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

Question

I need the steps used to figure out the equation.

Answer 1

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 m^th term in the sequence S_m is given by the formula

 

S_m = (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 

 

S_m = (2^{(m - 2[m/2])})*(2^([m/2]+1)-1)

 

===========================

 

So the 2010^th 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.

Answer 2

Here is a link to an article that points out a bug in Wolfram Alpha related to a comic on xkcd.

http://www.explainxkcd.com/wiki/index.php/1292:_Pi_vs._Tau

Answer 3

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 k^th term via the recursion relation given by

 

J_k = J_k-1 + 1   for odd k

J_k = 2J_k-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.

Answer 4

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:

2 | 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.