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Is it possible to find a closed form using modular arithmetic?
$begingroup$
Look at this sequence:
$$a_n=left{1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, cdots right}$$
This is a periodic sequence. Repeating block is $left{1, 0, 0, 2right}.$
Here is my closed form:
$$a_n = frac 14((-1)^n + (2 + i) (-i)^n + (2 - i) i^n + 3)$$
where, $i^2=-1$
Is it possible to find a closed form using modular arithmetic?
I mean for example,
$Amod 3+Bmod 4 cdots cdots cdots$
Is it possible?
sequences-and-series number-theory elementary-number-theory modular-arithmetic closed-form
asked Jan 24 at 23:02
ElementaryElementary
360111
$endgroup$
add a comment |
$begingroup$
Look at this sequence:
$$a_n=left{1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, cdots right}$$
This is a periodic sequence. Repeating block is $left{1, 0, 0, 2right}.$
Here is my closed form:
$$a_n = frac 14((-1)^n + (2 + i) (-i)^n + (2 - i) i^n + 3)$$
where, $i^2=-1$
Is it possible to find a closed form using modular arithmetic?
I mean for example,
$Amod 3+Bmod 4 cdots cdots cdots$
Is it possible?
sequences-and-series number-theory elementary-number-theory modular-arithmetic closed-form
asked Jan 24 at 23:02
ElementaryElementary
360111
$endgroup$
add a comment |
$begingroup$
Look at this sequence:
$$a_n=left{1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, cdots right}$$
This is a periodic sequence. Repeating block is $left{1, 0, 0, 2right}.$
Here is my closed form:
$$a_n = frac 14((-1)^n + (2 + i) (-i)^n + (2 - i) i^n + 3)$$
where, $i^2=-1$
Is it possible to find a closed form using modular arithmetic?
I mean for example,
$Amod 3+Bmod 4 cdots cdots cdots$
Is it possible?
sequences-and-series number-theory elementary-number-theory modular-arithmetic closed-form
asked Jan 24 at 23:02
ElementaryElementary
360111
$endgroup$
Look at this sequence:
$$a_n=left{1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, cdots right}$$
This is a periodic sequence. Repeating block is $left{1, 0, 0, 2right}.$
Here is my closed form:
$$a_n = frac 14((-1)^n + (2 + i) (-i)^n + (2 - i) i^n + 3)$$
where, $i^2=-1$
Is it possible to find a closed form using modular arithmetic?
I mean for example,
$Amod 3+Bmod 4 cdots cdots cdots$
Is it possible?
sequences-and-series number-theory elementary-number-theory modular-arithmetic closed-form
sequences-and-series number-theory elementary-number-theory modular-arithmetic closed-form
asked Jan 24 at 23:02
ElementaryElementary
360111
asked Jan 24 at 23:02
ElementaryElementary
360111
edited Jan 25 at 7:25
Elementary
asked Jan 24 at 23:02
ElementaryElementary
360111
asked Jan 24 at 23:02
ElementaryElementary
360111
asked Jan 24 at 23:02
ElementaryElementary
360111
360111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You want a function of $n$ that behaves as follows:
$f(n) = begin{cases} 1 text{ if } n mod 4 = 1\ 0 text{ if } n mod 4 = 2,3\ 2 text{ if } n mod 4 = 0end{cases}$
Let's denote $n mod 4$ by $n'$. The polynomial $n'(n'-2)(n'-3)$ is $0$ when $n'=0,2,3$ and takes the value $2$ when $n'=1$. So this suggests
$frac 1 2 n'(n'-2)(n'-3)$
should be part of a closed form expression. Similarly $(n'-1)(n'-2)(n'-3)$ is $0$ when $n'=1,2,3$ and takes the value $-6$ when $n'=0$. So this suggests
$-frac 1 3 (n'-1)(n'-2)(n'-3)$
should be part of the expression too. So we have
$f(n) = frac 1 2 n'(n'-2)(n'-3) - frac 1 3 (n'-1)(n'-2)(n'-3)\
= frac 1 6 (n'+2)(n'-2)(n'-3) \
= frac 1 6 (n'^3 -3n'^2 -4n'+12) text{ where } n'=nmod 4$
answered Jan 25 at 10:16
gandalf61gandalf61
9,109825
$endgroup$
$begingroup$
That polynomial is the one interpolating $(1,1),(2,0),(3,0),(4,2)$.
$endgroup$
– lhf
Jan 25 at 10:19
add a comment |
$begingroup$
Using modular-arithmetic is quite similar to using floor functions so you can also use that instead. Here is a possible solution:
$$
f(n) = frac{1}{6}left((n-4lfloor n/4rfloor)^3 -3(n-4lfloor n/4rfloor)^2 - 4(n-lfloor n/4 rfloor) + 12 right)
$$
In this case $n-4lfloor n/4 rfloor$ would be the same as $npmod 4$.
answered Jan 25 at 9:47
Yong Hao NgYong Hao Ng
3,6141222
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
You want a function of $n$ that behaves as follows:
$f(n) = begin{cases} 1 text{ if } n mod 4 = 1\ 0 text{ if } n mod 4 = 2,3\ 2 text{ if } n mod 4 = 0end{cases}$
Let's denote $n mod 4$ by $n'$. The polynomial $n'(n'-2)(n'-3)$ is $0$ when $n'=0,2,3$ and takes the value $2$ when $n'=1$. So this suggests
$frac 1 2 n'(n'-2)(n'-3)$
should be part of a closed form expression. Similarly $(n'-1)(n'-2)(n'-3)$ is $0$ when $n'=1,2,3$ and takes the value $-6$ when $n'=0$. So this suggests
$-frac 1 3 (n'-1)(n'-2)(n'-3)$
should be part of the expression too. So we have
$f(n) = frac 1 2 n'(n'-2)(n'-3) - frac 1 3 (n'-1)(n'-2)(n'-3)\
= frac 1 6 (n'+2)(n'-2)(n'-3) \
= frac 1 6 (n'^3 -3n'^2 -4n'+12) text{ where } n'=nmod 4$
answered Jan 25 at 10:16
gandalf61gandalf61
9,109825
$endgroup$
$begingroup$
That polynomial is the one interpolating $(1,1),(2,0),(3,0),(4,2)$.
$endgroup$
– lhf
Jan 25 at 10:19
add a comment |
$begingroup$
You want a function of $n$ that behaves as follows:
$f(n) = begin{cases} 1 text{ if } n mod 4 = 1\ 0 text{ if } n mod 4 = 2,3\ 2 text{ if } n mod 4 = 0end{cases}$
Let's denote $n mod 4$ by $n'$. The polynomial $n'(n'-2)(n'-3)$ is $0$ when $n'=0,2,3$ and takes the value $2$ when $n'=1$. So this suggests
$frac 1 2 n'(n'-2)(n'-3)$
should be part of a closed form expression. Similarly $(n'-1)(n'-2)(n'-3)$ is $0$ when $n'=1,2,3$ and takes the value $-6$ when $n'=0$. So this suggests
$-frac 1 3 (n'-1)(n'-2)(n'-3)$
should be part of the expression too. So we have
$f(n) = frac 1 2 n'(n'-2)(n'-3) - frac 1 3 (n'-1)(n'-2)(n'-3)\
= frac 1 6 (n'+2)(n'-2)(n'-3) \
= frac 1 6 (n'^3 -3n'^2 -4n'+12) text{ where } n'=nmod 4$
answered Jan 25 at 10:16
gandalf61gandalf61
9,109825
$endgroup$
$begingroup$
That polynomial is the one interpolating $(1,1),(2,0),(3,0),(4,2)$.
$endgroup$
– lhf
Jan 25 at 10:19
add a comment |
$begingroup$
You want a function of $n$ that behaves as follows:
$f(n) = begin{cases} 1 text{ if } n mod 4 = 1\ 0 text{ if } n mod 4 = 2,3\ 2 text{ if } n mod 4 = 0end{cases}$
Let's denote $n mod 4$ by $n'$. The polynomial $n'(n'-2)(n'-3)$ is $0$ when $n'=0,2,3$ and takes the value $2$ when $n'=1$. So this suggests
$frac 1 2 n'(n'-2)(n'-3)$
should be part of a closed form expression. Similarly $(n'-1)(n'-2)(n'-3)$ is $0$ when $n'=1,2,3$ and takes the value $-6$ when $n'=0$. So this suggests
$-frac 1 3 (n'-1)(n'-2)(n'-3)$
should be part of the expression too. So we have
$f(n) = frac 1 2 n'(n'-2)(n'-3) - frac 1 3 (n'-1)(n'-2)(n'-3)\
= frac 1 6 (n'+2)(n'-2)(n'-3) \
= frac 1 6 (n'^3 -3n'^2 -4n'+12) text{ where } n'=nmod 4$
answered Jan 25 at 10:16
gandalf61gandalf61
9,109825
$endgroup$
You want a function of $n$ that behaves as follows:
$f(n) = begin{cases} 1 text{ if } n mod 4 = 1\ 0 text{ if } n mod 4 = 2,3\ 2 text{ if } n mod 4 = 0end{cases}$
Let's denote $n mod 4$ by $n'$. The polynomial $n'(n'-2)(n'-3)$ is $0$ when $n'=0,2,3$ and takes the value $2$ when $n'=1$. So this suggests
$frac 1 2 n'(n'-2)(n'-3)$
should be part of a closed form expression. Similarly $(n'-1)(n'-2)(n'-3)$ is $0$ when $n'=1,2,3$ and takes the value $-6$ when $n'=0$. So this suggests
$-frac 1 3 (n'-1)(n'-2)(n'-3)$
should be part of the expression too. So we have
$f(n) = frac 1 2 n'(n'-2)(n'-3) - frac 1 3 (n'-1)(n'-2)(n'-3)\
= frac 1 6 (n'+2)(n'-2)(n'-3) \
= frac 1 6 (n'^3 -3n'^2 -4n'+12) text{ where } n'=nmod 4$
answered Jan 25 at 10:16
gandalf61gandalf61
9,109825
answered Jan 25 at 10:16
gandalf61gandalf61
9,109825
answered Jan 25 at 10:16
gandalf61gandalf61
9,109825
answered Jan 25 at 10:16
gandalf61gandalf61
9,109825
9,109825
$begingroup$
That polynomial is the one interpolating $(1,1),(2,0),(3,0),(4,2)$.
$endgroup$
– lhf
Jan 25 at 10:19
add a comment |
$begingroup$
That polynomial is the one interpolating $(1,1),(2,0),(3,0),(4,2)$.
$endgroup$
– lhf
Jan 25 at 10:19
$begingroup$
That polynomial is the one interpolating $(1,1),(2,0),(3,0),(4,2)$.
$endgroup$
– lhf
Jan 25 at 10:19
$begingroup$
That polynomial is the one interpolating $(1,1),(2,0),(3,0),(4,2)$.
$endgroup$
– lhf
Jan 25 at 10:19
add a comment |
$begingroup$
Using modular-arithmetic is quite similar to using floor functions so you can also use that instead. Here is a possible solution:
$$
f(n) = frac{1}{6}left((n-4lfloor n/4rfloor)^3 -3(n-4lfloor n/4rfloor)^2 - 4(n-lfloor n/4 rfloor) + 12 right)
$$
In this case $n-4lfloor n/4 rfloor$ would be the same as $npmod 4$.
answered Jan 25 at 9:47
Yong Hao NgYong Hao Ng
3,6141222
$endgroup$
add a comment |
$begingroup$
Using modular-arithmetic is quite similar to using floor functions so you can also use that instead. Here is a possible solution:
$$
f(n) = frac{1}{6}left((n-4lfloor n/4rfloor)^3 -3(n-4lfloor n/4rfloor)^2 - 4(n-lfloor n/4 rfloor) + 12 right)
$$
In this case $n-4lfloor n/4 rfloor$ would be the same as $npmod 4$.
answered Jan 25 at 9:47
Yong Hao NgYong Hao Ng
3,6141222
$endgroup$
add a comment |
$begingroup$
Using modular-arithmetic is quite similar to using floor functions so you can also use that instead. Here is a possible solution:
$$
f(n) = frac{1}{6}left((n-4lfloor n/4rfloor)^3 -3(n-4lfloor n/4rfloor)^2 - 4(n-lfloor n/4 rfloor) + 12 right)
$$
In this case $n-4lfloor n/4 rfloor$ would be the same as $npmod 4$.
answered Jan 25 at 9:47
Yong Hao NgYong Hao Ng
3,6141222
$endgroup$
Using modular-arithmetic is quite similar to using floor functions so you can also use that instead. Here is a possible solution:
$$
f(n) = frac{1}{6}left((n-4lfloor n/4rfloor)^3 -3(n-4lfloor n/4rfloor)^2 - 4(n-lfloor n/4 rfloor) + 12 right)
$$
In this case $n-4lfloor n/4 rfloor$ would be the same as $npmod 4$.
answered Jan 25 at 9:47
Yong Hao NgYong Hao Ng
3,6141222
answered Jan 25 at 9:47
Yong Hao NgYong Hao Ng
3,6141222
answered Jan 25 at 9:47
Yong Hao NgYong Hao Ng
3,6141222
answered Jan 25 at 9:47
Yong Hao NgYong Hao Ng
3,6141222
3,6141222
add a comment |
add a comment |
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