What is the relation in this sequence












3












$begingroup$


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







share|cite|improve this question









New contributor




Shobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    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
















3












$begingroup$


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







share|cite|improve this question









New contributor




Shobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    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














3












3








3


3



$begingroup$


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







share|cite|improve this question









New contributor




Shobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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






share|cite|improve this question









New contributor




Shobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Shobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited yesterday









YuiTo Cheng

746322




746322






New contributor




Shobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









ShobiShobi

1185




1185




New contributor




Shobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Shobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Shobi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    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








  • 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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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



















6












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the code part :)
    $endgroup$
    – Shobi
    yesterday











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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
















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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














3












3








3





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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














  • 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











6












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the code part :)
    $endgroup$
    – Shobi
    yesterday
















6












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the code part :)
    $endgroup$
    – Shobi
    yesterday














6












6








6





$begingroup$

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.






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$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









Yves DaoustYves Daoust

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  • $begingroup$
    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




$begingroup$
Thanks for the code part :)
$endgroup$
– Shobi
yesterday










Shobi is a new contributor. Be nice, and check out our Code of Conduct.










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