MathDB

Consider a sequence whose first term is a given positive integer

N > 1

. Consider the prime factorization of

N

. If

N

is a power of 2, the sequence consists of a single term:

N

. Otherwise, the second term of the sequence is obtained by replacing the largest prime factor

p

N

with

p + 1

in the prime factorization. If the new number is not a power of 2, we repeat the same procedure with it, remembering to factor it again into primes. If it is a power of 2, the numerical sequence ends. And so on.
For example, if the first term of the sequence is

N = 300 = 2^2 \cdot 3 \cdot 5^2

, since its largest prime factor is

p = 5

, the second term is

2^2 \cdot 3 \cdot (5 + 1)^2 = 2^4 \cdot 3^3

. Repeating the procedure, the largest prime factor of the second term is

p = 3

, so the third term is

2^4 \cdot (3 + 1)^3 = 2^{10}

. Since we obtained a power of 2, the sequence has 3 terms:

2^2 \cdot 3 \cdot 5^2

2^4 \cdot 3^3

, and

2^{10}

.
a) How many terms does the sequence have if the first term is

N = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23

? b) Show that if a prime factor

p

leaves a remainder of 1 when divided by 3, then

\frac{p+1}{2}

is an integer that also leaves a remainder of 1 when divided by 3. c) Present an initial term

N

less than 1,000,000 (one million) such that the sequence starting from

N

has exactly 11 terms.

Problem Statement