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Does $ sum_i=1^n n^k_i $ determines $(k_1,…,k_n)$?

5

1

$begingroup$

Let $k_1,...,k_ninmathbbN$. Does the power sum
$$
sum_i=1^n n^k_i
$$
uniquely determines the $n$-tuple $(k_1,...,k_n)$?

Remark: In the case $n=2$, this is true. However, when trying to generalize to an arbitrarily sized sum, it doesn't hold. For example,
$$
2^0+2^0+2^2=2^1+2^1+2^1.
$$
I then thought of a fixed sum size, equal to the considered base, but I don't know exactly how to argue or build this bijection. The motivation behind this question comes from trying to determine the $n$-tuple $(k_1,...,k_n)$ from the sum
$$
sum_i=1^n g(k_i),
$$
where $g$ is some constructible function.

sequences-and-series number-theory summation power-series

share|cite|improve this question

asked Mar 18 at 19:22

sam wolfesam wolfe

820525

$endgroup$

add a comment | 

5

1

$begingroup$

Let $k_1,...,k_ninmathbbN$. Does the power sum
$$
sum_i=1^n n^k_i
$$
uniquely determines the $n$-tuple $(k_1,...,k_n)$?

Remark: In the case $n=2$, this is true. However, when trying to generalize to an arbitrarily sized sum, it doesn't hold. For example,
$$
2^0+2^0+2^2=2^1+2^1+2^1.
$$
I then thought of a fixed sum size, equal to the considered base, but I don't know exactly how to argue or build this bijection. The motivation behind this question comes from trying to determine the $n$-tuple $(k_1,...,k_n)$ from the sum
$$
sum_i=1^n g(k_i),
$$
where $g$ is some constructible function.

sequences-and-series number-theory summation power-series

share|cite|improve this question

asked Mar 18 at 19:22

sam wolfesam wolfe

820525

$endgroup$

add a comment | 

$begingroup$

Let $k_1,...,k_ninmathbbN$. Does the power sum
$$
sum_i=1^n n^k_i
$$
uniquely determines the $n$-tuple $(k_1,...,k_n)$?

Remark: In the case $n=2$, this is true. However, when trying to generalize to an arbitrarily sized sum, it doesn't hold. For example,
$$
2^0+2^0+2^2=2^1+2^1+2^1.
$$
I then thought of a fixed sum size, equal to the considered base, but I don't know exactly how to argue or build this bijection. The motivation behind this question comes from trying to determine the $n$-tuple $(k_1,...,k_n)$ from the sum
$$
sum_i=1^n g(k_i),
$$
where $g$ is some constructible function.

sequences-and-series number-theory summation power-series

share|cite|improve this question

asked Mar 18 at 19:22

sam wolfesam wolfe

820525

$endgroup$

share|cite|improve this question

asked Mar 18 at 19:22

sam wolfesam wolfe

820525

share|cite|improve this question

asked Mar 18 at 19:22

sam wolfesam wolfe

820525

asked Mar 18 at 19:22

sam wolfesam wolfe

820525

asked Mar 18 at 19:22

sam wolfesam wolfe

820525

1 Answer
1

active

oldest

votes

3

$begingroup$

Obviously you can only hope to determine the $n$-tuple up to permutation, and you can only hope to determine even that when $n>1$. So I'll assume that's what you want. And as you see, it doesn't work in general when the number of $k_i$ is not equal to the base of the exponent.

Then the answer is yes, if you know $n$. It's a corollary of the uniqueness of base-$n$ representations. You can read off the $k_i$ from the digits of the number $s = sum_i=1^n n^k_i$ written base $n$. The $j^smalltextth$ digit ($j=0$ is the first digit) will tell you how many $k_i$ are equal to $j$, unless there is exactly one $1$ and no other nonzero digits in $s$, in which case all the $k_i$ are equal to $j-1$ where $j$ is the place where the $1$ occurs.

share|cite|improve this answer

edited Mar 18 at 20:00

answered Mar 18 at 19:34

cspruncsprun

2,755210

$endgroup$

add a comment | 

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3

$begingroup$

Obviously you can only hope to determine the $n$-tuple up to permutation, and you can only hope to determine even that when $n>1$. So I'll assume that's what you want. And as you see, it doesn't work in general when the number of $k_i$ is not equal to the base of the exponent.

Then the answer is yes, if you know $n$. It's a corollary of the uniqueness of base-$n$ representations. You can read off the $k_i$ from the digits of the number $s = sum_i=1^n n^k_i$ written base $n$. The $j^smalltextth$ digit ($j=0$ is the first digit) will tell you how many $k_i$ are equal to $j$, unless there is exactly one $1$ and no other nonzero digits in $s$, in which case all the $k_i$ are equal to $j-1$ where $j$ is the place where the $1$ occurs.

share|cite|improve this answer

edited Mar 18 at 20:00

answered Mar 18 at 19:34

cspruncsprun

2,755210

$endgroup$

add a comment | 

3

$begingroup$

Obviously you can only hope to determine the $n$-tuple up to permutation, and you can only hope to determine even that when $n>1$. So I'll assume that's what you want. And as you see, it doesn't work in general when the number of $k_i$ is not equal to the base of the exponent.

Then the answer is yes, if you know $n$. It's a corollary of the uniqueness of base-$n$ representations. You can read off the $k_i$ from the digits of the number $s = sum_i=1^n n^k_i$ written base $n$. The $j^smalltextth$ digit ($j=0$ is the first digit) will tell you how many $k_i$ are equal to $j$, unless there is exactly one $1$ and no other nonzero digits in $s$, in which case all the $k_i$ are equal to $j-1$ where $j$ is the place where the $1$ occurs.

share|cite|improve this answer

edited Mar 18 at 20:00

answered Mar 18 at 19:34

cspruncsprun

2,755210

$endgroup$

add a comment | 

$begingroup$

Obviously you can only hope to determine the $n$-tuple up to permutation, and you can only hope to determine even that when $n>1$. So I'll assume that's what you want. And as you see, it doesn't work in general when the number of $k_i$ is not equal to the base of the exponent.

Then the answer is yes, if you know $n$. It's a corollary of the uniqueness of base-$n$ representations. You can read off the $k_i$ from the digits of the number $s = sum_i=1^n n^k_i$ written base $n$. The $j^smalltextth$ digit ($j=0$ is the first digit) will tell you how many $k_i$ are equal to $j$, unless there is exactly one $1$ and no other nonzero digits in $s$, in which case all the $k_i$ are equal to $j-1$ where $j$ is the place where the $1$ occurs.

share|cite|improve this answer

edited Mar 18 at 20:00

answered Mar 18 at 19:34

cspruncsprun

2,755210

$endgroup$

share|cite|improve this answer

edited Mar 18 at 20:00

answered Mar 18 at 19:34

cspruncsprun

2,755210

answered Mar 18 at 19:34

cspruncsprun

2,755210

answered Mar 18 at 19:34

cspruncsprun

2,755210