# Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.21

21. Conjecture a formula for the nth term of ${a_n}$ if the first ten terms of this sequence are as follows.
a) 3, 11,19,27,35,43,51,59,67,75
b) 5, 7, 11, 19, 35,67, 131,259,515, 1027
c) 1,0,0, 1,0,0,0,0,1,0
d) 1, 3,4,7,11,18,29,47,76,123

Solution:

(a) 3,11,19,27,35,43,51,59,67,75

We can see that this sequence is neither in arithmetic progression nor in geometric Progression. We use the method of inspection to find the nth term of this sequence.

$3=8(1)-5$
$11=8(2)-5$
$19=8(3)-5$
$27=8(4)-5$
$35=8(5)-5$
$43=8(6)-5$
$51=8(7)-5$
$59=8(8)-5$
$67=8(9)-5$
$75=8(10)-5$

Therefore we have for $n=1,2,3,...,10$
$a_n=8n-5$

(b) 5, 7,11,19,35,67,131,259,515,1027

We can see that this sequence is neither in arithmetic progression nor in geometric Progression. We use the method of inspection to find the nth term of this sequence.

$5=2+3$
$7=2^2+3$
$11=2^3+3$
$19=2^4+3$
$35=2^5+3$
$67=2^6+3$
$131=2^7+3$
$259=2^8+3$
$515=2^9+3$
$1027=2^{10}+3$

Therefore we have for $n=1,2,3,...,10$
$a_n=2^n+3$

(c) 1,0,0,1,0,0,0,0,1,0
We can see that this sequence is neither in arithmetic progression nor in geometric Progression. It might strike to use greatest integer function here. We observe that For perfect squares the value turns out to be 1 otherwise it is 0. So, we use the method of inspection to find the nth term of this sequence.
$1=[[\sqrt{1}]/\sqrt{1}$
$0=[[\sqrt{2}]/\sqrt{2}$
$=[1/\sqrt{2}]$

$0=[[\sqrt{3}]/\sqrt{3}]$
$=[1/\sqrt{3}]$
$1=[[\sqrt{4}]/\sqrt{4}]$
$=[2/2]$

$0=[[\sqrt{5}]/\sqrt{5}]$
$=[2/\sqrt{5}]$
$0=[[\sqrt{6}]/\sqrt{6}]$
$=[2/\sqrt{6}]$

$0=[[\sqrt{7}]/\sqrt{7}]$
$=[2/\sqrt{7}]$

$0=[[\sqrt{8}]/\sqrt{8}]$
$=[2/2 \sqrt{2}]$
$=[1/\sqrt{2}]$

$1=[[\sqrt{9}]/\sqrt{9}]$
$=[3/3]$

$0=[[\sqrt{10}]/\sqrt{10}]$
$=[3/\sqrt{10}]$

Therefore we have for $n=1,2,...10$

$a_n=[[\sqrt{n}]/\sqrt{n}]$

(d) 1, 3,4,7,11,18,29,47,76,123

We define Lucas numbers recursively by

$L_n = L_{n-1} + L_{n-2}$
For $n \geq 3$ with
$L_1 =1$
$L_2 =3$
Now, we use the recursive formula to find the successive numbers.
That is,

$L_3 = L_2 + L_1$
$= 3 + 1$
$= 4$

$L_4 = L_3 + L_2$
$= 4 + 3$
$= 7$

$L_5 = L_4 + L_3$
$= 7 + 4$
$= 11$

$L_6 = L_5 + L_4$
$= 11 + 7$
$= 18$

$L_7 = L_6 + L_5$
$= 18 + 11$
$= 29$

$L_8 = L_7 + L_6$
$= 29 + 18$
$= 47$

$L_9 = L_8 + L_7$
$= 47 + 29$
$= 76$

$L_10 = L_9 + L_8$
$= 76 + 47$
$= 123$

Replacing $L_n$ with $a_n$ we have
$a_n = a_{n-1} + a_{n-2}$

For $n \geq 3$ with
$a_1 = 1$
$a_2 = 3$