Mixing
Convergence of $a_n/n$ where $a_0=1$ and $a_n=a_{n/2}+a_{n/3}+a_{n/6}$
$begingroup$
Given $a_0=1$ and:$$a_n=a_{frac{n}{2}}+a_{frac{n}{3}}+a_{frac{n}{6}}$$Find convergence or divergence of $frac{a_n}{n}$.
Such a weird problem; I don't know how to attack it. I'm also fairly certain I typed it out correctly.
real-analysis sequences-and-series
edited Jan 11 at 22:06
Did
248k23223460
asked Mar 26 '12 at 5:04
user27572user27572
112
$endgroup$
|
show 5 more comments
$begingroup$
Given $a_0=1$ and:$$a_n=a_{frac{n}{2}}+a_{frac{n}{3}}+a_{frac{n}{6}}$$Find convergence or divergence of $frac{a_n}{n}$.
Such a weird problem; I don't know how to attack it. I'm also fairly certain I typed it out correctly.
real-analysis sequences-and-series
edited Jan 11 at 22:06
Did
248k23223460
asked Mar 26 '12 at 5:04
user27572user27572
112
$endgroup$
$begingroup$
I am told this has to do with stochastic processes and probability theory. I do not actually know if this is the case.
$endgroup$
– user27572
Mar 26 '12 at 5:10
1
$begingroup$
How is $frac{n}{6}$ defined when $n$ isn't a multiple of $6$?
$endgroup$
– Steven Stadnicki
Mar 26 '12 at 5:26
$begingroup$
The sequence needs more than just $a_0$ to be well-defined. You need to know beforehand the values of $a_n$ for all $n$ which are not multiples of $6$.
$endgroup$
– Antonio Vargas
Mar 26 '12 at 5:30
2
$begingroup$
@user27572, then the problem is silly exactly as it was given.
$endgroup$
– Antonio Vargas
Mar 26 '12 at 5:36
1
$begingroup$
Alright, so I'm going to work under the assumption that the professor handed out a mistyped question, and everyone except me figured this out. :( In which case, thanks guys! It was still very helpful to understand why this question was demonstrably unsolvable. I believe my professor might've meant $a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$, as specified by @Henry.
$endgroup$
– user27572
Mar 26 '12 at 19:23
|
show 5 more comments
$begingroup$
Given $a_0=1$ and:$$a_n=a_{frac{n}{2}}+a_{frac{n}{3}}+a_{frac{n}{6}}$$Find convergence or divergence of $frac{a_n}{n}$.
Such a weird problem; I don't know how to attack it. I'm also fairly certain I typed it out correctly.
real-analysis sequences-and-series
edited Jan 11 at 22:06
Did
248k23223460
asked Mar 26 '12 at 5:04
user27572user27572
112
$endgroup$
Given $a_0=1$ and:$$a_n=a_{frac{n}{2}}+a_{frac{n}{3}}+a_{frac{n}{6}}$$Find convergence or divergence of $frac{a_n}{n}$.
Such a weird problem; I don't know how to attack it. I'm also fairly certain I typed it out correctly.
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Jan 11 at 22:06
Did
248k23223460
asked Mar 26 '12 at 5:04
user27572user27572
112
edited Jan 11 at 22:06
Did
248k23223460
asked Mar 26 '12 at 5:04
user27572user27572
112
edited Jan 11 at 22:06
Did
248k23223460
edited Jan 11 at 22:06
Did
248k23223460
edited Jan 11 at 22:06
Did
248k23223460
248k23223460
asked Mar 26 '12 at 5:04
user27572user27572
112
asked Mar 26 '12 at 5:04
user27572user27572
112
asked Mar 26 '12 at 5:04
user27572user27572
112
112
$begingroup$
I am told this has to do with stochastic processes and probability theory. I do not actually know if this is the case.
$endgroup$
– user27572
Mar 26 '12 at 5:10
1
$begingroup$
How is $frac{n}{6}$ defined when $n$ isn't a multiple of $6$?
$endgroup$
– Steven Stadnicki
Mar 26 '12 at 5:26
$begingroup$
The sequence needs more than just $a_0$ to be well-defined. You need to know beforehand the values of $a_n$ for all $n$ which are not multiples of $6$.
$endgroup$
– Antonio Vargas
Mar 26 '12 at 5:30
2
$begingroup$
@user27572, then the problem is silly exactly as it was given.
$endgroup$
– Antonio Vargas
Mar 26 '12 at 5:36
1
$begingroup$
Alright, so I'm going to work under the assumption that the professor handed out a mistyped question, and everyone except me figured this out. :( In which case, thanks guys! It was still very helpful to understand why this question was demonstrably unsolvable. I believe my professor might've meant $a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$, as specified by @Henry.
$endgroup$
– user27572
Mar 26 '12 at 19:23
|
show 5 more comments
$begingroup$
I am told this has to do with stochastic processes and probability theory. I do not actually know if this is the case.
$endgroup$
– user27572
Mar 26 '12 at 5:10
1
$begingroup$
How is $frac{n}{6}$ defined when $n$ isn't a multiple of $6$?
$endgroup$
– Steven Stadnicki
Mar 26 '12 at 5:26
$begingroup$
The sequence needs more than just $a_0$ to be well-defined. You need to know beforehand the values of $a_n$ for all $n$ which are not multiples of $6$.
$endgroup$
– Antonio Vargas
Mar 26 '12 at 5:30
2
$begingroup$
@user27572, then the problem is silly exactly as it was given.
$endgroup$
– Antonio Vargas
Mar 26 '12 at 5:36
1
$begingroup$
Alright, so I'm going to work under the assumption that the professor handed out a mistyped question, and everyone except me figured this out. :( In which case, thanks guys! It was still very helpful to understand why this question was demonstrably unsolvable. I believe my professor might've meant $a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$, as specified by @Henry.
$endgroup$
– user27572
Mar 26 '12 at 19:23
$begingroup$
I am told this has to do with stochastic processes and probability theory. I do not actually know if this is the case.
$endgroup$
– user27572
Mar 26 '12 at 5:10
$begingroup$
I am told this has to do with stochastic processes and probability theory. I do not actually know if this is the case.
$endgroup$
– user27572
Mar 26 '12 at 5:10
1
1
$begingroup$
How is $frac{n}{6}$ defined when $n$ isn't a multiple of $6$?
$endgroup$
– Steven Stadnicki
Mar 26 '12 at 5:26
$begingroup$
How is $frac{n}{6}$ defined when $n$ isn't a multiple of $6$?
$endgroup$
– Steven Stadnicki
Mar 26 '12 at 5:26
$begingroup$
The sequence needs more than just $a_0$ to be well-defined. You need to know beforehand the values of $a_n$ for all $n$ which are not multiples of $6$.
$endgroup$
– Antonio Vargas
Mar 26 '12 at 5:30
$begingroup$
The sequence needs more than just $a_0$ to be well-defined. You need to know beforehand the values of $a_n$ for all $n$ which are not multiples of $6$.
$endgroup$
– Antonio Vargas
Mar 26 '12 at 5:30
2
2
$begingroup$
@user27572, then the problem is silly exactly as it was given.
$endgroup$
– Antonio Vargas
Mar 26 '12 at 5:36
$begingroup$
@user27572, then the problem is silly exactly as it was given.
$endgroup$
– Antonio Vargas
Mar 26 '12 at 5:36
1
1
$begingroup$
Alright, so I'm going to work under the assumption that the professor handed out a mistyped question, and everyone except me figured this out. :( In which case, thanks guys! It was still very helpful to understand why this question was demonstrably unsolvable. I believe my professor might've meant $a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$, as specified by @Henry.
$endgroup$
– user27572
Mar 26 '12 at 19:23
$begingroup$
Alright, so I'm going to work under the assumption that the professor handed out a mistyped question, and everyone except me figured this out. :( In which case, thanks guys! It was still very helpful to understand why this question was demonstrably unsolvable. I believe my professor might've meant $a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$, as specified by @Henry.
$endgroup$
– user27572
Mar 26 '12 at 19:23
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
On the assumption that the question is really $$a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$$ starting with $a_0=1$ (otherwise how do you work out $a_1$?), this is OEIS A007731 and was covered by P. Erdos, A. Hildebrand, A. Odlyzko, P. Pudaite and B. Reznick, The asymptotic behavior of a family of sequences, Pacific J. Math., 126 (1987), pp. 227-241.
They showed that $dfrac{a_n}{n}$ converges to $dfrac{12}{log_e 432} approx 1.97744865$, though the coveregence is very slow: $frac{a_n}{n}$ is about $1.8430$ when $n=432^5-1$ but about $2.1175$ when $n=432^5$.
answered Mar 26 '12 at 8:43
HenryHenry
100k480165
$endgroup$
$begingroup$
I'm almost certain that this was what my professor meant, although I have the paper right in front of me, and it says otherwise. ಠ_ಠ Very helpful, @Henry , thanks!
$endgroup$
– user27572
Mar 26 '12 at 19:24
add a comment |
$begingroup$
Caveat: This answer is concerned with the problem as stated by the OP, that is, with sequence(s) defined by $a_0=1$ and $a_n=a_{n/2}+a_{n/3}+a_{n/6}$ for every $ngeqslant1$ such that the RHS of the equality makes sense.
user27572: Are you absolutely sure that this problem does not have a solution as given?
Absolutely sure. As explained by @Antonio in the comments the initial condition $a_0=1$ is not enough to determine $(a_n)_{ngeqslant0}$ since one needs to choose the value of $a_n$ for every $n$ not a multiple of $6$.
Obviously, each sequence defined by $a_0=1$ and $a_n=beta n$ for some fixed $beta$ and for every nonzero $n$, is a solution. For this choice, the sequence $(b_n)_{ngeqslant1}$ defined by $b_n=a_n/n$ is convergent, to $beta$.
Here is a different, explicit, choice of $a_n$ for $n$ a power of $2$ and for $n$ a power of $3$, which yields a divergent subsequence $(b_n)_n$ when $n$ is restricted to the products of a power of $2$ by a power of $3$, and, a fortiori, a divergent complete sequence $(b_n)_{ngeqslant1}$. Define
$$
a_n=r^{i+j}quadtext{if}quad n=2^i3^j,
$$
for some fixed $r$.
A one-line computation shows that this solves the desired recursion as soon as $r$ solves $r^2=2r+1$, hence, for example, for $r=1+sqrt2$.
Now, $b_nto+infty$ when $ntoinfty$ while $n$ is a power of $2$ because $rgt2$ and $b_nto0$ when $ntoinfty$ while $n$ is a power of $3$ because $rlt3$. Furthermore, for every finite positive $beta$, there exists a sequence $(n_k)_k$ such that every $n_k$ is the product of a power of $2$ by a power of $3$ and such that $b_{n_k}tobeta$ when $ktoinfty$. Thus, the set of limit points of the sequence $(b_n)_{n}$ when $n$ is restricted to the products of a power of $2$ by a power of $3$ is $[0,+infty]$.
More generally, $a_n=betacdotprodlimits_pr_p^{i_p}$ for $n=prodlimits_pp^{i_p}$, where the products are over the set of primes $p$, yields a solution for every family of weights $(r_p)_p$ such that $r_2r_3=r_2+r_3+1$.
The (admissible) choice $r_p=p$ for every prime $p$ yields $a_n=beta n$. This is the only case when $(b_n)_{ngeqslant1}$ converges. The (admissible) choice $r_2=r_3=1+sqrt2$ yields the solution above.
Note: The recursion $a_n=a_{lfloor n/2rfloor}+a_{lfloor n/3rfloor}+a_{lfloor n/6rfloor}$ is a different story...
answered Mar 26 '12 at 6:25
DidDid
248k23223460
$endgroup$
$begingroup$
I have to conclude that this was not what my professor meant, but nonetheless I was very interested in a question/solution in the general vein of what I thought my professor meant originally. Cool stuff.
$endgroup$
– user27572
Mar 26 '12 at 19:28
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
On the assumption that the question is really $$a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$$ starting with $a_0=1$ (otherwise how do you work out $a_1$?), this is OEIS A007731 and was covered by P. Erdos, A. Hildebrand, A. Odlyzko, P. Pudaite and B. Reznick, The asymptotic behavior of a family of sequences, Pacific J. Math., 126 (1987), pp. 227-241.
They showed that $dfrac{a_n}{n}$ converges to $dfrac{12}{log_e 432} approx 1.97744865$, though the coveregence is very slow: $frac{a_n}{n}$ is about $1.8430$ when $n=432^5-1$ but about $2.1175$ when $n=432^5$.
answered Mar 26 '12 at 8:43
HenryHenry
100k480165
$endgroup$
$begingroup$
I'm almost certain that this was what my professor meant, although I have the paper right in front of me, and it says otherwise. ಠ_ಠ Very helpful, @Henry , thanks!
$endgroup$
– user27572
Mar 26 '12 at 19:24
add a comment |
$begingroup$
On the assumption that the question is really $$a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$$ starting with $a_0=1$ (otherwise how do you work out $a_1$?), this is OEIS A007731 and was covered by P. Erdos, A. Hildebrand, A. Odlyzko, P. Pudaite and B. Reznick, The asymptotic behavior of a family of sequences, Pacific J. Math., 126 (1987), pp. 227-241.
They showed that $dfrac{a_n}{n}$ converges to $dfrac{12}{log_e 432} approx 1.97744865$, though the coveregence is very slow: $frac{a_n}{n}$ is about $1.8430$ when $n=432^5-1$ but about $2.1175$ when $n=432^5$.
answered Mar 26 '12 at 8:43
HenryHenry
100k480165
$endgroup$
$begingroup$
I'm almost certain that this was what my professor meant, although I have the paper right in front of me, and it says otherwise. ಠ_ಠ Very helpful, @Henry , thanks!
$endgroup$
– user27572
Mar 26 '12 at 19:24
add a comment |
$begingroup$
On the assumption that the question is really $$a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$$ starting with $a_0=1$ (otherwise how do you work out $a_1$?), this is OEIS A007731 and was covered by P. Erdos, A. Hildebrand, A. Odlyzko, P. Pudaite and B. Reznick, The asymptotic behavior of a family of sequences, Pacific J. Math., 126 (1987), pp. 227-241.
They showed that $dfrac{a_n}{n}$ converges to $dfrac{12}{log_e 432} approx 1.97744865$, though the coveregence is very slow: $frac{a_n}{n}$ is about $1.8430$ when $n=432^5-1$ but about $2.1175$ when $n=432^5$.
answered Mar 26 '12 at 8:43
HenryHenry
100k480165
$endgroup$
On the assumption that the question is really $$a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$$ starting with $a_0=1$ (otherwise how do you work out $a_1$?), this is OEIS A007731 and was covered by P. Erdos, A. Hildebrand, A. Odlyzko, P. Pudaite and B. Reznick, The asymptotic behavior of a family of sequences, Pacific J. Math., 126 (1987), pp. 227-241.
They showed that $dfrac{a_n}{n}$ converges to $dfrac{12}{log_e 432} approx 1.97744865$, though the coveregence is very slow: $frac{a_n}{n}$ is about $1.8430$ when $n=432^5-1$ but about $2.1175$ when $n=432^5$.
answered Mar 26 '12 at 8:43
HenryHenry
100k480165
answered Mar 26 '12 at 8:43
HenryHenry
100k480165
answered Mar 26 '12 at 8:43
HenryHenry
100k480165
answered Mar 26 '12 at 8:43
HenryHenry
100k480165
100k480165
$begingroup$
I'm almost certain that this was what my professor meant, although I have the paper right in front of me, and it says otherwise. ಠ_ಠ Very helpful, @Henry , thanks!
$endgroup$
– user27572
Mar 26 '12 at 19:24
add a comment |
$begingroup$
I'm almost certain that this was what my professor meant, although I have the paper right in front of me, and it says otherwise. ಠ_ಠ Very helpful, @Henry , thanks!
$endgroup$
– user27572
Mar 26 '12 at 19:24
$begingroup$
I'm almost certain that this was what my professor meant, although I have the paper right in front of me, and it says otherwise. ಠ_ಠ Very helpful, @Henry , thanks!
$endgroup$
– user27572
Mar 26 '12 at 19:24
$begingroup$
I'm almost certain that this was what my professor meant, although I have the paper right in front of me, and it says otherwise. ಠ_ಠ Very helpful, @Henry , thanks!
$endgroup$
– user27572
Mar 26 '12 at 19:24
add a comment |
$begingroup$
Caveat: This answer is concerned with the problem as stated by the OP, that is, with sequence(s) defined by $a_0=1$ and $a_n=a_{n/2}+a_{n/3}+a_{n/6}$ for every $ngeqslant1$ such that the RHS of the equality makes sense.
user27572: Are you absolutely sure that this problem does not have a solution as given?
Absolutely sure. As explained by @Antonio in the comments the initial condition $a_0=1$ is not enough to determine $(a_n)_{ngeqslant0}$ since one needs to choose the value of $a_n$ for every $n$ not a multiple of $6$.
Obviously, each sequence defined by $a_0=1$ and $a_n=beta n$ for some fixed $beta$ and for every nonzero $n$, is a solution. For this choice, the sequence $(b_n)_{ngeqslant1}$ defined by $b_n=a_n/n$ is convergent, to $beta$.
Here is a different, explicit, choice of $a_n$ for $n$ a power of $2$ and for $n$ a power of $3$, which yields a divergent subsequence $(b_n)_n$ when $n$ is restricted to the products of a power of $2$ by a power of $3$, and, a fortiori, a divergent complete sequence $(b_n)_{ngeqslant1}$. Define
$$
a_n=r^{i+j}quadtext{if}quad n=2^i3^j,
$$
for some fixed $r$.
A one-line computation shows that this solves the desired recursion as soon as $r$ solves $r^2=2r+1$, hence, for example, for $r=1+sqrt2$.
Now, $b_nto+infty$ when $ntoinfty$ while $n$ is a power of $2$ because $rgt2$ and $b_nto0$ when $ntoinfty$ while $n$ is a power of $3$ because $rlt3$. Furthermore, for every finite positive $beta$, there exists a sequence $(n_k)_k$ such that every $n_k$ is the product of a power of $2$ by a power of $3$ and such that $b_{n_k}tobeta$ when $ktoinfty$. Thus, the set of limit points of the sequence $(b_n)_{n}$ when $n$ is restricted to the products of a power of $2$ by a power of $3$ is $[0,+infty]$.
More generally, $a_n=betacdotprodlimits_pr_p^{i_p}$ for $n=prodlimits_pp^{i_p}$, where the products are over the set of primes $p$, yields a solution for every family of weights $(r_p)_p$ such that $r_2r_3=r_2+r_3+1$.
The (admissible) choice $r_p=p$ for every prime $p$ yields $a_n=beta n$. This is the only case when $(b_n)_{ngeqslant1}$ converges. The (admissible) choice $r_2=r_3=1+sqrt2$ yields the solution above.
Note: The recursion $a_n=a_{lfloor n/2rfloor}+a_{lfloor n/3rfloor}+a_{lfloor n/6rfloor}$ is a different story...
answered Mar 26 '12 at 6:25
DidDid
248k23223460
$endgroup$
$begingroup$
I have to conclude that this was not what my professor meant, but nonetheless I was very interested in a question/solution in the general vein of what I thought my professor meant originally. Cool stuff.
$endgroup$
– user27572
Mar 26 '12 at 19:28
add a comment |
$begingroup$
Caveat: This answer is concerned with the problem as stated by the OP, that is, with sequence(s) defined by $a_0=1$ and $a_n=a_{n/2}+a_{n/3}+a_{n/6}$ for every $ngeqslant1$ such that the RHS of the equality makes sense.
user27572: Are you absolutely sure that this problem does not have a solution as given?
Absolutely sure. As explained by @Antonio in the comments the initial condition $a_0=1$ is not enough to determine $(a_n)_{ngeqslant0}$ since one needs to choose the value of $a_n$ for every $n$ not a multiple of $6$.
Obviously, each sequence defined by $a_0=1$ and $a_n=beta n$ for some fixed $beta$ and for every nonzero $n$, is a solution. For this choice, the sequence $(b_n)_{ngeqslant1}$ defined by $b_n=a_n/n$ is convergent, to $beta$.
Here is a different, explicit, choice of $a_n$ for $n$ a power of $2$ and for $n$ a power of $3$, which yields a divergent subsequence $(b_n)_n$ when $n$ is restricted to the products of a power of $2$ by a power of $3$, and, a fortiori, a divergent complete sequence $(b_n)_{ngeqslant1}$. Define
$$
a_n=r^{i+j}quadtext{if}quad n=2^i3^j,
$$
for some fixed $r$.
A one-line computation shows that this solves the desired recursion as soon as $r$ solves $r^2=2r+1$, hence, for example, for $r=1+sqrt2$.
Now, $b_nto+infty$ when $ntoinfty$ while $n$ is a power of $2$ because $rgt2$ and $b_nto0$ when $ntoinfty$ while $n$ is a power of $3$ because $rlt3$. Furthermore, for every finite positive $beta$, there exists a sequence $(n_k)_k$ such that every $n_k$ is the product of a power of $2$ by a power of $3$ and such that $b_{n_k}tobeta$ when $ktoinfty$. Thus, the set of limit points of the sequence $(b_n)_{n}$ when $n$ is restricted to the products of a power of $2$ by a power of $3$ is $[0,+infty]$.
More generally, $a_n=betacdotprodlimits_pr_p^{i_p}$ for $n=prodlimits_pp^{i_p}$, where the products are over the set of primes $p$, yields a solution for every family of weights $(r_p)_p$ such that $r_2r_3=r_2+r_3+1$.
The (admissible) choice $r_p=p$ for every prime $p$ yields $a_n=beta n$. This is the only case when $(b_n)_{ngeqslant1}$ converges. The (admissible) choice $r_2=r_3=1+sqrt2$ yields the solution above.
Note: The recursion $a_n=a_{lfloor n/2rfloor}+a_{lfloor n/3rfloor}+a_{lfloor n/6rfloor}$ is a different story...
answered Mar 26 '12 at 6:25
DidDid
248k23223460
$endgroup$
$begingroup$
I have to conclude that this was not what my professor meant, but nonetheless I was very interested in a question/solution in the general vein of what I thought my professor meant originally. Cool stuff.
$endgroup$
– user27572
Mar 26 '12 at 19:28
add a comment |
$begingroup$
Caveat: This answer is concerned with the problem as stated by the OP, that is, with sequence(s) defined by $a_0=1$ and $a_n=a_{n/2}+a_{n/3}+a_{n/6}$ for every $ngeqslant1$ such that the RHS of the equality makes sense.
user27572: Are you absolutely sure that this problem does not have a solution as given?
Absolutely sure. As explained by @Antonio in the comments the initial condition $a_0=1$ is not enough to determine $(a_n)_{ngeqslant0}$ since one needs to choose the value of $a_n$ for every $n$ not a multiple of $6$.
Obviously, each sequence defined by $a_0=1$ and $a_n=beta n$ for some fixed $beta$ and for every nonzero $n$, is a solution. For this choice, the sequence $(b_n)_{ngeqslant1}$ defined by $b_n=a_n/n$ is convergent, to $beta$.
Here is a different, explicit, choice of $a_n$ for $n$ a power of $2$ and for $n$ a power of $3$, which yields a divergent subsequence $(b_n)_n$ when $n$ is restricted to the products of a power of $2$ by a power of $3$, and, a fortiori, a divergent complete sequence $(b_n)_{ngeqslant1}$. Define
$$
a_n=r^{i+j}quadtext{if}quad n=2^i3^j,
$$
for some fixed $r$.
A one-line computation shows that this solves the desired recursion as soon as $r$ solves $r^2=2r+1$, hence, for example, for $r=1+sqrt2$.
Now, $b_nto+infty$ when $ntoinfty$ while $n$ is a power of $2$ because $rgt2$ and $b_nto0$ when $ntoinfty$ while $n$ is a power of $3$ because $rlt3$. Furthermore, for every finite positive $beta$, there exists a sequence $(n_k)_k$ such that every $n_k$ is the product of a power of $2$ by a power of $3$ and such that $b_{n_k}tobeta$ when $ktoinfty$. Thus, the set of limit points of the sequence $(b_n)_{n}$ when $n$ is restricted to the products of a power of $2$ by a power of $3$ is $[0,+infty]$.
More generally, $a_n=betacdotprodlimits_pr_p^{i_p}$ for $n=prodlimits_pp^{i_p}$, where the products are over the set of primes $p$, yields a solution for every family of weights $(r_p)_p$ such that $r_2r_3=r_2+r_3+1$.
The (admissible) choice $r_p=p$ for every prime $p$ yields $a_n=beta n$. This is the only case when $(b_n)_{ngeqslant1}$ converges. The (admissible) choice $r_2=r_3=1+sqrt2$ yields the solution above.
Note: The recursion $a_n=a_{lfloor n/2rfloor}+a_{lfloor n/3rfloor}+a_{lfloor n/6rfloor}$ is a different story...
answered Mar 26 '12 at 6:25
DidDid
248k23223460
$endgroup$
Caveat: This answer is concerned with the problem as stated by the OP, that is, with sequence(s) defined by $a_0=1$ and $a_n=a_{n/2}+a_{n/3}+a_{n/6}$ for every $ngeqslant1$ such that the RHS of the equality makes sense.
user27572: Are you absolutely sure that this problem does not have a solution as given?
Absolutely sure. As explained by @Antonio in the comments the initial condition $a_0=1$ is not enough to determine $(a_n)_{ngeqslant0}$ since one needs to choose the value of $a_n$ for every $n$ not a multiple of $6$.
Obviously, each sequence defined by $a_0=1$ and $a_n=beta n$ for some fixed $beta$ and for every nonzero $n$, is a solution. For this choice, the sequence $(b_n)_{ngeqslant1}$ defined by $b_n=a_n/n$ is convergent, to $beta$.
Here is a different, explicit, choice of $a_n$ for $n$ a power of $2$ and for $n$ a power of $3$, which yields a divergent subsequence $(b_n)_n$ when $n$ is restricted to the products of a power of $2$ by a power of $3$, and, a fortiori, a divergent complete sequence $(b_n)_{ngeqslant1}$. Define
$$
a_n=r^{i+j}quadtext{if}quad n=2^i3^j,
$$
for some fixed $r$.
A one-line computation shows that this solves the desired recursion as soon as $r$ solves $r^2=2r+1$, hence, for example, for $r=1+sqrt2$.
Now, $b_nto+infty$ when $ntoinfty$ while $n$ is a power of $2$ because $rgt2$ and $b_nto0$ when $ntoinfty$ while $n$ is a power of $3$ because $rlt3$. Furthermore, for every finite positive $beta$, there exists a sequence $(n_k)_k$ such that every $n_k$ is the product of a power of $2$ by a power of $3$ and such that $b_{n_k}tobeta$ when $ktoinfty$. Thus, the set of limit points of the sequence $(b_n)_{n}$ when $n$ is restricted to the products of a power of $2$ by a power of $3$ is $[0,+infty]$.
More generally, $a_n=betacdotprodlimits_pr_p^{i_p}$ for $n=prodlimits_pp^{i_p}$, where the products are over the set of primes $p$, yields a solution for every family of weights $(r_p)_p$ such that $r_2r_3=r_2+r_3+1$.
The (admissible) choice $r_p=p$ for every prime $p$ yields $a_n=beta n$. This is the only case when $(b_n)_{ngeqslant1}$ converges. The (admissible) choice $r_2=r_3=1+sqrt2$ yields the solution above.
Note: The recursion $a_n=a_{lfloor n/2rfloor}+a_{lfloor n/3rfloor}+a_{lfloor n/6rfloor}$ is a different story...
answered Mar 26 '12 at 6:25
DidDid
248k23223460
edited Mar 26 '12 at 11:03
answered Mar 26 '12 at 6:25
DidDid
248k23223460
answered Mar 26 '12 at 6:25
DidDid
248k23223460
answered Mar 26 '12 at 6:25
DidDid
248k23223460
248k23223460
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I have to conclude that this was not what my professor meant, but nonetheless I was very interested in a question/solution in the general vein of what I thought my professor meant originally. Cool stuff.
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– user27572
Mar 26 '12 at 19:28
add a comment |
$begingroup$
I have to conclude that this was not what my professor meant, but nonetheless I was very interested in a question/solution in the general vein of what I thought my professor meant originally. Cool stuff.
$endgroup$
– user27572
Mar 26 '12 at 19:28
$begingroup$
I have to conclude that this was not what my professor meant, but nonetheless I was very interested in a question/solution in the general vein of what I thought my professor meant originally. Cool stuff.
$endgroup$
– user27572
Mar 26 '12 at 19:28
$begingroup$
I have to conclude that this was not what my professor meant, but nonetheless I was very interested in a question/solution in the general vein of what I thought my professor meant originally. Cool stuff.
$endgroup$
– user27572
Mar 26 '12 at 19:28
add a comment |
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I am told this has to do with stochastic processes and probability theory. I do not actually know if this is the case.
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– user27572
Mar 26 '12 at 5:10
1
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How is $frac{n}{6}$ defined when $n$ isn't a multiple of $6$?
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– Steven Stadnicki
Mar 26 '12 at 5:26
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The sequence needs more than just $a_0$ to be well-defined. You need to know beforehand the values of $a_n$ for all $n$ which are not multiples of $6$.
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– Antonio Vargas
Mar 26 '12 at 5:30
2
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@user27572, then the problem is silly exactly as it was given.
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– Antonio Vargas
Mar 26 '12 at 5:36
1
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Alright, so I'm going to work under the assumption that the professor handed out a mistyped question, and everyone except me figured this out. :( In which case, thanks guys! It was still very helpful to understand why this question was demonstrably unsolvable. I believe my professor might've meant $a_n=a_{lfloorfrac{n}2rfloor}+a_{lfloorfrac{n}3rfloor}+a_{lfloorfrac{n}6rfloor}$, as specified by @Henry.
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– user27572
Mar 26 '12 at 19:23