Integer sequence

lm6485

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For each of these lists of integers, provide a simple formula or rule that generates the terms of an integer sequence that begins with the given list. Assuming that your formula or rule is correct, determine the next three terms of the sequence.

a) 3,6,11,18,27,38,51,66,83,102,...
b)7,11,15,19,23,27,31,35,39,43,...
c) 1,10,11,100,101,110,111,1000,1001,1010,1011,...
d) 1,2,2,2,3,3,3,3,3,5,5,5,5,5,5,5,...
e)0,2,8,26,80,242,728,2186,6560,19682
 
lm6485 said:
For each of these lists of integers, provide a simple formula or rule that generates the terms of an integer sequence that begins with the given list. Assuming that your formula or rule is correct, determine the next three terms of the sequence.


a) 3,6,11,18,27,38,51,66,83,102,...

Look at it a little and this one may pop out at you. But, have you ever used the Method of Finite Differences?.

Note, if we construct a table, subtract the top numbers (6-3=3, 11-6=5,18-11=7, 27-18=9, 38-27=11), and place that difference under those we subtracted, we get:

\(\displaystyle 3 \;\ \;\ 6 \;\ \;\ 11 \;\ \;\ 18 \;\ \;\ 27 \;\ \;\38\)

\(\displaystyle \;\ 3 \;\ \;\ 5 \;\ \;\ 7 \;\ \;\ \;\ 9 \;\ \;\ 11\)

\(\displaystyle \;\ \;\ 2 \;\ \;\ 2 \;\ \;\ \;\ 2 \;\ \;\ \;\ 2\)

we have three rows, with all the numbers the same in the bottom row.

Now, we know this is a quadratic.

Using \(\displaystyle ax^{2}+bx+c\), let x=1and we get \(\displaystyle a+b+c\)

Let x=2, we get \(\displaystyle 4a+2b+c\)

Let x=3, we get \(\displaystyle 9a+3b+c\)

Subtract as before

\(\displaystyle a+b+c \;\ \;\ \;\ 4a+2b+c \;\ \;\ \;\ 9a+3b+c\)

..........\(\displaystyle 3a+b\)........... \(\displaystyle 5a+b\)

.........................\(\displaystyle 2a\)

Now, set the first entry in each row equal to the corresponding first number in the pyramid:

\(\displaystyle a+b+c=3\)

\(\displaystyle 3a+b=3\)

\(\displaystyle 2a=2\)

Solve the system and we get \(\displaystyle a=1, \;\ b=0, \;\ c=2\)

The formula for the sequence is \(\displaystyle \boxed{x^{2}+2}\)

Enter in x=1and we get 3, enter in x=2 and we get 6, and so on. Try it with other sequences.
 
According question b)

For each of these lists of integers, provide a simple formula or rule that generates the terms of an integer sequence that begins with the given list. Assuming that your formula or rule is correct, determine the next three terms of the sequence.

a) 3,6,11,18,27,38,51,66,83,102,...
b)7,11,15,19,23,27,31,35,39,43,...
c) 1,10,11,100,101,110,111,1000,1001,1010,1011,...
d) 1,2,2,2,3,3,3,3,3,5,5,5,5,5,5,5,...
e)0,2,8,26,80,242,728,2186,6560,19682

Hello,can anyone please help me to solve b) Question on how to get the formula:p

Really appreciate ur help :)
 
Hello, lm6485!

For each of these lists of integers, provide a simple formula that generates the terms of the sequence.
Assuming that your formula or rule is correct, determine the next three terms of the sequence.

\(\displaystyle a)\;3,6,11,18,27,38,51,66,83,102\, \hdots\)
Already explained.



\(\displaystyle b)\;7,11,15,19,23,27,31,35,39,43\,\hdots\)
You really can't get this one?

Hint: The numbers "go up by 4".



\(\displaystyle c)\;1,10,11,100,101,110,111,1000,1001,1010,1011\,\hdots\)
This one is sneaky.

These are the natural numbers {1, 2, 3, 4, ... } written in base-two.

The next three are: .1100, 1101, 1110.



\(\displaystyle d)\;1,2,2,2,3,3,3,3,3,5,5,5,5,5,5,5\,\hdots\)

The sequence is: .one 1, three 2's, five 3's, . . .

I assume they are followed by: .seven 4's, nine 5's, eleven 6's, and so on.




\(\displaystyle e)\;0,2,8,26,80,242,728,2186,6560,19682\,\hdots\)

I don't understand this one. . The zero doesn't fit; ignore it!


I found a strange pattern . . .

. . \(\displaystyle \begin{array}{ccccc}n & a_n\times f && a_{n+1} \\ \hline \\ 1 & 2 \times \frac{4}{1} &=& 8 \\ \\ 2 & 8 \times\frac{13}{4} &=& 26 \\ \\ 3 & 26 \times \frac{40}{13} &=& 80 \\ \\ 4 & 80 \times \frac{121}{40} &=& 242 \\ \\ 5 & 242 \times\frac{364}{121} &=& 728 \end{array}\)


If we know the \(\displaystyle n^{th}\) term, \(\displaystyle a_n\), we can find the next term, \(\displaystyle a_{n+1}\)
. . by multiplying a particular fraction \(\displaystyle f.\)

And that fraction is: .\(\displaystyle f \:=\:\dfrac{3^{n+1}-1}{3^n-1}\)


Hence: .\(\displaystyle a_{n+1} \;=\;\dfrac{3^{n+1}-1}{3^n-1}\cdot a_n\)


So I've found a recursive relation for the sequence,
. . but not a closed form.
 
Soroban, that last one just looks like the powers of 3 minus 1.

\(\displaystyle 3^{0}-1=0\)

\(\displaystyle 3^{1}-1=2\)

\(\displaystyle 3^{2}-1=8\)

\(\displaystyle 3^{3}-1=26\)

\(\displaystyle 3^{4}-1=80\)

\(\displaystyle 3^{5}-1=242\)
.
.
.
.
\(\displaystyle 3^{9}-1=19682\)
 
Hello, lm6485!

Already explained.



You really can't get this one?

Hint: The numbers "go up by 4".



This one is sneaky.

These are the natural numbers {1, 2, 3, 4, ... } written in base-two.

The next three are: .1100, 1101, 1110.




The sequence is: .one 1, three 2's, five 3's, . . .

I assume they are followed by: .seven 4's, nine 5's, eleven 6's, and so on.





I don't understand this one. . The zero doesn't fit; ignore it!


I found a strange pattern . . .

. . \(\displaystyle \begin{array}{ccccc}n & a_n\times f && a_{n+1} \\ \hline \\ 1 & 2 \times \frac{4}{1} &=& 8 \\ \\ 2 & 8 \times\frac{13}{4} &=& 26 \\ \\ 3 & 26 \times \frac{40}{13} &=& 80 \\ \\ 4 & 80 \times \frac{121}{40} &=& 242 \\ \\ 5 & 242 \times\frac{364}{121} &=& 728 \end{array}\)


If we know the \(\displaystyle n^{th}\) term, \(\displaystyle a_n\), we can find the next term, \(\displaystyle a_{n+1}\)
. . by multiplying a particular fraction \(\displaystyle f.\)

And that fraction is: .\(\displaystyle f \:=\:\dfrac{3^{n+1}-1}{3^n-1}\)


Hence: .\(\displaystyle a_{n+1} \;=\;\dfrac{3^{n+1}-1}{3^n-1}\cdot a_n\)


So I've found a recursive relation for the sequence,
. . but not a closed form.





thank u for ur explanation,btw yes question b increse up by 4..:)..anyway I got confuse when the question ask me to write in formula and rule..hihi..ok now I understand this question...really appreciate ur help..:)
 
Recall Fibonacci's sequence: \(\displaystyle f_0=0, f_1=1\) and \(\displaystyle f_{k+2} = f_{k+1}+f_k\).

The sequence then looks like: 0,1,1,2,3,5,8,13,21,...

Pretty sure d is:

(one 1)(three 2's)(five 3s)(seven 5s)(nine 8s)(eleven 13s),...

i.e. the nth grouping will contain 2n-1 copies of \(\displaystyle f_{n+2}\), the (n+2)nd fibonacci number. But given such a short list, hard to tell.
 
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