Sequence problem

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Raiyan Jamil
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Sequence problem

Unread post by Raiyan Jamil » Fri Nov 07, 2014 10:02 pm

What is the 68th term of the following sequence and why-

1,4,5,16,17,20...
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tanmoy
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Re: Sequence problem

Unread post by tanmoy » Sat Nov 08, 2014 6:33 pm

The $68$th term of the sequence is $4112$.
The terms of the sequence are either power of $4$ or the sum of the powers of $4$,such as:The first term is $4^{0}$,the 2nd term is $4^{1}$ ,the 3rd term is $4^{0}+4^{1}$ etc.We can get $2^{n+1}-1$ terms by using $o-n$th powers of $4$.So,we can get $2^{6}-1=63$ terms by using $0-5$th powers of $4$.The $63$th term is $4^{0}+4^{1}+4^{2}+4^{3}+4^{4}+4^{5}=1365$.The $64$th term is $4^{6}=4096$.The $65$th term is $4096+1=4097$,the $66$th term is $4096+4=4100$,the $67$th term is $4096+1+4=4101$ and the $68$th term is $4096+16=4112$
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Nirjhor
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Re: Sequence problem

Unread post by Nirjhor » Sat Nov 08, 2014 8:06 pm

Convert the terms to base $4$. You'll get the sequence of consecutive binary naturals.

So to find the $n$-th term, convert $n$ to binary, consider the result to be in base $4$ and convert to decimal.
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