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Determine the 2010th term for the following sequence: 1,2,3,6,7,14,15,30,31,62,63

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4 Answers

There are multiple answers to this question because a finite number of terms do not precisely determine the sequence.  For example, one could fit a polynomial f(x) of degree 10 to this sequence so that f(i) was equal to the ith term.  
 
However here is another approach
 
First lets let [x] be the greatest integer in x so that, for example
[1/2] = 0, [5/2] = 2, and so on.  Note that if k is an even number [k/2] = k/2, and if k is an odd number [k/2] = (k-1)/2 , and [0] = 0.
 
I will assume the sequence starts with S0
then the mth term in the sequence Sm is given by the formula
 
Sm = (2(m - 2[m/2]))∑[i≥0, i≤[m/2])2i
 
The way this formula works is that
(1) for even numbers m (including zero) the leading factor becomes 1, and for odd numbers it's 2. 
(2) for even numbers the summation goes from 0 to m/2; for odd numbers it goes from 0 to (m-1)/2
 
So, for example, the 7th number in the sequence would be 2*(23+22+21+20) = 30;
and the 10th number in the sequence would be  1*(25+24+23+22+21+20) = 63
 
The sum of the powers of 2 from 0 to [m/2] can also be expressed as 2([m/2]+1)-1, so we could rewrite the formula as 
 
Sm = (2(m - 2[m/2]))*(2([m/2]+1)-1)
 
===========================
 
So the 2010th term requested in the problem, since we defined the first term in the sequence to be S0, would be S2009
 
S2009  = 2*(21005-1)  = 21006-2
 
I couldn't find any calculator that would compute this number without rounding.  It's a number with about 303 digits.

Comments

685765508599211085406992031398401158759299079491541508764000248557024672719959118395646962442045349201660590667234013968119772982843080987903012964780708787451812337588750783066948774723991753080189067657794974398949244241113521123786594812548932026532556574571938698730267509225767960757581162756440062
But I haven't verified the accuracy.
Thanks Ryan!   I forgot about trying Wolfram.
Machine precision in nowhere near that good. Anything after to 20th value is almost definitely going to be meaningless
Again, I haven't personally verified the result, but I would not dismiss it out of hand. There are peer-reviewed published values of numbers such as pi computed to a million digits or more. Algorithms can be developed that do not rely merely on the accuracy of floating point operations.
that is true.  I guess wolfram alpha could define a data type with an arbitrary bit length to get any precision needed.  Not sure if that is done or not, but would be interesting to know
The number has 303 digits, the correct number.  I would bet that it's correct.  I've found some errors on Wolfram before, but they generally have been minor.
It is interesting to note that Wolfram's result for 2^1006 ends in the digits 64 which is 2 more than the last two digits of the result for (2^1006)-2. (62)
First of all, 1006 - 2 = 1004, and 1004/4 = 251.  This is significant because it means 21006 ends in the digit 4, which means 21006 - 2 ends in the digit 2.  So I claim that the Wolfram answer ends in the correct digit.
Note that the number 21006 = 21004 *24,  and 1004/4 = 251.   This means that 21006 ends in the digit 4, which means that the Wolfram answer ends with the correct digit.

Comment

Determine the 2010th term for the following sequence: 1,2,3,6,7,14,15,30,31,62,63

Let’s look at the odd terms, 1,3,7,…

k n t
1 0 1 = 1
3 1 3 = 2 + 1
5 2 7 = 4 + 2 + 1
7 3 15 = 8 + 4 + 2 + 1

k n t = 2^(n)+2^(n-1)+…+2+1
= 2^(n+1)-1, it’s a binary number and geometric series!

We want to know the 2010th number in the given sequence;
i.e., k = 2n+1 = 2010
2n = 2009
n = 2009/2 which isn’t evenly divisible,
so we want the even k term that’s 1 less than for k = 2011:

k = 2n+1 = 2011
2n = 2010
n = 1005
t = 2^(1005+1)-1
t-1 = 2^(1006)-2

So the 2010th number in the series is 2^(1006)-2.
 
2^(1006) = 10^(1006 log(2)) ≈ 6.857655085992232 * 10^302

Here’s another construct where this sequence occurs:

| 1 1 1  1   1   1 ...    1
        2 6 14 30 62 ...  R-1
     1 3 7 15 31 63 ... | R = p(2)

p(x) = x^n + x^(n-1) + x^(n-2) + … + x + 1

and p(2) is our binary number.
Here is a link to an article that points out a bug in Wolfram Alpha related to a comic on xkcd.

Comments

Haha nice, I personally wouldn't trust any computer calculation out past ~20 digits, and there are very few applications that actually require that level of precision.  I believe they calculate pi out to thousands of digits one at a time, therefore not ever needing to store the entire number out to full precision.

Comment

This is a little hard to answer without knowing exactly what level of math this question is intended for.  But, you can solve for the kth term via the recursion relation given by
 
Jk = Jk-1 + 1   for odd k
Jk = 2Jk-1   for even k
with J0 = 0, and k>0
 
using the above, we get
 
J1 = 1
J2 = 2
J3 = 3
J4 = 6
J5 = 7
J6 = 14
 
and so forth....
 
This recursive algorithm is trivial to implement numerically by writing a simple script in MATLAB, Python, or writing a simple C code. I get a value of 6.8577×10302 (to 4 digits of precision).
 
For a problem like this, there is no real trick to getting the equation other than staring at the numbers and trying to find out a pattern. Once you have the pattern, the trick is to figure out a way to express it mathematically, which again really just sitting and thinking it through.  The best way to get good at this type of problem is to just do as many as you can to get used to looking at sequences of numbers and finding patterns.  This is a real case of practice makes perfect.