How to find the nth term in the following sequence: 1,1,2,2,4,4,8,8,16,16 The Next CEO of Stack OverflowHow to interpret the OEIS function for the “even fractal sequence” A103391 (1, 2, 2, 3, 2, 4, 3, 5, …)What will be nth term of the following sequence?How to find the nth term of this sequence?Number of possible ordered sequencesFind nth term of sequenceHow can i find the decimal values with a list of integers?Given a sequence find nth termFind nth term for below sequenceProve $lim_ntoinftyU_n = 1$ given $0 lt U_n - 1over U_nlt 1over n$ and $U_n>0$How to find the nth term in quadratic sequence?
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How to find the nth term in the following sequence: 1,1,2,2,4,4,8,8,16,16
The Next CEO of Stack OverflowHow to interpret the OEIS function for the “even fractal sequence” A103391 (1, 2, 2, 3, 2, 4, 3, 5, …)What will be nth term of the following sequence?How to find the nth term of this sequence?Number of possible ordered sequencesFind nth term of sequenceHow can i find the decimal values with a list of integers?Given a sequence find nth termFind nth term for below sequenceProve $lim_ntoinftyU_n = 1$ given $0 lt U_n - 1over U_nlt 1over n$ and $U_n>0$How to find the nth term in quadratic sequence?
$begingroup$
I'm having difficulty in finding the formula for the sequence above, when I put this in wolframalpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.
I'm thinking I'll most likely need to have a condition for even numbers and another for non even numbers.
Any help would be highly appreciated.
sequences-and-series
asked 23 mins ago
AnonymousAnonymous
131
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm having difficulty in finding the formula for the sequence above, when I put this in wolframalpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.
I'm thinking I'll most likely need to have a condition for even numbers and another for non even numbers.
Any help would be highly appreciated.
sequences-and-series
asked 23 mins ago
AnonymousAnonymous
131
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
How about using the floor function?
$endgroup$
– John. P
19 mins ago
add a comment |
$begingroup$
I'm having difficulty in finding the formula for the sequence above, when I put this in wolframalpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.
I'm thinking I'll most likely need to have a condition for even numbers and another for non even numbers.
Any help would be highly appreciated.
sequences-and-series
asked 23 mins ago
AnonymousAnonymous
131
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm having difficulty in finding the formula for the sequence above, when I put this in wolframalpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.
I'm thinking I'll most likely need to have a condition for even numbers and another for non even numbers.
Any help would be highly appreciated.
sequences-and-series
sequences-and-series
asked 23 mins ago
AnonymousAnonymous
131
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 23 mins ago
AnonymousAnonymous
131
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 23 mins ago
AnonymousAnonymous
131
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 23 mins ago
AnonymousAnonymous
131
asked 23 mins ago
AnonymousAnonymous
131
131
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
How about using the floor function?
$endgroup$
– John. P
19 mins ago
add a comment |
1
$begingroup$
How about using the floor function?
$endgroup$
– John. P
19 mins ago
1
1
$begingroup$
How about using the floor function?
$endgroup$
– John. P
19 mins ago
$begingroup$
How about using the floor function?
$endgroup$
– John. P
19 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
These are just powers of two. So: $2^lfloor n / 2rfloor$
answered 19 mins ago
FlowersFlowers
653410
$endgroup$
$begingroup$
I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
$endgroup$
– Anonymous
6 mins ago
$begingroup$
This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
$endgroup$
– Flowers
3 mins ago
add a comment |
$begingroup$
Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
$$a_n = 2 a_n - 2, qquad a_0 = a_1 = 1.$$
Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^n - 2$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt2$. So, the general solution is
$$a_n = A (sqrt2)^n + B(-sqrt2)^n = (sqrt2)^n [A + B(-1)^n] .$$
Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.
answered 3 mins ago
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor fracn2 rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.
So, the sequence is given (for appropriate indexing) by $$color#df0000boxeda_n := 2^lfloor n / 2 rfloor .$$
answered 17 mins ago
TravisTravis
63.8k769151
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
These are just powers of two. So: $2^lfloor n / 2rfloor$
answered 19 mins ago
FlowersFlowers
653410
$endgroup$
$begingroup$
I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
$endgroup$
– Anonymous
6 mins ago
$begingroup$
This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
$endgroup$
– Flowers
3 mins ago
add a comment |
$begingroup$
These are just powers of two. So: $2^lfloor n / 2rfloor$
answered 19 mins ago
FlowersFlowers
653410
$endgroup$
$begingroup$
I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
$endgroup$
– Anonymous
6 mins ago
$begingroup$
This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
$endgroup$
– Flowers
3 mins ago
add a comment |
$begingroup$
These are just powers of two. So: $2^lfloor n / 2rfloor$
answered 19 mins ago
FlowersFlowers
653410
$endgroup$
These are just powers of two. So: $2^lfloor n / 2rfloor$
answered 19 mins ago
FlowersFlowers
653410
answered 19 mins ago
FlowersFlowers
653410
answered 19 mins ago
FlowersFlowers
653410
answered 19 mins ago
FlowersFlowers
653410
653410
$begingroup$
I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
$endgroup$
– Anonymous
6 mins ago
$begingroup$
This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
$endgroup$
– Flowers
3 mins ago
add a comment |
$begingroup$
I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
$endgroup$
– Anonymous
6 mins ago
$begingroup$
This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
$endgroup$
– Flowers
3 mins ago
$begingroup$
I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
$endgroup$
– Anonymous
6 mins ago
$begingroup$
I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
$endgroup$
– Anonymous
6 mins ago
$begingroup$
This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
$endgroup$
– Flowers
3 mins ago
$begingroup$
This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
$endgroup$
– Flowers
3 mins ago
add a comment |
$begingroup$
Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
$$a_n = 2 a_n - 2, qquad a_0 = a_1 = 1.$$
Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^n - 2$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt2$. So, the general solution is
$$a_n = A (sqrt2)^n + B(-sqrt2)^n = (sqrt2)^n [A + B(-1)^n] .$$
Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.
answered 3 mins ago
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
$$a_n = 2 a_n - 2, qquad a_0 = a_1 = 1.$$
Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^n - 2$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt2$. So, the general solution is
$$a_n = A (sqrt2)^n + B(-sqrt2)^n = (sqrt2)^n [A + B(-1)^n] .$$
Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.
answered 3 mins ago
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
$$a_n = 2 a_n - 2, qquad a_0 = a_1 = 1.$$
Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^n - 2$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt2$. So, the general solution is
$$a_n = A (sqrt2)^n + B(-sqrt2)^n = (sqrt2)^n [A + B(-1)^n] .$$
Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.
answered 3 mins ago
TravisTravis
63.8k769151
$endgroup$
Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
$$a_n = 2 a_n - 2, qquad a_0 = a_1 = 1.$$
Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^n - 2$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt2$. So, the general solution is
$$a_n = A (sqrt2)^n + B(-sqrt2)^n = (sqrt2)^n [A + B(-1)^n] .$$
Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.
answered 3 mins ago
TravisTravis
63.8k769151
answered 3 mins ago
TravisTravis
63.8k769151
answered 3 mins ago
TravisTravis
63.8k769151
answered 3 mins ago
TravisTravis
63.8k769151
63.8k769151
add a comment |
add a comment |
$begingroup$
The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor fracn2 rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.
So, the sequence is given (for appropriate indexing) by $$color#df0000boxeda_n := 2^lfloor n / 2 rfloor .$$
answered 17 mins ago
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor fracn2 rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.
So, the sequence is given (for appropriate indexing) by $$color#df0000boxeda_n := 2^lfloor n / 2 rfloor .$$
answered 17 mins ago
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor fracn2 rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.
So, the sequence is given (for appropriate indexing) by $$color#df0000boxeda_n := 2^lfloor n / 2 rfloor .$$
answered 17 mins ago
TravisTravis
63.8k769151
$endgroup$
The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor fracn2 rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.
So, the sequence is given (for appropriate indexing) by $$color#df0000boxeda_n := 2^lfloor n / 2 rfloor .$$
answered 17 mins ago
TravisTravis
63.8k769151
answered 17 mins ago
TravisTravis
63.8k769151
answered 17 mins ago
TravisTravis
63.8k769151
answered 17 mins ago
TravisTravis
63.8k769151
63.8k769151
add a comment |
add a comment |
Anonymous is a new contributor. Be nice, and check out our Code of Conduct.
Anonymous is a new contributor. Be nice, and check out our Code of Conduct.
Anonymous is a new contributor. Be nice, and check out our Code of Conduct.
Anonymous is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
How about using the floor function?
$endgroup$
– John. P
19 mins ago