+sum_{k=1}^{n-1}frac1{k(k+1)}
=frac1n+sum_{k=1}^{n-1}left(frac1k-frac1{k+1}right)=1.
$$
answered Nov 25 '18 at 22:09
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Nov 25 '18 at 22:09
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Nov 25 '18 at 22:09
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Nov 25 '18 at 22:09
Lord Shark the UnknownLord Shark the Unknown
102k1059132
102k1059132
$begingroup$
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
$endgroup$
– Brian Hopkins
Nov 26 '18 at 15:13
add a comment |
$begingroup$
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
$endgroup$
– Brian Hopkins
Nov 26 '18 at 15:13
$begingroup$
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
$endgroup$
– Brian Hopkins
Nov 26 '18 at 15:13
$begingroup$
Very quick and clean, thanks. Writing your permutation in "long form," $(2,3,ldots,n,1)$ always works. See a follow-up question on MathOverflow that includes data on how many permutations satisfy this sum condition.
$endgroup$
– Brian Hopkins
Nov 26 '18 at 15:13
add a comment |
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