Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.24

24. Find three different formulas or rules for the terms of a sequence a_n if the first three terms of this sequence are 2, 3, 6.

Solution: We consider the first three terms of a sequence 2,3,6.
The simplest form of this sequence is
a_n=2^{n}-(n-1)
For n=1,2,3,...
We can even use recursive formula for this sequence, that is,
a_1=2
a_2=3
a_n=a_{n-1}.a_{n-2}
For n \geq 3
We can also formulate the nth term of this sequence by inserting factorials and doing inspection.
2=\frac{(1+1)!}{2^0}
3=\frac{(2+1)!}{2^1}
6=\frac{(3+1)!}{2^2}
Therefore, we have for n=1,2,3,...
a_n=\frac{(n+1)!}{2^{n-1}}

Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.23

23. Find three different formulas or rules for the terms of a sequence {a_n}  if the first three terms of this sequence are  1,2,4.

Solution: We consider the first three terms of a sequence as 1,2,4

The simplest form of this sequence is

a_n=2^{n-1}
For n=1,2,3,...
We can even use recursive formula for this sequence, that is,
a_1=1
a_2=2
a_n=a_{n-1}+2a_{n-2}
For n \geq 3
We can also formulate the nth term of this sequence by inspection.
1=\frac{1^2-1+2}{2}
2=\frac{2^2-2+2}{2}
4=\frac{3^2-3+2}{2}
Therefore, we have for n=1,2,3,...
a_n=\frac{n^2-n+2}{2}

Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.22

22. Conjecture a formula for the nth term of {a_n}  if the first ten terms of this sequence are as follows.

a) 2, 6, 18, 54, 162, 486, 1458, 4374, 13122, 39366
b) 1, 1, 0, 1, 1, 0, 1, 1, 0, 1
c) 1, 2, 3, 5, 7, 10, 13, 17, 21, 26
d) 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365

Solution:

(a) 2, 6, 18, 54,162,486,1458,4374,13122,39366

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.

2=2(3^0)
6=2(3^1)
18=2(3^2)
54=2(3^3)
162=2(3^4)
486=2(3^5)
1458=2(3^6)
4374=2(3^7)
13122=2(3^8)
39366=2(3^9)

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

a_n=2(3^{n-1})

(b) 1,1,0,1,1,0,1,1,0,1

This sequence contains 0’s and 1’s only. We notice that 0 occurs in third multiple term of the sequence.
Therefore, we have for n=1,2,...,10

a_n = \left\{\begin{matrix}0; n=3m&for& m=1,2,3 & \\1; otherwise & \end{matrix}\right.

(c) 1, 2, 3,5,7,10,13,17,21,26

We can see that this sequence is neither in arithmetic progression nor in geometric Progression. We split the sequence into odd and even terms and use the method of inspection to find the nth term of this sequence. Now, for 1,3,7,13,21 we have

1=1^2-0
3=2^2-1
7=3^2-2
13=4^2-3
21=5^2-4

So the odd termed sequence is of the form

a_{2n-1} = n^2-(n-1)
Where n=1,2,3,4,5
Similarly, for 2,5,10,17,26 we have

2=1^2+1
5=2^2+1
10=3^2+1
17=4^2+1
26=5^2+1

So the even termed sequence is of the form

a_{2n}=n^2+1
where n=1,2,3,4,5
Therefore, we have for n=1,2,…,10

a_n = \left\{\begin{matrix} a_{2m-1} = m^2-(m-1),m=1,2,3,4,5 & \\ a_{2m} = m^2 + 1,m=1,2,3,4,5 & \end{matrix}\right.

(d) 3, 5,11,21,43,85,171,341,683,1365

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-(-1)^2
11=2.5-(-1)^3
21=2.11-(-1)^4
43=2.21-(-1)^5
85=2.43-(-1)^6
171=2.85-(-1)^7
341=2.171-(-1)^8
683=2.341-(-1)^9
1365=2.683-(-1)^10

Therefore we have for n=1,2,….,10

a_1=3,a_n=2a_{n-1}-(-1)^n, n \geq 2

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