Question

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.

$\begin{array}{r}3,6,11,18,27,38,51,66,83,102,\dots \\ 7,11,15,19,23,27,31,35,39,43,\dots \\ 1,10,11,100,101,110,111,1000,1001,1010,1011,\dots \\ 1,2,2,2,3,3,3,3,3,5,5,5,5,5,5,5,\dots \\ 0,2,8,26,80,242,728,2186,6560,19682,\dots \mid \\ 1,3,15,105,945,10395,135135,2027025,34459425,\dots \\ 1,0,0,1,1,1,0,0,0,0,1,1,1,1,1,\dots \\ 2,4,16,256,65536,4294967296,\dots \end{array}$

Short answer

The following are the next three terms of the sequence:

$\begin{array}{r}123,146,171\\ 47,57,55\\ 1100,1101,1110\\ 8,8,8\\ 59048,177146,531440\\ 654729075,13749310575,316234143225\\ 0,0,0\\ 18446744073709551616\\ 340282366920938463463374607431768211456\\ 1157920892373161954235709850086879078532699846656405640394575840079131296339936\end{array}$

Step-by-step solution

Step 1:

1. Given: 3,6,11,18,27,38,51,66,83,102,.....

We note is the previous term increased by 3,11 is the previous term increased by 5, 18 is the previous term increased by 7, and so on.

Thus, each term is the previous term increased by 2n + 1

$\begin{array}{r}{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}+2\mathrm{n}-1\\ {\mathrm{a}}_1=3\end{array}$

If the pattern continues, then the ${11}^{\mathrm{th}},{12}^{\mathrm{th}},\mathrm{and}{13}^{\mathrm{th}}$term become

$\begin{array}{r}{\mathrm{a}}_{11}={\mathrm{a}}_{10}+2\left(11\right)-1=102+22-1=123\\ {\mathrm{a}}_{12}={\mathrm{a}}_{11}+2\left(12\right)-1=123+24-1=146\\ {\mathrm{a}}_{13}={\mathrm{a}}_{12}+2\left(13\right)-1=146+26-1=171\end{array}$

Thus, the next three terms in the sequence are then123 , 146 and 171

Step 2:

Given: 7, 11, 15, 19, 23, 27, 31, 35, 39, 43,.....

We note each term is the previous term increased by 4:

${\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}+1}+4$

If the pattern continues, then the ${11}^{\mathrm{th}},{12}^{\mathrm{th}},\mathrm{and}{13}^{\mathrm{th}}$term become

$\begin{array}{r}{a}_{11}=a_{10}+4=43+4=47\\ a_{12}=a_{11}+4=47+4=51\\ a_{13}=a_{12}+4=51+4=55\end{array}$

Thus, the next three terms in the sequence are then 47, 51 and 55.

Step 3:

Given:$1,10,11,100,101,110,111,1000,1001,1010,1011,\dots$

We note that the given sequence are the positive integers in increasing order in their binary form.

If this pattern continues, the next terms would then be 12, 13 and 14 in binary form (since 1011 corresponds with 11 ).

$1,10,11,100,101,110,111,1000,1001,1010,1011,\dots$

Thus, the next three terms in the sequence are then 1100, 1101 and 1110.

Step 4:

Given:$1,2,2,2,3,3,3,3,3,5,5,5,5,5,5,5,\dots$

We note the integers are repeated once, thrice, times, times, etc.

Thus, the following integer will be repeated times.

We note that each integer is the sum of the previous two integers (not taking into account the repetition of the integers),

Since 1 + 2 = 3 and 2 + 3 = 5

The next integer would then be 3 + 5 = 8

role="math" localid="1668669987802" $1,2,2,2,3,3,3,3,3,5,5,5,5,5,5,5,8,8,8,8,8,8,8,8,8,\dots \dots$

Thus, the next three terms in the sequence are then8, 8and8.

Step 5:

Given: $0,2,8,26,80,242,728,2186,6560,19682,\dots$

The powers of $3\text{are}1,3,9,27,81,243,729,2186,6561,19682,\dots$

We note that the terms in the given sequence are the powers of 3 decreased by 1

${\mathrm{a}}_{\mathrm{n}}=3^{\mathrm{n}-1}-1$

If this pattern continues, we would then expect the ${11}^{\mathrm{th}},{12}^{\mathrm{th}},\mathrm{and}{13}^{\mathrm{th}}$term to be:

$\begin{array}{r}{\mathrm{a}}_{11}={\mathrm{a}}^{11-1}-1=3^{10}-1=59049-1=59048\\ {\mathrm{a}}_{12}={\mathrm{a}}^{12-1}-1=3^{11}-1=177147-1=177146\\ {\mathrm{a}}_{13}={\mathrm{a}}^{13-1}-1=3^{12}-1=531441-1=531440\end{array}$

Thus, the next three terms in the sequence are then 59048, 177146and 531440

Step 6:

Given: $1,3,15,105,945,10395,135135,2027025,34459425,\dots$

The second term is the first term multiplied by 3

The third term is the second term multiplied by 5

The fourth term is the third term multiplied by 7

The fifth term is the fourth term multiplied by 9

And so on.

${\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}.\left(2\mathrm{n}-1\right)$

If the pattern continues, the ${10}^{\mathrm{th}}{11}^{\mathrm{th}},\mathrm{and}{12}^{\mathrm{th}}$term become:

$\begin{array}{r}{\mathrm{a}}_{10}={\mathrm{a}}_9\cdot (2\cdot 10-1)=34459425\cdot 19=654729075\\ {\mathrm{a}}_{11}={\mathrm{a}}_{10}\cdot (2\cdot 11-1)=654729075\cdot 21=13749310575\\ {\mathrm{a}}_{12}={\mathrm{a}}_{11}\cdot (2\cdot 12-1)=13749310575\cdot 23=316234143225\end{array}$

Thus, the next three terms in the sequence are then$654729075,13749310575$and 316234143225 .

Step 7:

Given:$1,0,0,1,1,1,0,0,0,0,1,1,1,1,1,\dots$

The sequence alternates and .

The number of times each term is repeated increases

The sequence starts with 1 one, then 2 zeros, then 3 ones, then 4 zeros, then 5 ones.

Assuming the pattern continues, we then expect 6 zeros.

$1,0,0,1,1,1,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,\dots$

Thus, the next three terms are all zeros.

Step 8:

Given: $2,4,16,256,65536,4294967296,\dots$

We note that each term is the previous term squared.

$a_n=a_{n-1}^2$

Assuming the pattern continues, we then expect the $7^{\mathrm{th}}{8}^{\mathrm{th}},\mathrm{and}{9}^{\mathrm{th}}$, term of the series to be:

$\begin{array}{r}{\mathrm{a}}_7={\mathrm{a}}_{7-1}^2={\mathrm{a}}_6^2={4294967296}^2=18446744073709551616\\ {\mathrm{a}}_8={\mathrm{a}}_{8-1}^2={\mathrm{a}}_7^2={18446744073709551616}^2=340282366920938463463374607431768211456\\ {\mathrm{a}}_9={\mathrm{a}}_{9-1}^2={\mathrm{a}}_8^2={340282366920938463463374607431768211456}^2\\ =1157920892373161954235709850086879078532699846656405640394575840079131296339936\end{array}$