What is the relation in this sequence
$begingroup$
I was attending a coding competition and this question came up
There is a box containing n chocolates. You can either take 1 chocolate or 3 chocolates at a time until its empty. So in total how many ways you can empty the chocolate for a given n.
Eg: n = 3
ways you can take out is 2 ie:
111
3
ways you can take out when n=4
is 3 ie:
1111
31
13
likewise, I have calculated manually for n=5 -> 4 etc etc..
so the result sequence comes up like 2,3,4,6,9...
I am not able to figure out what kind of relation it has or how do I solve it. I'm not able to figure out what kind of problem it is as well (Permutation, combinations, series?). let me know if I need to post any more details.
sequences-and-series permutations contest-math combinations
New contributor
$endgroup$
add a comment |
$begingroup$
I was attending a coding competition and this question came up
There is a box containing n chocolates. You can either take 1 chocolate or 3 chocolates at a time until its empty. So in total how many ways you can empty the chocolate for a given n.
Eg: n = 3
ways you can take out is 2 ie:
111
3
ways you can take out when n=4
is 3 ie:
1111
31
13
likewise, I have calculated manually for n=5 -> 4 etc etc..
so the result sequence comes up like 2,3,4,6,9...
I am not able to figure out what kind of relation it has or how do I solve it. I'm not able to figure out what kind of problem it is as well (Permutation, combinations, series?). let me know if I need to post any more details.
sequences-and-series permutations contest-math combinations
New contributor
$endgroup$
$begingroup$
So you're looking at the number of positive integers solutions to $$ 3x + y = k $$ for a fixed value for $k$. Maybe something like this can help: math.stackexchange.com/questions/29019/… and handakafunda.com/…
$endgroup$
– Matti P.
yesterday
2
$begingroup$
@MattiP. I think $x,y$ should be non-negative integers, not be positive integers. And the order is also important. For example, if $k=4$, then $(x,y)=(1,1)$ or $(0,4)$. But in the case $(x,y)=(1,1)$, $31$ and $13$ is not the same way...
$endgroup$
– Doyun Nam
yesterday
$begingroup$
yes. the order is important.
$endgroup$
– Shobi
yesterday
$begingroup$
The OEIS is always available as a last (or first) resort. Entering your sequence 2, 3, 4, 6, 9 gives A000930 which includes lots of (potentially) useful stuff. The Formulas section may be of use to you.
$endgroup$
– Peter Phipps
yesterday
$begingroup$
Thanks, Peter. I actually tried some online sequence solutions, but they didn't gave me much of idea. but this one I am encountering it for the first time its super usefull
$endgroup$
– Shobi
22 hours ago
add a comment |
$begingroup$
I was attending a coding competition and this question came up
There is a box containing n chocolates. You can either take 1 chocolate or 3 chocolates at a time until its empty. So in total how many ways you can empty the chocolate for a given n.
Eg: n = 3
ways you can take out is 2 ie:
111
3
ways you can take out when n=4
is 3 ie:
1111
31
13
likewise, I have calculated manually for n=5 -> 4 etc etc..
so the result sequence comes up like 2,3,4,6,9...
I am not able to figure out what kind of relation it has or how do I solve it. I'm not able to figure out what kind of problem it is as well (Permutation, combinations, series?). let me know if I need to post any more details.
sequences-and-series permutations contest-math combinations
New contributor
$endgroup$
I was attending a coding competition and this question came up
There is a box containing n chocolates. You can either take 1 chocolate or 3 chocolates at a time until its empty. So in total how many ways you can empty the chocolate for a given n.
Eg: n = 3
ways you can take out is 2 ie:
111
3
ways you can take out when n=4
is 3 ie:
1111
31
13
likewise, I have calculated manually for n=5 -> 4 etc etc..
so the result sequence comes up like 2,3,4,6,9...
I am not able to figure out what kind of relation it has or how do I solve it. I'm not able to figure out what kind of problem it is as well (Permutation, combinations, series?). let me know if I need to post any more details.
sequences-and-series permutations contest-math combinations
sequences-and-series permutations contest-math combinations
New contributor
New contributor
edited yesterday
YuiTo Cheng
746322
746322
New contributor
asked yesterday
ShobiShobi
1185
1185
New contributor
New contributor
$begingroup$
So you're looking at the number of positive integers solutions to $$ 3x + y = k $$ for a fixed value for $k$. Maybe something like this can help: math.stackexchange.com/questions/29019/… and handakafunda.com/…
$endgroup$
– Matti P.
yesterday
2
$begingroup$
@MattiP. I think $x,y$ should be non-negative integers, not be positive integers. And the order is also important. For example, if $k=4$, then $(x,y)=(1,1)$ or $(0,4)$. But in the case $(x,y)=(1,1)$, $31$ and $13$ is not the same way...
$endgroup$
– Doyun Nam
yesterday
$begingroup$
yes. the order is important.
$endgroup$
– Shobi
yesterday
$begingroup$
The OEIS is always available as a last (or first) resort. Entering your sequence 2, 3, 4, 6, 9 gives A000930 which includes lots of (potentially) useful stuff. The Formulas section may be of use to you.
$endgroup$
– Peter Phipps
yesterday
$begingroup$
Thanks, Peter. I actually tried some online sequence solutions, but they didn't gave me much of idea. but this one I am encountering it for the first time its super usefull
$endgroup$
– Shobi
22 hours ago
add a comment |
$begingroup$
So you're looking at the number of positive integers solutions to $$ 3x + y = k $$ for a fixed value for $k$. Maybe something like this can help: math.stackexchange.com/questions/29019/… and handakafunda.com/…
$endgroup$
– Matti P.
yesterday
2
$begingroup$
@MattiP. I think $x,y$ should be non-negative integers, not be positive integers. And the order is also important. For example, if $k=4$, then $(x,y)=(1,1)$ or $(0,4)$. But in the case $(x,y)=(1,1)$, $31$ and $13$ is not the same way...
$endgroup$
– Doyun Nam
yesterday
$begingroup$
yes. the order is important.
$endgroup$
– Shobi
yesterday
$begingroup$
The OEIS is always available as a last (or first) resort. Entering your sequence 2, 3, 4, 6, 9 gives A000930 which includes lots of (potentially) useful stuff. The Formulas section may be of use to you.
$endgroup$
– Peter Phipps
yesterday
$begingroup$
Thanks, Peter. I actually tried some online sequence solutions, but they didn't gave me much of idea. but this one I am encountering it for the first time its super usefull
$endgroup$
– Shobi
22 hours ago
$begingroup$
So you're looking at the number of positive integers solutions to $$ 3x + y = k $$ for a fixed value for $k$. Maybe something like this can help: math.stackexchange.com/questions/29019/… and handakafunda.com/…
$endgroup$
– Matti P.
yesterday
$begingroup$
So you're looking at the number of positive integers solutions to $$ 3x + y = k $$ for a fixed value for $k$. Maybe something like this can help: math.stackexchange.com/questions/29019/… and handakafunda.com/…
$endgroup$
– Matti P.
yesterday
2
2
$begingroup$
@MattiP. I think $x,y$ should be non-negative integers, not be positive integers. And the order is also important. For example, if $k=4$, then $(x,y)=(1,1)$ or $(0,4)$. But in the case $(x,y)=(1,1)$, $31$ and $13$ is not the same way...
$endgroup$
– Doyun Nam
yesterday
$begingroup$
@MattiP. I think $x,y$ should be non-negative integers, not be positive integers. And the order is also important. For example, if $k=4$, then $(x,y)=(1,1)$ or $(0,4)$. But in the case $(x,y)=(1,1)$, $31$ and $13$ is not the same way...
$endgroup$
– Doyun Nam
yesterday
$begingroup$
yes. the order is important.
$endgroup$
– Shobi
yesterday
$begingroup$
yes. the order is important.
$endgroup$
– Shobi
yesterday
$begingroup$
The OEIS is always available as a last (or first) resort. Entering your sequence 2, 3, 4, 6, 9 gives A000930 which includes lots of (potentially) useful stuff. The Formulas section may be of use to you.
$endgroup$
– Peter Phipps
yesterday
$begingroup$
The OEIS is always available as a last (or first) resort. Entering your sequence 2, 3, 4, 6, 9 gives A000930 which includes lots of (potentially) useful stuff. The Formulas section may be of use to you.
$endgroup$
– Peter Phipps
yesterday
$begingroup$
Thanks, Peter. I actually tried some online sequence solutions, but they didn't gave me much of idea. but this one I am encountering it for the first time its super usefull
$endgroup$
– Shobi
22 hours ago
$begingroup$
Thanks, Peter. I actually tried some online sequence solutions, but they didn't gave me much of idea. but this one I am encountering it for the first time its super usefull
$endgroup$
– Shobi
22 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think this problem is one of the problems related to recurrence relation.
Let $a_n$ be the number of ways.
For example, $a_1 =1$, $a_2 = 1$, $a_3=2$, $a_4=3$ as you mentioned above.
Let's think of the number of ways for $n$. Obviously, it is $a_n$.
If we take 1 chocolate at the first time, then the number of remaining chocolates is $n-1$.
And the number of ways for $n-1$ is $a_{n-1}$.
If we take 3 chocolate at the first time, then the number of remaining chocolates is $n-3$.
And the number of ways for $n-3$ is $a_{n-3}$.
Therefore, for $n geq 4$,
$$a_n = a_{n-1} + a_{n-3}.$$
The initial value is given as above.
$endgroup$
1
$begingroup$
wonderful. I have calculated values till n=10 manually. and this formula seems to agree with our findings. Thanks man :)
$endgroup$
– Shobi
yesterday
add a comment |
$begingroup$
As we can take one or three, we have the recurrence
$$c_n=c_{n-1}+c_{n-3}$$
with the initial condition
$$c_2=c_1=c_0=1,$$ as with $n<3$ there is a single option.
It is an easy matter to program this using a recursive function, preferably with memoization. Anon-recursive solution is also possible and preferable, storing the values in an array.
C= [1, 1, 1]
for n in range(3, 20):
C.append(C[n-1] + C[n-3])
1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872
Now let the roots of the caracteristic polynomial, $r^3-r^2-1$ be $r_0, r_1, overline r_1$ (there is a pair of conjugates). The general solution of the recurrence is
$$c_n=ar_0^n+br_1^n+overline boverline r_1^n,$$
where constants are determined by the system of equations
$$a+b+overline b=1,\ar_0+br_1+overline boverline r_1=1,\ar_0^2+br_1^2+overline boverline r_1^2=1.\$$
As one can check from the numerical values, the ratio of two successive terms quickly tends to the constant $1.465571231876768$, which is the real root (it has the largest modulus).
By numerical experimentation, it appears that the simple formula yields exact values in a large range:
$$c_n=[0.6114919919508175cdot1.465571231876768^n]$$
where the brackets denote rounding to the nearest integer.
$endgroup$
$begingroup$
Thanks for the code part :)
$endgroup$
– Shobi
yesterday
add a comment |
Your Answer
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2 Answers
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2 Answers
2
active
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oldest
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$begingroup$
I think this problem is one of the problems related to recurrence relation.
Let $a_n$ be the number of ways.
For example, $a_1 =1$, $a_2 = 1$, $a_3=2$, $a_4=3$ as you mentioned above.
Let's think of the number of ways for $n$. Obviously, it is $a_n$.
If we take 1 chocolate at the first time, then the number of remaining chocolates is $n-1$.
And the number of ways for $n-1$ is $a_{n-1}$.
If we take 3 chocolate at the first time, then the number of remaining chocolates is $n-3$.
And the number of ways for $n-3$ is $a_{n-3}$.
Therefore, for $n geq 4$,
$$a_n = a_{n-1} + a_{n-3}.$$
The initial value is given as above.
$endgroup$
1
$begingroup$
wonderful. I have calculated values till n=10 manually. and this formula seems to agree with our findings. Thanks man :)
$endgroup$
– Shobi
yesterday
add a comment |
$begingroup$
I think this problem is one of the problems related to recurrence relation.
Let $a_n$ be the number of ways.
For example, $a_1 =1$, $a_2 = 1$, $a_3=2$, $a_4=3$ as you mentioned above.
Let's think of the number of ways for $n$. Obviously, it is $a_n$.
If we take 1 chocolate at the first time, then the number of remaining chocolates is $n-1$.
And the number of ways for $n-1$ is $a_{n-1}$.
If we take 3 chocolate at the first time, then the number of remaining chocolates is $n-3$.
And the number of ways for $n-3$ is $a_{n-3}$.
Therefore, for $n geq 4$,
$$a_n = a_{n-1} + a_{n-3}.$$
The initial value is given as above.
$endgroup$
1
$begingroup$
wonderful. I have calculated values till n=10 manually. and this formula seems to agree with our findings. Thanks man :)
$endgroup$
– Shobi
yesterday
add a comment |
$begingroup$
I think this problem is one of the problems related to recurrence relation.
Let $a_n$ be the number of ways.
For example, $a_1 =1$, $a_2 = 1$, $a_3=2$, $a_4=3$ as you mentioned above.
Let's think of the number of ways for $n$. Obviously, it is $a_n$.
If we take 1 chocolate at the first time, then the number of remaining chocolates is $n-1$.
And the number of ways for $n-1$ is $a_{n-1}$.
If we take 3 chocolate at the first time, then the number of remaining chocolates is $n-3$.
And the number of ways for $n-3$ is $a_{n-3}$.
Therefore, for $n geq 4$,
$$a_n = a_{n-1} + a_{n-3}.$$
The initial value is given as above.
$endgroup$
I think this problem is one of the problems related to recurrence relation.
Let $a_n$ be the number of ways.
For example, $a_1 =1$, $a_2 = 1$, $a_3=2$, $a_4=3$ as you mentioned above.
Let's think of the number of ways for $n$. Obviously, it is $a_n$.
If we take 1 chocolate at the first time, then the number of remaining chocolates is $n-1$.
And the number of ways for $n-1$ is $a_{n-1}$.
If we take 3 chocolate at the first time, then the number of remaining chocolates is $n-3$.
And the number of ways for $n-3$ is $a_{n-3}$.
Therefore, for $n geq 4$,
$$a_n = a_{n-1} + a_{n-3}.$$
The initial value is given as above.
edited yesterday
answered yesterday
Doyun NamDoyun Nam
63619
63619
1
$begingroup$
wonderful. I have calculated values till n=10 manually. and this formula seems to agree with our findings. Thanks man :)
$endgroup$
– Shobi
yesterday
add a comment |
1
$begingroup$
wonderful. I have calculated values till n=10 manually. and this formula seems to agree with our findings. Thanks man :)
$endgroup$
– Shobi
yesterday
1
1
$begingroup$
wonderful. I have calculated values till n=10 manually. and this formula seems to agree with our findings. Thanks man :)
$endgroup$
– Shobi
yesterday
$begingroup$
wonderful. I have calculated values till n=10 manually. and this formula seems to agree with our findings. Thanks man :)
$endgroup$
– Shobi
yesterday
add a comment |
$begingroup$
As we can take one or three, we have the recurrence
$$c_n=c_{n-1}+c_{n-3}$$
with the initial condition
$$c_2=c_1=c_0=1,$$ as with $n<3$ there is a single option.
It is an easy matter to program this using a recursive function, preferably with memoization. Anon-recursive solution is also possible and preferable, storing the values in an array.
C= [1, 1, 1]
for n in range(3, 20):
C.append(C[n-1] + C[n-3])
1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872
Now let the roots of the caracteristic polynomial, $r^3-r^2-1$ be $r_0, r_1, overline r_1$ (there is a pair of conjugates). The general solution of the recurrence is
$$c_n=ar_0^n+br_1^n+overline boverline r_1^n,$$
where constants are determined by the system of equations
$$a+b+overline b=1,\ar_0+br_1+overline boverline r_1=1,\ar_0^2+br_1^2+overline boverline r_1^2=1.\$$
As one can check from the numerical values, the ratio of two successive terms quickly tends to the constant $1.465571231876768$, which is the real root (it has the largest modulus).
By numerical experimentation, it appears that the simple formula yields exact values in a large range:
$$c_n=[0.6114919919508175cdot1.465571231876768^n]$$
where the brackets denote rounding to the nearest integer.
$endgroup$
$begingroup$
Thanks for the code part :)
$endgroup$
– Shobi
yesterday
add a comment |
$begingroup$
As we can take one or three, we have the recurrence
$$c_n=c_{n-1}+c_{n-3}$$
with the initial condition
$$c_2=c_1=c_0=1,$$ as with $n<3$ there is a single option.
It is an easy matter to program this using a recursive function, preferably with memoization. Anon-recursive solution is also possible and preferable, storing the values in an array.
C= [1, 1, 1]
for n in range(3, 20):
C.append(C[n-1] + C[n-3])
1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872
Now let the roots of the caracteristic polynomial, $r^3-r^2-1$ be $r_0, r_1, overline r_1$ (there is a pair of conjugates). The general solution of the recurrence is
$$c_n=ar_0^n+br_1^n+overline boverline r_1^n,$$
where constants are determined by the system of equations
$$a+b+overline b=1,\ar_0+br_1+overline boverline r_1=1,\ar_0^2+br_1^2+overline boverline r_1^2=1.\$$
As one can check from the numerical values, the ratio of two successive terms quickly tends to the constant $1.465571231876768$, which is the real root (it has the largest modulus).
By numerical experimentation, it appears that the simple formula yields exact values in a large range:
$$c_n=[0.6114919919508175cdot1.465571231876768^n]$$
where the brackets denote rounding to the nearest integer.
$endgroup$
$begingroup$
Thanks for the code part :)
$endgroup$
– Shobi
yesterday
add a comment |
$begingroup$
As we can take one or three, we have the recurrence
$$c_n=c_{n-1}+c_{n-3}$$
with the initial condition
$$c_2=c_1=c_0=1,$$ as with $n<3$ there is a single option.
It is an easy matter to program this using a recursive function, preferably with memoization. Anon-recursive solution is also possible and preferable, storing the values in an array.
C= [1, 1, 1]
for n in range(3, 20):
C.append(C[n-1] + C[n-3])
1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872
Now let the roots of the caracteristic polynomial, $r^3-r^2-1$ be $r_0, r_1, overline r_1$ (there is a pair of conjugates). The general solution of the recurrence is
$$c_n=ar_0^n+br_1^n+overline boverline r_1^n,$$
where constants are determined by the system of equations
$$a+b+overline b=1,\ar_0+br_1+overline boverline r_1=1,\ar_0^2+br_1^2+overline boverline r_1^2=1.\$$
As one can check from the numerical values, the ratio of two successive terms quickly tends to the constant $1.465571231876768$, which is the real root (it has the largest modulus).
By numerical experimentation, it appears that the simple formula yields exact values in a large range:
$$c_n=[0.6114919919508175cdot1.465571231876768^n]$$
where the brackets denote rounding to the nearest integer.
$endgroup$
As we can take one or three, we have the recurrence
$$c_n=c_{n-1}+c_{n-3}$$
with the initial condition
$$c_2=c_1=c_0=1,$$ as with $n<3$ there is a single option.
It is an easy matter to program this using a recursive function, preferably with memoization. Anon-recursive solution is also possible and preferable, storing the values in an array.
C= [1, 1, 1]
for n in range(3, 20):
C.append(C[n-1] + C[n-3])
1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872
Now let the roots of the caracteristic polynomial, $r^3-r^2-1$ be $r_0, r_1, overline r_1$ (there is a pair of conjugates). The general solution of the recurrence is
$$c_n=ar_0^n+br_1^n+overline boverline r_1^n,$$
where constants are determined by the system of equations
$$a+b+overline b=1,\ar_0+br_1+overline boverline r_1=1,\ar_0^2+br_1^2+overline boverline r_1^2=1.\$$
As one can check from the numerical values, the ratio of two successive terms quickly tends to the constant $1.465571231876768$, which is the real root (it has the largest modulus).
By numerical experimentation, it appears that the simple formula yields exact values in a large range:
$$c_n=[0.6114919919508175cdot1.465571231876768^n]$$
where the brackets denote rounding to the nearest integer.
edited yesterday
answered yesterday
Yves DaoustYves Daoust
126k672226
126k672226
$begingroup$
Thanks for the code part :)
$endgroup$
– Shobi
yesterday
add a comment |
$begingroup$
Thanks for the code part :)
$endgroup$
– Shobi
yesterday
$begingroup$
Thanks for the code part :)
$endgroup$
– Shobi
yesterday
$begingroup$
Thanks for the code part :)
$endgroup$
– Shobi
yesterday
add a comment |
Shobi is a new contributor. Be nice, and check out our Code of Conduct.
Shobi is a new contributor. Be nice, and check out our Code of Conduct.
Shobi is a new contributor. Be nice, and check out our Code of Conduct.
Shobi is a new contributor. Be nice, and check out our Code of Conduct.
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So you're looking at the number of positive integers solutions to $$ 3x + y = k $$ for a fixed value for $k$. Maybe something like this can help: math.stackexchange.com/questions/29019/… and handakafunda.com/…
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– Matti P.
yesterday
2
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@MattiP. I think $x,y$ should be non-negative integers, not be positive integers. And the order is also important. For example, if $k=4$, then $(x,y)=(1,1)$ or $(0,4)$. But in the case $(x,y)=(1,1)$, $31$ and $13$ is not the same way...
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– Doyun Nam
yesterday
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yes. the order is important.
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– Shobi
yesterday
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The OEIS is always available as a last (or first) resort. Entering your sequence 2, 3, 4, 6, 9 gives A000930 which includes lots of (potentially) useful stuff. The Formulas section may be of use to you.
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– Peter Phipps
yesterday
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Thanks, Peter. I actually tried some online sequence solutions, but they didn't gave me much of idea. but this one I am encountering it for the first time its super usefull
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– Shobi
22 hours ago