Can you find a formula or rule for the nth term of a sequence related to the prime numbers or prime factorizations so that the initial terms of the sequence have these values?
a) 2, 2, 3, 5, 5, 7, 7, 11, 11, 11, 11, 13, 13.....
b) 0, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6.........
c) 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1....
d) 1, -1, -1, 0, -1, 1, -1, 0, 0, 1, -1, 0,-1, 1, 1.....
e) 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0.......
f) 4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369....
Step 1 :
In this problem we have to find a formula or rule for the nth term of a sequence.
Given the sequence 2, 2, 3, 5, 5, 7, 7, 11, 11, 11, 11, 13, 13.....Rule of term is
Where “p” is the set of prime number.
Step 2 :
(b) given the sequence 0, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6.........
The rule of term is
Where is the
prime.
is the least prime factor of n.
Step 3 :
(c) given the sequence is 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1....
The rule of term is
