Mixing
Algebraic Closed Form for $sum_{n=1}^{k}left( n- 3 lfloor frac{n-1}{3} rfloorright)$
$begingroup$
Let's look at the following sequence:
$a_n=left{1,2,3,1,2,3,1,2,3,1,2,3,...right}$
I'm trying to calculate:
$$sum_{n=1}^{k} a_n$$
Attempts:
I have a Closed Form for this sequence.
$$a_n=n- 3 bigglfloor frac{n-1}{3} biggrfloor$$
The problem is, I'm looking for a closed form for this summation:
$$sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$
Is it possible?
algebra-precalculus summation closed-form floor-function
edited Jan 8 at 9:28
DavidG
2,1851720
asked Jan 8 at 9:12
ElementaryElementary
364110
$endgroup$
add a comment |
$begingroup$
Let's look at the following sequence:
$a_n=left{1,2,3,1,2,3,1,2,3,1,2,3,...right}$
I'm trying to calculate:
$$sum_{n=1}^{k} a_n$$
Attempts:
I have a Closed Form for this sequence.
$$a_n=n- 3 bigglfloor frac{n-1}{3} biggrfloor$$
The problem is, I'm looking for a closed form for this summation:
$$sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$
Is it possible?
algebra-precalculus summation closed-form floor-function
edited Jan 8 at 9:28
DavidG
2,1851720
asked Jan 8 at 9:12
ElementaryElementary
364110
$endgroup$
$begingroup$
There is a closed form with a bunch of floor functions.
$endgroup$
– Claude Leibovici
Jan 8 at 9:52
$begingroup$
You recieved 5 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:49
$begingroup$
@5xum I've been upvote before. Now, I accept. By the way, If you add the most recent result to your answer, it will be perfect. Regards.
$endgroup$
– Elementary
Jan 15 at 9:03
add a comment |
$begingroup$
Let's look at the following sequence:
$a_n=left{1,2,3,1,2,3,1,2,3,1,2,3,...right}$
I'm trying to calculate:
$$sum_{n=1}^{k} a_n$$
Attempts:
I have a Closed Form for this sequence.
$$a_n=n- 3 bigglfloor frac{n-1}{3} biggrfloor$$
The problem is, I'm looking for a closed form for this summation:
$$sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$
Is it possible?
algebra-precalculus summation closed-form floor-function
edited Jan 8 at 9:28
DavidG
2,1851720
asked Jan 8 at 9:12
ElementaryElementary
364110
$endgroup$
Let's look at the following sequence:
$a_n=left{1,2,3,1,2,3,1,2,3,1,2,3,...right}$
I'm trying to calculate:
$$sum_{n=1}^{k} a_n$$
Attempts:
I have a Closed Form for this sequence.
$$a_n=n- 3 bigglfloor frac{n-1}{3} biggrfloor$$
The problem is, I'm looking for a closed form for this summation:
$$sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$
Is it possible?
algebra-precalculus summation closed-form floor-function
algebra-precalculus summation closed-form floor-function
edited Jan 8 at 9:28
DavidG
2,1851720
asked Jan 8 at 9:12
ElementaryElementary
364110
edited Jan 8 at 9:28
DavidG
2,1851720
asked Jan 8 at 9:12
ElementaryElementary
364110
edited Jan 8 at 9:28
DavidG
2,1851720
edited Jan 8 at 9:28
DavidG
2,1851720
edited Jan 8 at 9:28
DavidG
2,1851720
2,1851720
asked Jan 8 at 9:12
ElementaryElementary
364110
asked Jan 8 at 9:12
ElementaryElementary
364110
asked Jan 8 at 9:12
ElementaryElementary
364110
364110
$begingroup$
There is a closed form with a bunch of floor functions.
$endgroup$
– Claude Leibovici
Jan 8 at 9:52
$begingroup$
You recieved 5 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:49
$begingroup$
@5xum I've been upvote before. Now, I accept. By the way, If you add the most recent result to your answer, it will be perfect. Regards.
$endgroup$
– Elementary
Jan 15 at 9:03
add a comment |
$begingroup$
There is a closed form with a bunch of floor functions.
$endgroup$
– Claude Leibovici
Jan 8 at 9:52
$begingroup$
You recieved 5 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:49
$begingroup$
@5xum I've been upvote before. Now, I accept. By the way, If you add the most recent result to your answer, it will be perfect. Regards.
$endgroup$
– Elementary
Jan 15 at 9:03
$begingroup$
There is a closed form with a bunch of floor functions.
$endgroup$
– Claude Leibovici
Jan 8 at 9:52
$begingroup$
There is a closed form with a bunch of floor functions.
$endgroup$
– Claude Leibovici
Jan 8 at 9:52
$begingroup$
You recieved 5 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:49
$begingroup$
You recieved 5 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:49
$begingroup$
@5xum I've been upvote before. Now, I accept. By the way, If you add the most recent result to your answer, it will be perfect. Regards.
$endgroup$
– Elementary
Jan 15 at 9:03
$begingroup$
@5xum I've been upvote before. Now, I accept. By the way, If you add the most recent result to your answer, it will be perfect. Regards.
$endgroup$
– Elementary
Jan 15 at 9:03
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Writing down a couple of the sums:
$$1,3,6,7,9,12,13,15,18,dots$$
and comparing that to the sequence$$1,3,5,7,9cdots$$
gives you a clue that the difference between the arithmetic sequence and the sequence you want to describe is simply $$0,0,1,0,0,1,0,0,1,0,0,1dots$$ which is a sequence you can describe in closed form in a way similar to $a_n$.
That is, you can see that $$sum_{i=n}^k a_n = 2k-1 + b_k$$
where $b_k$ is equal to $1$ if $k$ is divisible by $3$ and $0$ otherwise.
You can express $b_n$ algebraically by taking $a_n$ and any function for which $f(1)=f(2)=0$ and $f(3)=1$, and you have $b_n=f(a_n)$.
I can't think of any "elegant" function $f$ at the moment, but a quadratic polynomial can surely do it, since we only have a restriction on three points. The quadratic polynomial that satisfies $f(1)=f(2)=0$ and $f(3)=1$ is $$f(x)=frac12x^2-frac32 x + 1.$$
Edit:
Thanks to BarryCipra, a nicer function (more in the spirit of your solution) for $b_k$ is
$$b_k = leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
answered Jan 8 at 9:19
5xum5xum
90.5k394161
$endgroup$
2
$begingroup$
How about $b_k=lfloor1+lfloor{kover3}rfloor-{kover3}rfloor$?
$endgroup$
– Barry Cipra
Jan 8 at 10:10
$begingroup$
@BarryCipra That works, yeah. I added it to my answer (hope you don't mind)
$endgroup$
– 5xum
Jan 8 at 10:22
$begingroup$
My pleasure. But you mean $b_k$, not $b_n$.
$endgroup$
– Barry Cipra
Jan 8 at 10:25
$begingroup$
And What exactly is the result? $sum_{i=1}^{n}a_n$
$endgroup$
– Elementary
Jan 8 at 11:24
$begingroup$
@Beginner, a complete answer is $$sum_{n=1}^ka_n=2k-1+leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$ Is that what your comment is asking for? If you like, it can be simplified to $$sum_{n=1}^ka_n=2k+leftlfloor leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
$endgroup$
– Barry Cipra
Jan 8 at 12:10
|
show 1 more comment
$begingroup$
Subtracting the "average" sequence
$$2,2,2,2,2,2,2,2,2,cdots$$ you get
$$-1,0,1,-1,0,1,-1,0,1,cdots$$
which sums as a periodic one
$$-1,-1,0,-1,-1,0,-1,-1,0,cdots$$
The latter can be expressed as
$$frac{(n-1)bmod3-nbmod 3-2}3.$$
So globally,
$$2n+frac{(n-1)bmod3-nbmod 3-2}3.$$
answered Jan 8 at 9:35
Yves DaoustYves Daoust
126k672225
$endgroup$
$begingroup$
(+1) , It would be great if we found a closed form that could be expressed by the floor function
$endgroup$
– Elementary
Jan 8 at 9:43
$begingroup$
@Beginner: modulo and floor are interchangeable, with simple arithmetic.
$endgroup$
– Yves Daoust
Jan 8 at 9:46
add a comment |
$begingroup$
If you consider$$b_k=sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$ you could see, after some simple manipulationd that it corresponds to the recurrence relation
$$b_k=b_{k-1}+b_{k-3}-b_{k-4}qquad text{with}qquad b_1=1, b_2=3, b_3=6, b_4=7$$ Using the conventional method you should end with
$$b_k=2k-frac{2}{3} left(1-cos left(frac{2 pi k}{3}right)right)$$
Edit
If you enjoy the floor function, using a CAS, I got for $b_k$ the small monster
$$frac{1}{2} left(k^2+k-3 leftlfloor frac{k}{3}rightrfloor ^2-3 leftlfloor
frac{k-2}{3}rightrfloor left(leftlfloor frac{k-2}{3}rightrfloor
+1right)-3 leftlfloor frac{k-1}{3}rightrfloor left(leftlfloor
frac{k-1}{3}rightrfloor +1right)+3 leftlfloor frac{k}{3}rightrfloor
right)$$
Which one do you prefer ?
answered Jan 8 at 9:47
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
$begingroup$
Of course, Floor function :)
$endgroup$
– Elementary
Jan 8 at 10:06
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@Beginner. Joke or serious ?
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– Claude Leibovici
Jan 8 at 10:09
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Both are beautiful. :) Thank you very much (+)
$endgroup$
– Elementary
Jan 8 at 10:34
add a comment |
$begingroup$
If $n equiv 0 pmod{3}$, i.e. say $n=3s$ (where $s geq 1$), then $a_n=3s-3lfloor s-frac{1}{3}rfloor=3s-3(s-1)=3$.
If $n equiv 1 pmod{3}$, i.e. say $n=3s+1$ (where $s geq 0$), then $a_n=3s+1-3lfloor srfloor=1$.
If $n equiv 2 pmod{3}$, i.e. say $n=3s+2$ (where $s geq 0$), then $a_n=3s+2-3lfloor s+frac{1}{3}rfloor=2$.
So depending on what $k$ is we can count the number of terms which are $1's$, $2's$ and $3's$.
For $k=1,2$, the sum will be $color{red}{1}$ and $color{red}{3}$, respectively. So for $k geq 3$, we do the following:
If $k=3t$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)=6t=color{blue}{2k}.$$
If $k=3t+1$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+1}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+1=6t+1=color{blue}{2k-1}.$$
Likewise we can get the expressions for $k=3t+2$ as
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+2}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+(1+2)=6t+3=color{blue}{2k-1}.$$
answered Jan 8 at 9:23
Anurag AAnurag A
26k12249
$endgroup$
$begingroup$
I think, if there is an exist a closed form, it should be expressed by floor function.
$endgroup$
– Elementary
Jan 8 at 9:26
$begingroup$
@Beginner you are summing up to $k$ terms, so the answer is a function of $k$. I am not sure what you mean by the closed form has to be in terms of floor function.
$endgroup$
– Anurag A
Jan 8 at 9:29
$begingroup$
No, I didn't do this.
$endgroup$
– Elementary
Jan 8 at 9:32
add a comment |
$begingroup$
Sorry can't comment on @5xum's answer. Building on the quadratic function we can derive:
$S_k=frac{1}{2}((k-3lfloorfrac{k}{3}rfloor)^2+k+9lfloorfrac{k}{3}rfloor)$, which is pretty neat.
To derive this, note that from @5xum's answer, $S_k=2k-1+frac{1}{2}(x^2 -3x+2)$ where $x=k-3lfloorfrac{k}{3}rfloor$. We can write $k=x+3lfloorfrac{k}{3}rfloor$, so that:
$$S_k=frac{1}{2}(x^2 -3x+2+4x+12lfloorfrac{k}{3}rfloor-2)=frac{1}{2}(x^2+x+12lfloorfrac{k}{3}rfloor). $$
Substituting $x$ back yields the result.
answered Jan 8 at 10:04
user46666user46666
335
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Writing down a couple of the sums:
$$1,3,6,7,9,12,13,15,18,dots$$
and comparing that to the sequence$$1,3,5,7,9cdots$$
gives you a clue that the difference between the arithmetic sequence and the sequence you want to describe is simply $$0,0,1,0,0,1,0,0,1,0,0,1dots$$ which is a sequence you can describe in closed form in a way similar to $a_n$.
That is, you can see that $$sum_{i=n}^k a_n = 2k-1 + b_k$$
where $b_k$ is equal to $1$ if $k$ is divisible by $3$ and $0$ otherwise.
You can express $b_n$ algebraically by taking $a_n$ and any function for which $f(1)=f(2)=0$ and $f(3)=1$, and you have $b_n=f(a_n)$.
I can't think of any "elegant" function $f$ at the moment, but a quadratic polynomial can surely do it, since we only have a restriction on three points. The quadratic polynomial that satisfies $f(1)=f(2)=0$ and $f(3)=1$ is $$f(x)=frac12x^2-frac32 x + 1.$$
Edit:
Thanks to BarryCipra, a nicer function (more in the spirit of your solution) for $b_k$ is
$$b_k = leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
answered Jan 8 at 9:19
5xum5xum
90.5k394161
$endgroup$
2
$begingroup$
How about $b_k=lfloor1+lfloor{kover3}rfloor-{kover3}rfloor$?
$endgroup$
– Barry Cipra
Jan 8 at 10:10
$begingroup$
@BarryCipra That works, yeah. I added it to my answer (hope you don't mind)
$endgroup$
– 5xum
Jan 8 at 10:22
$begingroup$
My pleasure. But you mean $b_k$, not $b_n$.
$endgroup$
– Barry Cipra
Jan 8 at 10:25
$begingroup$
And What exactly is the result? $sum_{i=1}^{n}a_n$
$endgroup$
– Elementary
Jan 8 at 11:24
$begingroup$
@Beginner, a complete answer is $$sum_{n=1}^ka_n=2k-1+leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$ Is that what your comment is asking for? If you like, it can be simplified to $$sum_{n=1}^ka_n=2k+leftlfloor leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
$endgroup$
– Barry Cipra
Jan 8 at 12:10
|
show 1 more comment
$begingroup$
Writing down a couple of the sums:
$$1,3,6,7,9,12,13,15,18,dots$$
and comparing that to the sequence$$1,3,5,7,9cdots$$
gives you a clue that the difference between the arithmetic sequence and the sequence you want to describe is simply $$0,0,1,0,0,1,0,0,1,0,0,1dots$$ which is a sequence you can describe in closed form in a way similar to $a_n$.
That is, you can see that $$sum_{i=n}^k a_n = 2k-1 + b_k$$
where $b_k$ is equal to $1$ if $k$ is divisible by $3$ and $0$ otherwise.
You can express $b_n$ algebraically by taking $a_n$ and any function for which $f(1)=f(2)=0$ and $f(3)=1$, and you have $b_n=f(a_n)$.
I can't think of any "elegant" function $f$ at the moment, but a quadratic polynomial can surely do it, since we only have a restriction on three points. The quadratic polynomial that satisfies $f(1)=f(2)=0$ and $f(3)=1$ is $$f(x)=frac12x^2-frac32 x + 1.$$
Edit:
Thanks to BarryCipra, a nicer function (more in the spirit of your solution) for $b_k$ is
$$b_k = leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
answered Jan 8 at 9:19
5xum5xum
90.5k394161
$endgroup$
2
$begingroup$
How about $b_k=lfloor1+lfloor{kover3}rfloor-{kover3}rfloor$?
$endgroup$
– Barry Cipra
Jan 8 at 10:10
$begingroup$
@BarryCipra That works, yeah. I added it to my answer (hope you don't mind)
$endgroup$
– 5xum
Jan 8 at 10:22
$begingroup$
My pleasure. But you mean $b_k$, not $b_n$.
$endgroup$
– Barry Cipra
Jan 8 at 10:25
$begingroup$
And What exactly is the result? $sum_{i=1}^{n}a_n$
$endgroup$
– Elementary
Jan 8 at 11:24
$begingroup$
@Beginner, a complete answer is $$sum_{n=1}^ka_n=2k-1+leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$ Is that what your comment is asking for? If you like, it can be simplified to $$sum_{n=1}^ka_n=2k+leftlfloor leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
$endgroup$
– Barry Cipra
Jan 8 at 12:10
|
show 1 more comment
$begingroup$
Writing down a couple of the sums:
$$1,3,6,7,9,12,13,15,18,dots$$
and comparing that to the sequence$$1,3,5,7,9cdots$$
gives you a clue that the difference between the arithmetic sequence and the sequence you want to describe is simply $$0,0,1,0,0,1,0,0,1,0,0,1dots$$ which is a sequence you can describe in closed form in a way similar to $a_n$.
That is, you can see that $$sum_{i=n}^k a_n = 2k-1 + b_k$$
where $b_k$ is equal to $1$ if $k$ is divisible by $3$ and $0$ otherwise.
You can express $b_n$ algebraically by taking $a_n$ and any function for which $f(1)=f(2)=0$ and $f(3)=1$, and you have $b_n=f(a_n)$.
I can't think of any "elegant" function $f$ at the moment, but a quadratic polynomial can surely do it, since we only have a restriction on three points. The quadratic polynomial that satisfies $f(1)=f(2)=0$ and $f(3)=1$ is $$f(x)=frac12x^2-frac32 x + 1.$$
Edit:
Thanks to BarryCipra, a nicer function (more in the spirit of your solution) for $b_k$ is
$$b_k = leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
answered Jan 8 at 9:19
5xum5xum
90.5k394161
$endgroup$
Writing down a couple of the sums:
$$1,3,6,7,9,12,13,15,18,dots$$
and comparing that to the sequence$$1,3,5,7,9cdots$$
gives you a clue that the difference between the arithmetic sequence and the sequence you want to describe is simply $$0,0,1,0,0,1,0,0,1,0,0,1dots$$ which is a sequence you can describe in closed form in a way similar to $a_n$.
That is, you can see that $$sum_{i=n}^k a_n = 2k-1 + b_k$$
where $b_k$ is equal to $1$ if $k$ is divisible by $3$ and $0$ otherwise.
You can express $b_n$ algebraically by taking $a_n$ and any function for which $f(1)=f(2)=0$ and $f(3)=1$, and you have $b_n=f(a_n)$.
I can't think of any "elegant" function $f$ at the moment, but a quadratic polynomial can surely do it, since we only have a restriction on three points. The quadratic polynomial that satisfies $f(1)=f(2)=0$ and $f(3)=1$ is $$f(x)=frac12x^2-frac32 x + 1.$$
Edit:
Thanks to BarryCipra, a nicer function (more in the spirit of your solution) for $b_k$ is
$$b_k = leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
answered Jan 8 at 9:19
5xum5xum
90.5k394161
edited Jan 8 at 14:14
answered Jan 8 at 9:19
5xum5xum
90.5k394161
answered Jan 8 at 9:19
5xum5xum
90.5k394161
answered Jan 8 at 9:19
5xum5xum
90.5k394161
90.5k394161
2
$begingroup$
How about $b_k=lfloor1+lfloor{kover3}rfloor-{kover3}rfloor$?
$endgroup$
– Barry Cipra
Jan 8 at 10:10
$begingroup$
@BarryCipra That works, yeah. I added it to my answer (hope you don't mind)
$endgroup$
– 5xum
Jan 8 at 10:22
$begingroup$
My pleasure. But you mean $b_k$, not $b_n$.
$endgroup$
– Barry Cipra
Jan 8 at 10:25
$begingroup$
And What exactly is the result? $sum_{i=1}^{n}a_n$
$endgroup$
– Elementary
Jan 8 at 11:24
$begingroup$
@Beginner, a complete answer is $$sum_{n=1}^ka_n=2k-1+leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$ Is that what your comment is asking for? If you like, it can be simplified to $$sum_{n=1}^ka_n=2k+leftlfloor leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
$endgroup$
– Barry Cipra
Jan 8 at 12:10
|
show 1 more comment
2
$begingroup$
How about $b_k=lfloor1+lfloor{kover3}rfloor-{kover3}rfloor$?
$endgroup$
– Barry Cipra
Jan 8 at 10:10
$begingroup$
@BarryCipra That works, yeah. I added it to my answer (hope you don't mind)
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– 5xum
Jan 8 at 10:22
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My pleasure. But you mean $b_k$, not $b_n$.
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– Barry Cipra
Jan 8 at 10:25
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And What exactly is the result? $sum_{i=1}^{n}a_n$
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– Elementary
Jan 8 at 11:24
$begingroup$
@Beginner, a complete answer is $$sum_{n=1}^ka_n=2k-1+leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$ Is that what your comment is asking for? If you like, it can be simplified to $$sum_{n=1}^ka_n=2k+leftlfloor leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
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– Barry Cipra
Jan 8 at 12:10
2
2
$begingroup$
How about $b_k=lfloor1+lfloor{kover3}rfloor-{kover3}rfloor$?
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– Barry Cipra
Jan 8 at 10:10
$begingroup$
How about $b_k=lfloor1+lfloor{kover3}rfloor-{kover3}rfloor$?
$endgroup$
– Barry Cipra
Jan 8 at 10:10
$begingroup$
@BarryCipra That works, yeah. I added it to my answer (hope you don't mind)
$endgroup$
– 5xum
Jan 8 at 10:22
$begingroup$
@BarryCipra That works, yeah. I added it to my answer (hope you don't mind)
$endgroup$
– 5xum
Jan 8 at 10:22
$begingroup$
My pleasure. But you mean $b_k$, not $b_n$.
$endgroup$
– Barry Cipra
Jan 8 at 10:25
$begingroup$
My pleasure. But you mean $b_k$, not $b_n$.
$endgroup$
– Barry Cipra
Jan 8 at 10:25
$begingroup$
And What exactly is the result? $sum_{i=1}^{n}a_n$
$endgroup$
– Elementary
Jan 8 at 11:24
$begingroup$
And What exactly is the result? $sum_{i=1}^{n}a_n$
$endgroup$
– Elementary
Jan 8 at 11:24
$begingroup$
@Beginner, a complete answer is $$sum_{n=1}^ka_n=2k-1+leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$ Is that what your comment is asking for? If you like, it can be simplified to $$sum_{n=1}^ka_n=2k+leftlfloor leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
$endgroup$
– Barry Cipra
Jan 8 at 12:10
$begingroup$
@Beginner, a complete answer is $$sum_{n=1}^ka_n=2k-1+leftlfloor 1 + leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$ Is that what your comment is asking for? If you like, it can be simplified to $$sum_{n=1}^ka_n=2k+leftlfloor leftlfloorfrac k3rightrfloor - frac k3rightrfloor$$
$endgroup$
– Barry Cipra
Jan 8 at 12:10
|
show 1 more comment
$begingroup$
Subtracting the "average" sequence
$$2,2,2,2,2,2,2,2,2,cdots$$ you get
$$-1,0,1,-1,0,1,-1,0,1,cdots$$
which sums as a periodic one
$$-1,-1,0,-1,-1,0,-1,-1,0,cdots$$
The latter can be expressed as
$$frac{(n-1)bmod3-nbmod 3-2}3.$$
So globally,
$$2n+frac{(n-1)bmod3-nbmod 3-2}3.$$
answered Jan 8 at 9:35
Yves DaoustYves Daoust
126k672225
$endgroup$
$begingroup$
(+1) , It would be great if we found a closed form that could be expressed by the floor function
$endgroup$
– Elementary
Jan 8 at 9:43
$begingroup$
@Beginner: modulo and floor are interchangeable, with simple arithmetic.
$endgroup$
– Yves Daoust
Jan 8 at 9:46
add a comment |
$begingroup$
Subtracting the "average" sequence
$$2,2,2,2,2,2,2,2,2,cdots$$ you get
$$-1,0,1,-1,0,1,-1,0,1,cdots$$
which sums as a periodic one
$$-1,-1,0,-1,-1,0,-1,-1,0,cdots$$
The latter can be expressed as
$$frac{(n-1)bmod3-nbmod 3-2}3.$$
So globally,
$$2n+frac{(n-1)bmod3-nbmod 3-2}3.$$
answered Jan 8 at 9:35
Yves DaoustYves Daoust
126k672225
$endgroup$
$begingroup$
(+1) , It would be great if we found a closed form that could be expressed by the floor function
$endgroup$
– Elementary
Jan 8 at 9:43
$begingroup$
@Beginner: modulo and floor are interchangeable, with simple arithmetic.
$endgroup$
– Yves Daoust
Jan 8 at 9:46
add a comment |
$begingroup$
Subtracting the "average" sequence
$$2,2,2,2,2,2,2,2,2,cdots$$ you get
$$-1,0,1,-1,0,1,-1,0,1,cdots$$
which sums as a periodic one
$$-1,-1,0,-1,-1,0,-1,-1,0,cdots$$
The latter can be expressed as
$$frac{(n-1)bmod3-nbmod 3-2}3.$$
So globally,
$$2n+frac{(n-1)bmod3-nbmod 3-2}3.$$
answered Jan 8 at 9:35
Yves DaoustYves Daoust
126k672225
$endgroup$
Subtracting the "average" sequence
$$2,2,2,2,2,2,2,2,2,cdots$$ you get
$$-1,0,1,-1,0,1,-1,0,1,cdots$$
which sums as a periodic one
$$-1,-1,0,-1,-1,0,-1,-1,0,cdots$$
The latter can be expressed as
$$frac{(n-1)bmod3-nbmod 3-2}3.$$
So globally,
$$2n+frac{(n-1)bmod3-nbmod 3-2}3.$$
answered Jan 8 at 9:35
Yves DaoustYves Daoust
126k672225
answered Jan 8 at 9:35
Yves DaoustYves Daoust
126k672225
answered Jan 8 at 9:35
Yves DaoustYves Daoust
126k672225
answered Jan 8 at 9:35
Yves DaoustYves Daoust
126k672225
126k672225
$begingroup$
(+1) , It would be great if we found a closed form that could be expressed by the floor function
$endgroup$
– Elementary
Jan 8 at 9:43
$begingroup$
@Beginner: modulo and floor are interchangeable, with simple arithmetic.
$endgroup$
– Yves Daoust
Jan 8 at 9:46
add a comment |
$begingroup$
(+1) , It would be great if we found a closed form that could be expressed by the floor function
$endgroup$
– Elementary
Jan 8 at 9:43
$begingroup$
@Beginner: modulo and floor are interchangeable, with simple arithmetic.
$endgroup$
– Yves Daoust
Jan 8 at 9:46
$begingroup$
(+1) , It would be great if we found a closed form that could be expressed by the floor function
$endgroup$
– Elementary
Jan 8 at 9:43
$begingroup$
(+1) , It would be great if we found a closed form that could be expressed by the floor function
$endgroup$
– Elementary
Jan 8 at 9:43
$begingroup$
@Beginner: modulo and floor are interchangeable, with simple arithmetic.
$endgroup$
– Yves Daoust
Jan 8 at 9:46
$begingroup$
@Beginner: modulo and floor are interchangeable, with simple arithmetic.
$endgroup$
– Yves Daoust
Jan 8 at 9:46
add a comment |
$begingroup$
If you consider$$b_k=sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$ you could see, after some simple manipulationd that it corresponds to the recurrence relation
$$b_k=b_{k-1}+b_{k-3}-b_{k-4}qquad text{with}qquad b_1=1, b_2=3, b_3=6, b_4=7$$ Using the conventional method you should end with
$$b_k=2k-frac{2}{3} left(1-cos left(frac{2 pi k}{3}right)right)$$
Edit
If you enjoy the floor function, using a CAS, I got for $b_k$ the small monster
$$frac{1}{2} left(k^2+k-3 leftlfloor frac{k}{3}rightrfloor ^2-3 leftlfloor
frac{k-2}{3}rightrfloor left(leftlfloor frac{k-2}{3}rightrfloor
+1right)-3 leftlfloor frac{k-1}{3}rightrfloor left(leftlfloor
frac{k-1}{3}rightrfloor +1right)+3 leftlfloor frac{k}{3}rightrfloor
right)$$
Which one do you prefer ?
answered Jan 8 at 9:47
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
$begingroup$
Of course, Floor function :)
$endgroup$
– Elementary
Jan 8 at 10:06
$begingroup$
@Beginner. Joke or serious ?
$endgroup$
– Claude Leibovici
Jan 8 at 10:09
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Both are beautiful. :) Thank you very much (+)
$endgroup$
– Elementary
Jan 8 at 10:34
add a comment |
$begingroup$
If you consider$$b_k=sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$ you could see, after some simple manipulationd that it corresponds to the recurrence relation
$$b_k=b_{k-1}+b_{k-3}-b_{k-4}qquad text{with}qquad b_1=1, b_2=3, b_3=6, b_4=7$$ Using the conventional method you should end with
$$b_k=2k-frac{2}{3} left(1-cos left(frac{2 pi k}{3}right)right)$$
Edit
If you enjoy the floor function, using a CAS, I got for $b_k$ the small monster
$$frac{1}{2} left(k^2+k-3 leftlfloor frac{k}{3}rightrfloor ^2-3 leftlfloor
frac{k-2}{3}rightrfloor left(leftlfloor frac{k-2}{3}rightrfloor
+1right)-3 leftlfloor frac{k-1}{3}rightrfloor left(leftlfloor
frac{k-1}{3}rightrfloor +1right)+3 leftlfloor frac{k}{3}rightrfloor
right)$$
Which one do you prefer ?
answered Jan 8 at 9:47
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
$begingroup$
Of course, Floor function :)
$endgroup$
– Elementary
Jan 8 at 10:06
$begingroup$
@Beginner. Joke or serious ?
$endgroup$
– Claude Leibovici
Jan 8 at 10:09
$begingroup$
Both are beautiful. :) Thank you very much (+)
$endgroup$
– Elementary
Jan 8 at 10:34
add a comment |
$begingroup$
If you consider$$b_k=sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$ you could see, after some simple manipulationd that it corresponds to the recurrence relation
$$b_k=b_{k-1}+b_{k-3}-b_{k-4}qquad text{with}qquad b_1=1, b_2=3, b_3=6, b_4=7$$ Using the conventional method you should end with
$$b_k=2k-frac{2}{3} left(1-cos left(frac{2 pi k}{3}right)right)$$
Edit
If you enjoy the floor function, using a CAS, I got for $b_k$ the small monster
$$frac{1}{2} left(k^2+k-3 leftlfloor frac{k}{3}rightrfloor ^2-3 leftlfloor
frac{k-2}{3}rightrfloor left(leftlfloor frac{k-2}{3}rightrfloor
+1right)-3 leftlfloor frac{k-1}{3}rightrfloor left(leftlfloor
frac{k-1}{3}rightrfloor +1right)+3 leftlfloor frac{k}{3}rightrfloor
right)$$
Which one do you prefer ?
answered Jan 8 at 9:47
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
If you consider$$b_k=sum_{n=1}^{k}left( n- 3 bigglfloor frac{n-1}{3} biggrfloorright)$$ you could see, after some simple manipulationd that it corresponds to the recurrence relation
$$b_k=b_{k-1}+b_{k-3}-b_{k-4}qquad text{with}qquad b_1=1, b_2=3, b_3=6, b_4=7$$ Using the conventional method you should end with
$$b_k=2k-frac{2}{3} left(1-cos left(frac{2 pi k}{3}right)right)$$
Edit
If you enjoy the floor function, using a CAS, I got for $b_k$ the small monster
$$frac{1}{2} left(k^2+k-3 leftlfloor frac{k}{3}rightrfloor ^2-3 leftlfloor
frac{k-2}{3}rightrfloor left(leftlfloor frac{k-2}{3}rightrfloor
+1right)-3 leftlfloor frac{k-1}{3}rightrfloor left(leftlfloor
frac{k-1}{3}rightrfloor +1right)+3 leftlfloor frac{k}{3}rightrfloor
right)$$
Which one do you prefer ?
answered Jan 8 at 9:47
Claude LeiboviciClaude Leibovici
120k1157132
edited Jan 8 at 10:00
answered Jan 8 at 9:47
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 8 at 9:47
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 8 at 9:47
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
$begingroup$
Of course, Floor function :)
$endgroup$
– Elementary
Jan 8 at 10:06
$begingroup$
@Beginner. Joke or serious ?
$endgroup$
– Claude Leibovici
Jan 8 at 10:09
$begingroup$
Both are beautiful. :) Thank you very much (+)
$endgroup$
– Elementary
Jan 8 at 10:34
add a comment |
$begingroup$
Of course, Floor function :)
$endgroup$
– Elementary
Jan 8 at 10:06
$begingroup$
@Beginner. Joke or serious ?
$endgroup$
– Claude Leibovici
Jan 8 at 10:09
$begingroup$
Both are beautiful. :) Thank you very much (+)
$endgroup$
– Elementary
Jan 8 at 10:34
$begingroup$
Of course, Floor function :)
$endgroup$
– Elementary
Jan 8 at 10:06
$begingroup$
Of course, Floor function :)
$endgroup$
– Elementary
Jan 8 at 10:06
$begingroup$
@Beginner. Joke or serious ?
$endgroup$
– Claude Leibovici
Jan 8 at 10:09
$begingroup$
@Beginner. Joke or serious ?
$endgroup$
– Claude Leibovici
Jan 8 at 10:09
$begingroup$
Both are beautiful. :) Thank you very much (+)
$endgroup$
– Elementary
Jan 8 at 10:34
$begingroup$
Both are beautiful. :) Thank you very much (+)
$endgroup$
– Elementary
Jan 8 at 10:34
add a comment |
$begingroup$
If $n equiv 0 pmod{3}$, i.e. say $n=3s$ (where $s geq 1$), then $a_n=3s-3lfloor s-frac{1}{3}rfloor=3s-3(s-1)=3$.
If $n equiv 1 pmod{3}$, i.e. say $n=3s+1$ (where $s geq 0$), then $a_n=3s+1-3lfloor srfloor=1$.
If $n equiv 2 pmod{3}$, i.e. say $n=3s+2$ (where $s geq 0$), then $a_n=3s+2-3lfloor s+frac{1}{3}rfloor=2$.
So depending on what $k$ is we can count the number of terms which are $1's$, $2's$ and $3's$.
For $k=1,2$, the sum will be $color{red}{1}$ and $color{red}{3}$, respectively. So for $k geq 3$, we do the following:
If $k=3t$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)=6t=color{blue}{2k}.$$
If $k=3t+1$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+1}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+1=6t+1=color{blue}{2k-1}.$$
Likewise we can get the expressions for $k=3t+2$ as
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+2}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+(1+2)=6t+3=color{blue}{2k-1}.$$
answered Jan 8 at 9:23
Anurag AAnurag A
26k12249
$endgroup$
$begingroup$
I think, if there is an exist a closed form, it should be expressed by floor function.
$endgroup$
– Elementary
Jan 8 at 9:26
$begingroup$
@Beginner you are summing up to $k$ terms, so the answer is a function of $k$. I am not sure what you mean by the closed form has to be in terms of floor function.
$endgroup$
– Anurag A
Jan 8 at 9:29
$begingroup$
No, I didn't do this.
$endgroup$
– Elementary
Jan 8 at 9:32
add a comment |
$begingroup$
If $n equiv 0 pmod{3}$, i.e. say $n=3s$ (where $s geq 1$), then $a_n=3s-3lfloor s-frac{1}{3}rfloor=3s-3(s-1)=3$.
If $n equiv 1 pmod{3}$, i.e. say $n=3s+1$ (where $s geq 0$), then $a_n=3s+1-3lfloor srfloor=1$.
If $n equiv 2 pmod{3}$, i.e. say $n=3s+2$ (where $s geq 0$), then $a_n=3s+2-3lfloor s+frac{1}{3}rfloor=2$.
So depending on what $k$ is we can count the number of terms which are $1's$, $2's$ and $3's$.
For $k=1,2$, the sum will be $color{red}{1}$ and $color{red}{3}$, respectively. So for $k geq 3$, we do the following:
If $k=3t$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)=6t=color{blue}{2k}.$$
If $k=3t+1$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+1}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+1=6t+1=color{blue}{2k-1}.$$
Likewise we can get the expressions for $k=3t+2$ as
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+2}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+(1+2)=6t+3=color{blue}{2k-1}.$$
answered Jan 8 at 9:23
Anurag AAnurag A
26k12249
$endgroup$
$begingroup$
I think, if there is an exist a closed form, it should be expressed by floor function.
$endgroup$
– Elementary
Jan 8 at 9:26
$begingroup$
@Beginner you are summing up to $k$ terms, so the answer is a function of $k$. I am not sure what you mean by the closed form has to be in terms of floor function.
$endgroup$
– Anurag A
Jan 8 at 9:29
$begingroup$
No, I didn't do this.
$endgroup$
– Elementary
Jan 8 at 9:32
add a comment |
$begingroup$
If $n equiv 0 pmod{3}$, i.e. say $n=3s$ (where $s geq 1$), then $a_n=3s-3lfloor s-frac{1}{3}rfloor=3s-3(s-1)=3$.
If $n equiv 1 pmod{3}$, i.e. say $n=3s+1$ (where $s geq 0$), then $a_n=3s+1-3lfloor srfloor=1$.
If $n equiv 2 pmod{3}$, i.e. say $n=3s+2$ (where $s geq 0$), then $a_n=3s+2-3lfloor s+frac{1}{3}rfloor=2$.
So depending on what $k$ is we can count the number of terms which are $1's$, $2's$ and $3's$.
For $k=1,2$, the sum will be $color{red}{1}$ and $color{red}{3}$, respectively. So for $k geq 3$, we do the following:
If $k=3t$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)=6t=color{blue}{2k}.$$
If $k=3t+1$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+1}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+1=6t+1=color{blue}{2k-1}.$$
Likewise we can get the expressions for $k=3t+2$ as
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+2}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+(1+2)=6t+3=color{blue}{2k-1}.$$
answered Jan 8 at 9:23
Anurag AAnurag A
26k12249
$endgroup$
If $n equiv 0 pmod{3}$, i.e. say $n=3s$ (where $s geq 1$), then $a_n=3s-3lfloor s-frac{1}{3}rfloor=3s-3(s-1)=3$.
If $n equiv 1 pmod{3}$, i.e. say $n=3s+1$ (where $s geq 0$), then $a_n=3s+1-3lfloor srfloor=1$.
If $n equiv 2 pmod{3}$, i.e. say $n=3s+2$ (where $s geq 0$), then $a_n=3s+2-3lfloor s+frac{1}{3}rfloor=2$.
So depending on what $k$ is we can count the number of terms which are $1's$, $2's$ and $3's$.
For $k=1,2$, the sum will be $color{red}{1}$ and $color{red}{3}$, respectively. So for $k geq 3$, we do the following:
If $k=3t$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)=6t=color{blue}{2k}.$$
If $k=3t+1$ for $t geq 1$, then
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+1}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+1=6t+1=color{blue}{2k-1}.$$
Likewise we can get the expressions for $k=3t+2$ as
$$sum_{n=1}^kleft(n-3leftlfloor n-frac{1}{3}rightrfloorright)=sum_{n=1}^{3t+2}left(n-3leftlfloor n-frac{1}{3}rightrfloorright)=t(1+2+3)+(1+2)=6t+3=color{blue}{2k-1}.$$
answered Jan 8 at 9:23
Anurag AAnurag A
26k12249
edited Jan 8 at 9:48
answered Jan 8 at 9:23
Anurag AAnurag A
26k12249
answered Jan 8 at 9:23
Anurag AAnurag A
26k12249
answered Jan 8 at 9:23
Anurag AAnurag A
26k12249
26k12249
$begingroup$
I think, if there is an exist a closed form, it should be expressed by floor function.
$endgroup$
– Elementary
Jan 8 at 9:26
$begingroup$
@Beginner you are summing up to $k$ terms, so the answer is a function of $k$. I am not sure what you mean by the closed form has to be in terms of floor function.
$endgroup$
– Anurag A
Jan 8 at 9:29
$begingroup$
No, I didn't do this.
$endgroup$
– Elementary
Jan 8 at 9:32
add a comment |
$begingroup$
I think, if there is an exist a closed form, it should be expressed by floor function.
$endgroup$
– Elementary
Jan 8 at 9:26
$begingroup$
@Beginner you are summing up to $k$ terms, so the answer is a function of $k$. I am not sure what you mean by the closed form has to be in terms of floor function.
$endgroup$
– Anurag A
Jan 8 at 9:29
$begingroup$
No, I didn't do this.
$endgroup$
– Elementary
Jan 8 at 9:32
$begingroup$
I think, if there is an exist a closed form, it should be expressed by floor function.
$endgroup$
– Elementary
Jan 8 at 9:26
$begingroup$
I think, if there is an exist a closed form, it should be expressed by floor function.
$endgroup$
– Elementary
Jan 8 at 9:26
$begingroup$
@Beginner you are summing up to $k$ terms, so the answer is a function of $k$. I am not sure what you mean by the closed form has to be in terms of floor function.
$endgroup$
– Anurag A
Jan 8 at 9:29
$begingroup$
@Beginner you are summing up to $k$ terms, so the answer is a function of $k$. I am not sure what you mean by the closed form has to be in terms of floor function.
$endgroup$
– Anurag A
Jan 8 at 9:29
$begingroup$
No, I didn't do this.
$endgroup$
– Elementary
Jan 8 at 9:32
$begingroup$
No, I didn't do this.
$endgroup$
– Elementary
Jan 8 at 9:32
add a comment |
$begingroup$
Sorry can't comment on @5xum's answer. Building on the quadratic function we can derive:
$S_k=frac{1}{2}((k-3lfloorfrac{k}{3}rfloor)^2+k+9lfloorfrac{k}{3}rfloor)$, which is pretty neat.
To derive this, note that from @5xum's answer, $S_k=2k-1+frac{1}{2}(x^2 -3x+2)$ where $x=k-3lfloorfrac{k}{3}rfloor$. We can write $k=x+3lfloorfrac{k}{3}rfloor$, so that:
$$S_k=frac{1}{2}(x^2 -3x+2+4x+12lfloorfrac{k}{3}rfloor-2)=frac{1}{2}(x^2+x+12lfloorfrac{k}{3}rfloor). $$
Substituting $x$ back yields the result.
answered Jan 8 at 10:04
user46666user46666
335
$endgroup$
add a comment |
$begingroup$
Sorry can't comment on @5xum's answer. Building on the quadratic function we can derive:
$S_k=frac{1}{2}((k-3lfloorfrac{k}{3}rfloor)^2+k+9lfloorfrac{k}{3}rfloor)$, which is pretty neat.
To derive this, note that from @5xum's answer, $S_k=2k-1+frac{1}{2}(x^2 -3x+2)$ where $x=k-3lfloorfrac{k}{3}rfloor$. We can write $k=x+3lfloorfrac{k}{3}rfloor$, so that:
$$S_k=frac{1}{2}(x^2 -3x+2+4x+12lfloorfrac{k}{3}rfloor-2)=frac{1}{2}(x^2+x+12lfloorfrac{k}{3}rfloor). $$
Substituting $x$ back yields the result.
answered Jan 8 at 10:04
user46666user46666
335
$endgroup$
add a comment |
$begingroup$
Sorry can't comment on @5xum's answer. Building on the quadratic function we can derive:
$S_k=frac{1}{2}((k-3lfloorfrac{k}{3}rfloor)^2+k+9lfloorfrac{k}{3}rfloor)$, which is pretty neat.
To derive this, note that from @5xum's answer, $S_k=2k-1+frac{1}{2}(x^2 -3x+2)$ where $x=k-3lfloorfrac{k}{3}rfloor$. We can write $k=x+3lfloorfrac{k}{3}rfloor$, so that:
$$S_k=frac{1}{2}(x^2 -3x+2+4x+12lfloorfrac{k}{3}rfloor-2)=frac{1}{2}(x^2+x+12lfloorfrac{k}{3}rfloor). $$
Substituting $x$ back yields the result.
answered Jan 8 at 10:04
user46666user46666
335
$endgroup$
Sorry can't comment on @5xum's answer. Building on the quadratic function we can derive:
$S_k=frac{1}{2}((k-3lfloorfrac{k}{3}rfloor)^2+k+9lfloorfrac{k}{3}rfloor)$, which is pretty neat.
To derive this, note that from @5xum's answer, $S_k=2k-1+frac{1}{2}(x^2 -3x+2)$ where $x=k-3lfloorfrac{k}{3}rfloor$. We can write $k=x+3lfloorfrac{k}{3}rfloor$, so that:
$$S_k=frac{1}{2}(x^2 -3x+2+4x+12lfloorfrac{k}{3}rfloor-2)=frac{1}{2}(x^2+x+12lfloorfrac{k}{3}rfloor). $$
Substituting $x$ back yields the result.
answered Jan 8 at 10:04
user46666user46666
335
edited Jan 8 at 10:17
answered Jan 8 at 10:04
user46666user46666
335
answered Jan 8 at 10:04
user46666user46666
335
answered Jan 8 at 10:04
user46666user46666
335
335
add a comment |
add a comment |
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$begingroup$
There is a closed form with a bunch of floor functions.
$endgroup$
– Claude Leibovici
Jan 8 at 9:52
$begingroup$
You recieved 5 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:49
$begingroup$
@5xum I've been upvote before. Now, I accept. By the way, If you add the most recent result to your answer, it will be perfect. Regards.
$endgroup$
– Elementary
Jan 15 at 9:03