$r(n)$ is the sum of the remainders when $n$ is divided by $1,2…,n$ …
For a positive interger $n$, let $r(n)$ be the sum of the remainders when $n$ is divided by $1,2, ldots ,n$ respectively. Prove that $r(k)=r(k+1)$ for infinitely many positive intergers $k$.
Here's my try:
By the division algorithm we have $a=bq+r$ and $a-bq=r$. So $r(k)= k(k-1)-(2q_2 + 3q_3+ ldots kq_k$) and $r(k+1)= (k-1)(k+1) -(2p_2 + 3p_3+ ldots kp_k$) where $q_i = lfloor frac {k}{i} rfloor$ and $p_i= lfloor frac {k+1}{i} rfloor$
So if $r(k)=r(k+1)$ then $0= k-1 -( 2(p_2-q_2)+ 3(p_3-q_3) ldots + k(p_k - q_k))$ and I think that $p_n-q_n$ can be equal to $0$ or $1$ depending on if $k+1$ is or is not divisible by $n$. I don't know if my last claim is correct but I think it is.
I would like to receive any hints or help. Thanks in advance.
number-theory divisibility
asked Nov 30 '18 at 2:45
VmimiVmimi
361212
|
show 4 more comments
For a positive interger $n$, let $r(n)$ be the sum of the remainders when $n$ is divided by $1,2, ldots ,n$ respectively. Prove that $r(k)=r(k+1)$ for infinitely many positive intergers $k$.
Here's my try:
By the division algorithm we have $a=bq+r$ and $a-bq=r$. So $r(k)= k(k-1)-(2q_2 + 3q_3+ ldots kq_k$) and $r(k+1)= (k-1)(k+1) -(2p_2 + 3p_3+ ldots kp_k$) where $q_i = lfloor frac {k}{i} rfloor$ and $p_i= lfloor frac {k+1}{i} rfloor$
So if $r(k)=r(k+1)$ then $0= k-1 -( 2(p_2-q_2)+ 3(p_3-q_3) ldots + k(p_k - q_k))$ and I think that $p_n-q_n$ can be equal to $0$ or $1$ depending on if $k+1$ is or is not divisible by $n$. I don't know if my last claim is correct but I think it is.
I would like to receive any hints or help. Thanks in advance.
number-theory divisibility
asked Nov 30 '18 at 2:45
VmimiVmimi
361212
Hint: Use your favourite programming language to calculate and output the values of $r(k)$ up to 100. Find the instances where $r(k)=r(k+1)$. A pattern should emerge and you can possibly formulate a conjecture for an infinite sequence of such occurances, then try to prove that conjecture.
– Ingix
Nov 30 '18 at 11:32
I don't know much about programming
– Vmimi
Nov 30 '18 at 14:52
Then do it by hand until $k=10$. You should find 3 values, maybe that is enough to start the conjecture. I can give you the solution outright, if you want.
– Ingix
Nov 30 '18 at 14:55
I found that $r(2^k)=r(2^k-1)$ for $k=1,2$ and $3$, but couldn't deduce anything else.
– Vmimi
Dec 1 '18 at 19:57
Yes, that's what I had in mind. It's true for all $k$. You can use your formula for the difference $r(k+1)-r(k)$, with $p_n-q_n in {0,1}$ when you take into account the divisibility condition, which is easy to handle for $k+1=2^m$. There is however a small error in your formula, you have forgotten the term $(k+1)p_{k+1}=k+1$ in $r(k+1)$. If you need more tips, say so.
– Ingix
Dec 1 '18 at 21:51
|
show 4 more comments
For a positive interger $n$, let $r(n)$ be the sum of the remainders when $n$ is divided by $1,2, ldots ,n$ respectively. Prove that $r(k)=r(k+1)$ for infinitely many positive intergers $k$.
Here's my try:
By the division algorithm we have $a=bq+r$ and $a-bq=r$. So $r(k)= k(k-1)-(2q_2 + 3q_3+ ldots kq_k$) and $r(k+1)= (k-1)(k+1) -(2p_2 + 3p_3+ ldots kp_k$) where $q_i = lfloor frac {k}{i} rfloor$ and $p_i= lfloor frac {k+1}{i} rfloor$
So if $r(k)=r(k+1)$ then $0= k-1 -( 2(p_2-q_2)+ 3(p_3-q_3) ldots + k(p_k - q_k))$ and I think that $p_n-q_n$ can be equal to $0$ or $1$ depending on if $k+1$ is or is not divisible by $n$. I don't know if my last claim is correct but I think it is.
I would like to receive any hints or help. Thanks in advance.
number-theory divisibility
asked Nov 30 '18 at 2:45
VmimiVmimi
361212
For a positive interger $n$, let $r(n)$ be the sum of the remainders when $n$ is divided by $1,2, ldots ,n$ respectively. Prove that $r(k)=r(k+1)$ for infinitely many positive intergers $k$.
Here's my try:
By the division algorithm we have $a=bq+r$ and $a-bq=r$. So $r(k)= k(k-1)-(2q_2 + 3q_3+ ldots kq_k$) and $r(k+1)= (k-1)(k+1) -(2p_2 + 3p_3+ ldots kp_k$) where $q_i = lfloor frac {k}{i} rfloor$ and $p_i= lfloor frac {k+1}{i} rfloor$
So if $r(k)=r(k+1)$ then $0= k-1 -( 2(p_2-q_2)+ 3(p_3-q_3) ldots + k(p_k - q_k))$ and I think that $p_n-q_n$ can be equal to $0$ or $1$ depending on if $k+1$ is or is not divisible by $n$. I don't know if my last claim is correct but I think it is.
I would like to receive any hints or help. Thanks in advance.
number-theory divisibility
number-theory divisibility
asked Nov 30 '18 at 2:45
VmimiVmimi
361212
asked Nov 30 '18 at 2:45
VmimiVmimi
361212
edited Dec 1 '18 at 23:30
Vmimi
asked Nov 30 '18 at 2:45
VmimiVmimi
361212
asked Nov 30 '18 at 2:45
VmimiVmimi
361212
asked Nov 30 '18 at 2:45
VmimiVmimi
361212
361212
Hint: Use your favourite programming language to calculate and output the values of $r(k)$ up to 100. Find the instances where $r(k)=r(k+1)$. A pattern should emerge and you can possibly formulate a conjecture for an infinite sequence of such occurances, then try to prove that conjecture.
– Ingix
Nov 30 '18 at 11:32
I don't know much about programming
– Vmimi
Nov 30 '18 at 14:52
Then do it by hand until $k=10$. You should find 3 values, maybe that is enough to start the conjecture. I can give you the solution outright, if you want.
– Ingix
Nov 30 '18 at 14:55
I found that $r(2^k)=r(2^k-1)$ for $k=1,2$ and $3$, but couldn't deduce anything else.
– Vmimi
Dec 1 '18 at 19:57
Yes, that's what I had in mind. It's true for all $k$. You can use your formula for the difference $r(k+1)-r(k)$, with $p_n-q_n in {0,1}$ when you take into account the divisibility condition, which is easy to handle for $k+1=2^m$. There is however a small error in your formula, you have forgotten the term $(k+1)p_{k+1}=k+1$ in $r(k+1)$. If you need more tips, say so.
– Ingix
Dec 1 '18 at 21:51
|
show 4 more comments
Hint: Use your favourite programming language to calculate and output the values of $r(k)$ up to 100. Find the instances where $r(k)=r(k+1)$. A pattern should emerge and you can possibly formulate a conjecture for an infinite sequence of such occurances, then try to prove that conjecture.
– Ingix
Nov 30 '18 at 11:32
I don't know much about programming
– Vmimi
Nov 30 '18 at 14:52
Then do it by hand until $k=10$. You should find 3 values, maybe that is enough to start the conjecture. I can give you the solution outright, if you want.
– Ingix
Nov 30 '18 at 14:55
I found that $r(2^k)=r(2^k-1)$ for $k=1,2$ and $3$, but couldn't deduce anything else.
– Vmimi
Dec 1 '18 at 19:57
Yes, that's what I had in mind. It's true for all $k$. You can use your formula for the difference $r(k+1)-r(k)$, with $p_n-q_n in {0,1}$ when you take into account the divisibility condition, which is easy to handle for $k+1=2^m$. There is however a small error in your formula, you have forgotten the term $(k+1)p_{k+1}=k+1$ in $r(k+1)$. If you need more tips, say so.
– Ingix
Dec 1 '18 at 21:51
Hint: Use your favourite programming language to calculate and output the values of $r(k)$ up to 100. Find the instances where $r(k)=r(k+1)$. A pattern should emerge and you can possibly formulate a conjecture for an infinite sequence of such occurances, then try to prove that conjecture.
– Ingix
Nov 30 '18 at 11:32
Hint: Use your favourite programming language to calculate and output the values of $r(k)$ up to 100. Find the instances where $r(k)=r(k+1)$. A pattern should emerge and you can possibly formulate a conjecture for an infinite sequence of such occurances, then try to prove that conjecture.
– Ingix
Nov 30 '18 at 11:32
I don't know much about programming
– Vmimi
Nov 30 '18 at 14:52
I don't know much about programming
– Vmimi
Nov 30 '18 at 14:52
Then do it by hand until $k=10$. You should find 3 values, maybe that is enough to start the conjecture. I can give you the solution outright, if you want.
– Ingix
Nov 30 '18 at 14:55
Then do it by hand until $k=10$. You should find 3 values, maybe that is enough to start the conjecture. I can give you the solution outright, if you want.
– Ingix
Nov 30 '18 at 14:55
I found that $r(2^k)=r(2^k-1)$ for $k=1,2$ and $3$, but couldn't deduce anything else.
– Vmimi
Dec 1 '18 at 19:57
I found that $r(2^k)=r(2^k-1)$ for $k=1,2$ and $3$, but couldn't deduce anything else.
– Vmimi
Dec 1 '18 at 19:57
Yes, that's what I had in mind. It's true for all $k$. You can use your formula for the difference $r(k+1)-r(k)$, with $p_n-q_n in {0,1}$ when you take into account the divisibility condition, which is easy to handle for $k+1=2^m$. There is however a small error in your formula, you have forgotten the term $(k+1)p_{k+1}=k+1$ in $r(k+1)$. If you need more tips, say so.
– Ingix
Dec 1 '18 at 21:51
Yes, that's what I had in mind. It's true for all $k$. You can use your formula for the difference $r(k+1)-r(k)$, with $p_n-q_n in {0,1}$ when you take into account the divisibility condition, which is easy to handle for $k+1=2^m$. There is however a small error in your formula, you have forgotten the term $(k+1)p_{k+1}=k+1$ in $r(k+1)$. If you need more tips, say so.
– Ingix
Dec 1 '18 at 21:51
|
show 4 more comments
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Hint: Use your favourite programming language to calculate and output the values of $r(k)$ up to 100. Find the instances where $r(k)=r(k+1)$. A pattern should emerge and you can possibly formulate a conjecture for an infinite sequence of such occurances, then try to prove that conjecture.
– Ingix
Nov 30 '18 at 11:32
I don't know much about programming
– Vmimi
Nov 30 '18 at 14:52
Then do it by hand until $k=10$. You should find 3 values, maybe that is enough to start the conjecture. I can give you the solution outright, if you want.
– Ingix
Nov 30 '18 at 14:55
I found that $r(2^k)=r(2^k-1)$ for $k=1,2$ and $3$, but couldn't deduce anything else.
– Vmimi
Dec 1 '18 at 19:57
Yes, that's what I had in mind. It's true for all $k$. You can use your formula for the difference $r(k+1)-r(k)$, with $p_n-q_n in {0,1}$ when you take into account the divisibility condition, which is easy to handle for $k+1=2^m$. There is however a small error in your formula, you have forgotten the term $(k+1)p_{k+1}=k+1$ in $r(k+1)$. If you need more tips, say so.
– Ingix
Dec 1 '18 at 21:51