Finding a recurrence relation
I need to find a recurrence relation or an exact formula to the sequence
$$1,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288,ldots$$
Well, considering $a_0=1$, $a_1=2$ and $a_2=4$, the terms $a_n$ for $ngeq3$ is obtained by adding a power of $2$. In fact the sequence is:
$$1xrightarrow{+1}2xrightarrow{+2}4xrightarrow{+4}8xrightarrow{+4}12xrightarrow{+4}16xrightarrow{+8}24xrightarrow{+8}32xrightarrow{+8}40xrightarrow{+8}48xrightarrow{+16}64xrightarrow{+16}80xrightarrow{+16}96xrightarrow{+16}112xrightarrow{+16}128xrightarrow{+32}160ldots$$
and we will have $5+1$ addition of $32=2^5$ and then $6+1$ addition of $64=2^6$ and so on.
Any, solutions, suggestions,... please.
recurrence-relations
asked Dec 12 '18 at 7:26
QurultayQurultay
591313
add a comment |
I need to find a recurrence relation or an exact formula to the sequence
$$1,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288,ldots$$
Well, considering $a_0=1$, $a_1=2$ and $a_2=4$, the terms $a_n$ for $ngeq3$ is obtained by adding a power of $2$. In fact the sequence is:
$$1xrightarrow{+1}2xrightarrow{+2}4xrightarrow{+4}8xrightarrow{+4}12xrightarrow{+4}16xrightarrow{+8}24xrightarrow{+8}32xrightarrow{+8}40xrightarrow{+8}48xrightarrow{+16}64xrightarrow{+16}80xrightarrow{+16}96xrightarrow{+16}112xrightarrow{+16}128xrightarrow{+32}160ldots$$
and we will have $5+1$ addition of $32=2^5$ and then $6+1$ addition of $64=2^6$ and so on.
Any, solutions, suggestions,... please.
recurrence-relations
asked Dec 12 '18 at 7:26
QurultayQurultay
591313
You should have added +2 twice, not once. Thus your sequence would be $$1,2,4,6,10,14,18,26,34,42,50,66,ldots$$
– Yiorgos S. Smyrlis
Dec 12 '18 at 7:44
@YiorgosS.Smyrlis well, this is not my choice. This sequence comes from a project.
– Qurultay
Dec 12 '18 at 8:16
0,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288
– Satish Ramanathan
Dec 12 '18 at 8:18
should it be starting with 0. If so, +2 two times, +4 three times, +8 four time and +16 five times and hence forth.
– Satish Ramanathan
Dec 12 '18 at 8:19
add a comment |
I need to find a recurrence relation or an exact formula to the sequence
$$1,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288,ldots$$
Well, considering $a_0=1$, $a_1=2$ and $a_2=4$, the terms $a_n$ for $ngeq3$ is obtained by adding a power of $2$. In fact the sequence is:
$$1xrightarrow{+1}2xrightarrow{+2}4xrightarrow{+4}8xrightarrow{+4}12xrightarrow{+4}16xrightarrow{+8}24xrightarrow{+8}32xrightarrow{+8}40xrightarrow{+8}48xrightarrow{+16}64xrightarrow{+16}80xrightarrow{+16}96xrightarrow{+16}112xrightarrow{+16}128xrightarrow{+32}160ldots$$
and we will have $5+1$ addition of $32=2^5$ and then $6+1$ addition of $64=2^6$ and so on.
Any, solutions, suggestions,... please.
recurrence-relations
asked Dec 12 '18 at 7:26
QurultayQurultay
591313
I need to find a recurrence relation or an exact formula to the sequence
$$1,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288,ldots$$
Well, considering $a_0=1$, $a_1=2$ and $a_2=4$, the terms $a_n$ for $ngeq3$ is obtained by adding a power of $2$. In fact the sequence is:
$$1xrightarrow{+1}2xrightarrow{+2}4xrightarrow{+4}8xrightarrow{+4}12xrightarrow{+4}16xrightarrow{+8}24xrightarrow{+8}32xrightarrow{+8}40xrightarrow{+8}48xrightarrow{+16}64xrightarrow{+16}80xrightarrow{+16}96xrightarrow{+16}112xrightarrow{+16}128xrightarrow{+32}160ldots$$
and we will have $5+1$ addition of $32=2^5$ and then $6+1$ addition of $64=2^6$ and so on.
Any, solutions, suggestions,... please.
recurrence-relations
recurrence-relations
asked Dec 12 '18 at 7:26
QurultayQurultay
591313
asked Dec 12 '18 at 7:26
QurultayQurultay
591313
asked Dec 12 '18 at 7:26
QurultayQurultay
591313
asked Dec 12 '18 at 7:26
QurultayQurultay
591313
asked Dec 12 '18 at 7:26
QurultayQurultay
591313
591313
You should have added +2 twice, not once. Thus your sequence would be $$1,2,4,6,10,14,18,26,34,42,50,66,ldots$$
– Yiorgos S. Smyrlis
Dec 12 '18 at 7:44
@YiorgosS.Smyrlis well, this is not my choice. This sequence comes from a project.
– Qurultay
Dec 12 '18 at 8:16
0,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288
– Satish Ramanathan
Dec 12 '18 at 8:18
should it be starting with 0. If so, +2 two times, +4 three times, +8 four time and +16 five times and hence forth.
– Satish Ramanathan
Dec 12 '18 at 8:19
add a comment |
You should have added +2 twice, not once. Thus your sequence would be $$1,2,4,6,10,14,18,26,34,42,50,66,ldots$$
– Yiorgos S. Smyrlis
Dec 12 '18 at 7:44
@YiorgosS.Smyrlis well, this is not my choice. This sequence comes from a project.
– Qurultay
Dec 12 '18 at 8:16
0,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288
– Satish Ramanathan
Dec 12 '18 at 8:18
should it be starting with 0. If so, +2 two times, +4 three times, +8 four time and +16 five times and hence forth.
– Satish Ramanathan
Dec 12 '18 at 8:19
You should have added +2 twice, not once. Thus your sequence would be $$1,2,4,6,10,14,18,26,34,42,50,66,ldots$$
– Yiorgos S. Smyrlis
Dec 12 '18 at 7:44
You should have added +2 twice, not once. Thus your sequence would be $$1,2,4,6,10,14,18,26,34,42,50,66,ldots$$
– Yiorgos S. Smyrlis
Dec 12 '18 at 7:44
@YiorgosS.Smyrlis well, this is not my choice. This sequence comes from a project.
– Qurultay
Dec 12 '18 at 8:16
@YiorgosS.Smyrlis well, this is not my choice. This sequence comes from a project.
– Qurultay
Dec 12 '18 at 8:16
0,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288
– Satish Ramanathan
Dec 12 '18 at 8:18
0,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288
– Satish Ramanathan
Dec 12 '18 at 8:18
should it be starting with 0. If so, +2 two times, +4 three times, +8 four time and +16 five times and hence forth.
– Satish Ramanathan
Dec 12 '18 at 8:19
should it be starting with 0. If so, +2 two times, +4 three times, +8 four time and +16 five times and hence forth.
– Satish Ramanathan
Dec 12 '18 at 8:19
add a comment |
1 Answer
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To compute the $n^{th}$ term, find $k$ such that $frac{k(k-1)}{2} leq n leq frac{k(k+1)}{2}$. Then:
$x(n) = x(n-1) + 2^{k-1}$.
For example, if $n=9$, $k=4$ and $x(9) = x(8) + 8$.
Let's first look at the roots of $k^2-k -2n=0$. They are given by $r_1 = frac{1-sqrt{1+8n}}{2}$ and $r_2 = frac{1+sqrt{1+8n}}{2}$.
The inequality $frac{k(k-1)}{2} leq n$ implies $k^2-k-2n leq 0$, i.e. $(k-r_1)(k-r_2) leq 0$, which is true when $r_1 leq k leq r_2$. Since $k >0$ and $r_1<0$, the condition reduces to $0 < k leq r_2$. For example, when $n=9$, $0 < k leq 4.77$.
Now let's look a the roots of $k^2+k-2n=0$. They are given by $r_3 = frac{-1-sqrt{1+8n}}{2}$ and $r_4 = frac{-1+sqrt{1+8n}}{2}$.
The inequality $n leq frac{k(k+1)}{2}$ implies $k^2+k-2n geq 0$, i.e. $(k-r_3)(k-r_4) geq 0$, which is true when $k leq r_3$ or $k geq r_4$. Since $k>0$, this reduces to $k geq r_4$. For example, when $n=9$, $k geq 3.77$.
Combining the two, we have:
$frac{-1+sqrt{1+8n}}{2} leq k leq frac{1+sqrt{1+8n}}{2}$.
One may be able to write this as $k = text{round} left( frac{sqrt{1+8n}}{2}right)$ and $x(n) = x(n-1) + 2^{k-1}$. The ambiguity when $8n+1$ is a perfect square needs to be resolved (e.g. when $n=10$).
answered Dec 12 '18 at 8:35
Aditya DuaAditya Dua
86418
Does the series start with 0 instead of 1? In other words (x(1) = 0)?
– Satish Ramanathan
Dec 12 '18 at 9:06
Thank you Aditya Duo. But the ambiguity you mentioned, is the main problem. The value $8n+1$ is a perfect square exactly when we have the jump from of addition, i.e. when $n=3,6,10,15,21,ldots$.
– Qurultay
Dec 12 '18 at 9:15
@Qurultay Working through a few examples, it looks like rounding down (i.e. truncating) when $8n+1$ is a perfect square works, though I don't have a rigorous argument. For example, when $n=15$, $sqrt{1+8n}/2 = 5.5$, which is 5 if you round down (and it is the right answer).
– Aditya Dua
Dec 12 '18 at 17:04
add a comment |
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To compute the $n^{th}$ term, find $k$ such that $frac{k(k-1)}{2} leq n leq frac{k(k+1)}{2}$. Then:
$x(n) = x(n-1) + 2^{k-1}$.
For example, if $n=9$, $k=4$ and $x(9) = x(8) + 8$.
Let's first look at the roots of $k^2-k -2n=0$. They are given by $r_1 = frac{1-sqrt{1+8n}}{2}$ and $r_2 = frac{1+sqrt{1+8n}}{2}$.
The inequality $frac{k(k-1)}{2} leq n$ implies $k^2-k-2n leq 0$, i.e. $(k-r_1)(k-r_2) leq 0$, which is true when $r_1 leq k leq r_2$. Since $k >0$ and $r_1<0$, the condition reduces to $0 < k leq r_2$. For example, when $n=9$, $0 < k leq 4.77$.
Now let's look a the roots of $k^2+k-2n=0$. They are given by $r_3 = frac{-1-sqrt{1+8n}}{2}$ and $r_4 = frac{-1+sqrt{1+8n}}{2}$.
The inequality $n leq frac{k(k+1)}{2}$ implies $k^2+k-2n geq 0$, i.e. $(k-r_3)(k-r_4) geq 0$, which is true when $k leq r_3$ or $k geq r_4$. Since $k>0$, this reduces to $k geq r_4$. For example, when $n=9$, $k geq 3.77$.
Combining the two, we have:
$frac{-1+sqrt{1+8n}}{2} leq k leq frac{1+sqrt{1+8n}}{2}$.
One may be able to write this as $k = text{round} left( frac{sqrt{1+8n}}{2}right)$ and $x(n) = x(n-1) + 2^{k-1}$. The ambiguity when $8n+1$ is a perfect square needs to be resolved (e.g. when $n=10$).
answered Dec 12 '18 at 8:35
Aditya DuaAditya Dua
86418
Does the series start with 0 instead of 1? In other words (x(1) = 0)?
– Satish Ramanathan
Dec 12 '18 at 9:06
Thank you Aditya Duo. But the ambiguity you mentioned, is the main problem. The value $8n+1$ is a perfect square exactly when we have the jump from of addition, i.e. when $n=3,6,10,15,21,ldots$.
– Qurultay
Dec 12 '18 at 9:15
@Qurultay Working through a few examples, it looks like rounding down (i.e. truncating) when $8n+1$ is a perfect square works, though I don't have a rigorous argument. For example, when $n=15$, $sqrt{1+8n}/2 = 5.5$, which is 5 if you round down (and it is the right answer).
– Aditya Dua
Dec 12 '18 at 17:04
add a comment |
To compute the $n^{th}$ term, find $k$ such that $frac{k(k-1)}{2} leq n leq frac{k(k+1)}{2}$. Then:
$x(n) = x(n-1) + 2^{k-1}$.
For example, if $n=9$, $k=4$ and $x(9) = x(8) + 8$.
Let's first look at the roots of $k^2-k -2n=0$. They are given by $r_1 = frac{1-sqrt{1+8n}}{2}$ and $r_2 = frac{1+sqrt{1+8n}}{2}$.
The inequality $frac{k(k-1)}{2} leq n$ implies $k^2-k-2n leq 0$, i.e. $(k-r_1)(k-r_2) leq 0$, which is true when $r_1 leq k leq r_2$. Since $k >0$ and $r_1<0$, the condition reduces to $0 < k leq r_2$. For example, when $n=9$, $0 < k leq 4.77$.
Now let's look a the roots of $k^2+k-2n=0$. They are given by $r_3 = frac{-1-sqrt{1+8n}}{2}$ and $r_4 = frac{-1+sqrt{1+8n}}{2}$.
The inequality $n leq frac{k(k+1)}{2}$ implies $k^2+k-2n geq 0$, i.e. $(k-r_3)(k-r_4) geq 0$, which is true when $k leq r_3$ or $k geq r_4$. Since $k>0$, this reduces to $k geq r_4$. For example, when $n=9$, $k geq 3.77$.
Combining the two, we have:
$frac{-1+sqrt{1+8n}}{2} leq k leq frac{1+sqrt{1+8n}}{2}$.
One may be able to write this as $k = text{round} left( frac{sqrt{1+8n}}{2}right)$ and $x(n) = x(n-1) + 2^{k-1}$. The ambiguity when $8n+1$ is a perfect square needs to be resolved (e.g. when $n=10$).
answered Dec 12 '18 at 8:35
Aditya DuaAditya Dua
86418
Does the series start with 0 instead of 1? In other words (x(1) = 0)?
– Satish Ramanathan
Dec 12 '18 at 9:06
Thank you Aditya Duo. But the ambiguity you mentioned, is the main problem. The value $8n+1$ is a perfect square exactly when we have the jump from of addition, i.e. when $n=3,6,10,15,21,ldots$.
– Qurultay
Dec 12 '18 at 9:15
@Qurultay Working through a few examples, it looks like rounding down (i.e. truncating) when $8n+1$ is a perfect square works, though I don't have a rigorous argument. For example, when $n=15$, $sqrt{1+8n}/2 = 5.5$, which is 5 if you round down (and it is the right answer).
– Aditya Dua
Dec 12 '18 at 17:04
add a comment |
To compute the $n^{th}$ term, find $k$ such that $frac{k(k-1)}{2} leq n leq frac{k(k+1)}{2}$. Then:
$x(n) = x(n-1) + 2^{k-1}$.
For example, if $n=9$, $k=4$ and $x(9) = x(8) + 8$.
Let's first look at the roots of $k^2-k -2n=0$. They are given by $r_1 = frac{1-sqrt{1+8n}}{2}$ and $r_2 = frac{1+sqrt{1+8n}}{2}$.
The inequality $frac{k(k-1)}{2} leq n$ implies $k^2-k-2n leq 0$, i.e. $(k-r_1)(k-r_2) leq 0$, which is true when $r_1 leq k leq r_2$. Since $k >0$ and $r_1<0$, the condition reduces to $0 < k leq r_2$. For example, when $n=9$, $0 < k leq 4.77$.
Now let's look a the roots of $k^2+k-2n=0$. They are given by $r_3 = frac{-1-sqrt{1+8n}}{2}$ and $r_4 = frac{-1+sqrt{1+8n}}{2}$.
The inequality $n leq frac{k(k+1)}{2}$ implies $k^2+k-2n geq 0$, i.e. $(k-r_3)(k-r_4) geq 0$, which is true when $k leq r_3$ or $k geq r_4$. Since $k>0$, this reduces to $k geq r_4$. For example, when $n=9$, $k geq 3.77$.
Combining the two, we have:
$frac{-1+sqrt{1+8n}}{2} leq k leq frac{1+sqrt{1+8n}}{2}$.
One may be able to write this as $k = text{round} left( frac{sqrt{1+8n}}{2}right)$ and $x(n) = x(n-1) + 2^{k-1}$. The ambiguity when $8n+1$ is a perfect square needs to be resolved (e.g. when $n=10$).
answered Dec 12 '18 at 8:35
Aditya DuaAditya Dua
86418
To compute the $n^{th}$ term, find $k$ such that $frac{k(k-1)}{2} leq n leq frac{k(k+1)}{2}$. Then:
$x(n) = x(n-1) + 2^{k-1}$.
For example, if $n=9$, $k=4$ and $x(9) = x(8) + 8$.
Let's first look at the roots of $k^2-k -2n=0$. They are given by $r_1 = frac{1-sqrt{1+8n}}{2}$ and $r_2 = frac{1+sqrt{1+8n}}{2}$.
The inequality $frac{k(k-1)}{2} leq n$ implies $k^2-k-2n leq 0$, i.e. $(k-r_1)(k-r_2) leq 0$, which is true when $r_1 leq k leq r_2$. Since $k >0$ and $r_1<0$, the condition reduces to $0 < k leq r_2$. For example, when $n=9$, $0 < k leq 4.77$.
Now let's look a the roots of $k^2+k-2n=0$. They are given by $r_3 = frac{-1-sqrt{1+8n}}{2}$ and $r_4 = frac{-1+sqrt{1+8n}}{2}$.
The inequality $n leq frac{k(k+1)}{2}$ implies $k^2+k-2n geq 0$, i.e. $(k-r_3)(k-r_4) geq 0$, which is true when $k leq r_3$ or $k geq r_4$. Since $k>0$, this reduces to $k geq r_4$. For example, when $n=9$, $k geq 3.77$.
Combining the two, we have:
$frac{-1+sqrt{1+8n}}{2} leq k leq frac{1+sqrt{1+8n}}{2}$.
One may be able to write this as $k = text{round} left( frac{sqrt{1+8n}}{2}right)$ and $x(n) = x(n-1) + 2^{k-1}$. The ambiguity when $8n+1$ is a perfect square needs to be resolved (e.g. when $n=10$).
answered Dec 12 '18 at 8:35
Aditya DuaAditya Dua
86418
answered Dec 12 '18 at 8:35
Aditya DuaAditya Dua
86418
answered Dec 12 '18 at 8:35
Aditya DuaAditya Dua
86418
answered Dec 12 '18 at 8:35
Aditya DuaAditya Dua
86418
86418
Does the series start with 0 instead of 1? In other words (x(1) = 0)?
– Satish Ramanathan
Dec 12 '18 at 9:06
Thank you Aditya Duo. But the ambiguity you mentioned, is the main problem. The value $8n+1$ is a perfect square exactly when we have the jump from of addition, i.e. when $n=3,6,10,15,21,ldots$.
– Qurultay
Dec 12 '18 at 9:15
@Qurultay Working through a few examples, it looks like rounding down (i.e. truncating) when $8n+1$ is a perfect square works, though I don't have a rigorous argument. For example, when $n=15$, $sqrt{1+8n}/2 = 5.5$, which is 5 if you round down (and it is the right answer).
– Aditya Dua
Dec 12 '18 at 17:04
add a comment |
Does the series start with 0 instead of 1? In other words (x(1) = 0)?
– Satish Ramanathan
Dec 12 '18 at 9:06
Thank you Aditya Duo. But the ambiguity you mentioned, is the main problem. The value $8n+1$ is a perfect square exactly when we have the jump from of addition, i.e. when $n=3,6,10,15,21,ldots$.
– Qurultay
Dec 12 '18 at 9:15
@Qurultay Working through a few examples, it looks like rounding down (i.e. truncating) when $8n+1$ is a perfect square works, though I don't have a rigorous argument. For example, when $n=15$, $sqrt{1+8n}/2 = 5.5$, which is 5 if you round down (and it is the right answer).
– Aditya Dua
Dec 12 '18 at 17:04
Does the series start with 0 instead of 1? In other words (x(1) = 0)?
– Satish Ramanathan
Dec 12 '18 at 9:06
Does the series start with 0 instead of 1? In other words (x(1) = 0)?
– Satish Ramanathan
Dec 12 '18 at 9:06
Thank you Aditya Duo. But the ambiguity you mentioned, is the main problem. The value $8n+1$ is a perfect square exactly when we have the jump from of addition, i.e. when $n=3,6,10,15,21,ldots$.
– Qurultay
Dec 12 '18 at 9:15
Thank you Aditya Duo. But the ambiguity you mentioned, is the main problem. The value $8n+1$ is a perfect square exactly when we have the jump from of addition, i.e. when $n=3,6,10,15,21,ldots$.
– Qurultay
Dec 12 '18 at 9:15
@Qurultay Working through a few examples, it looks like rounding down (i.e. truncating) when $8n+1$ is a perfect square works, though I don't have a rigorous argument. For example, when $n=15$, $sqrt{1+8n}/2 = 5.5$, which is 5 if you round down (and it is the right answer).
– Aditya Dua
Dec 12 '18 at 17:04
@Qurultay Working through a few examples, it looks like rounding down (i.e. truncating) when $8n+1$ is a perfect square works, though I don't have a rigorous argument. For example, when $n=15$, $sqrt{1+8n}/2 = 5.5$, which is 5 if you round down (and it is the right answer).
– Aditya Dua
Dec 12 '18 at 17:04
add a comment |
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You should have added +2 twice, not once. Thus your sequence would be $$1,2,4,6,10,14,18,26,34,42,50,66,ldots$$
– Yiorgos S. Smyrlis
Dec 12 '18 at 7:44
@YiorgosS.Smyrlis well, this is not my choice. This sequence comes from a project.
– Qurultay
Dec 12 '18 at 8:16
0,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288
– Satish Ramanathan
Dec 12 '18 at 8:18
should it be starting with 0. If so, +2 two times, +4 three times, +8 four time and +16 five times and hence forth.
– Satish Ramanathan
Dec 12 '18 at 8:19