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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 15 mins ago
AnonymousAnonymous
111
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
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 15 mins ago
AnonymousAnonymous
111
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
11 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 15 mins ago
AnonymousAnonymous
111
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 15 mins ago
AnonymousAnonymous
111
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 15 mins ago
AnonymousAnonymous
111
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 15 mins ago
AnonymousAnonymous
111
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 15 mins ago
AnonymousAnonymous
111
asked 15 mins ago
AnonymousAnonymous
111
111
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.
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1
$begingroup$
How about using the floor function?
$endgroup$
– John. P
11 mins ago
add a comment |
1
$begingroup$
How about using the floor function?
$endgroup$
– John. P
11 mins ago
1
1
$begingroup$
How about using the floor function?
$endgroup$
– John. P
11 mins ago
$begingroup$
How about using the floor function?
$endgroup$
– John. P
11 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
These are just powers of two. So: $2^lfloor n / 2rfloor$
answered 11 mins ago
FlowersFlowers
638410
$endgroup$
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 9 mins ago
TravisTravis
63.8k769151
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
These are just powers of two. So: $2^lfloor n / 2rfloor$
answered 11 mins ago
FlowersFlowers
638410
$endgroup$
add a comment |
$begingroup$
These are just powers of two. So: $2^lfloor n / 2rfloor$
answered 11 mins ago
FlowersFlowers
638410
$endgroup$
add a comment |
$begingroup$
These are just powers of two. So: $2^lfloor n / 2rfloor$
answered 11 mins ago
FlowersFlowers
638410
$endgroup$
These are just powers of two. So: $2^lfloor n / 2rfloor$
answered 11 mins ago
FlowersFlowers
638410
answered 11 mins ago
FlowersFlowers
638410
answered 11 mins ago
FlowersFlowers
638410
answered 11 mins ago
FlowersFlowers
638410
638410
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 9 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 9 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 9 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 9 mins ago
TravisTravis
63.8k769151
answered 9 mins ago
TravisTravis
63.8k769151
answered 9 mins ago
TravisTravis
63.8k769151
answered 9 mins ago
TravisTravis
63.8k769151
63.8k769151
add a comment |
add a comment |
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1
$begingroup$
How about using the floor function?
$endgroup$
– John. P
11 mins ago