A369110 a(n) is the number of distinct elements appearing in the sequence formed by recursively applying A063655 when starting from n.

4, 3, 2, 1, 2, 2, 4, 3, 3, 5, 6, 5, 5, 4, 4, 4, 5, 4, 5, 4, 6, 6, 7, 6, 6, 5, 6, 7, 8, 7, 7, 6, 5, 6, 6, 6, 8, 7, 5, 6, 7, 6, 6, 5, 5, 7, 6, 5, 5, 5, 5, 6, 6, 5, 5, 5, 7, 8, 6, 5, 7, 6, 5, 5, 5, 6, 8, 7, 6, 6, 7, 6, 7, 6, 5, 8, 5, 6, 6, 5, 5, 7, 7, 6, 7, 6, 7, 6, 7, 6, 5, 7, 7, 6, 7, 5, 8, 7, 5

Offset: 1

Author

Adnan Baysal, Jan 13 2024

Keywords

• nonn
• easy

Comments

A063655(n) gives the smallest semiperimeter of an integral rectangle with area n, which is the same thing as the minimum sum of two positive integers whose product is n. In this sequence, A063655 is applied recursively until a cycle is found. Then the number of distinct elements appearing in this process is given as a(n). Note that it's conjectured that a cycle will be found at some point.

Conjecture: The cycle part of each sequence generated by the recursion is one of (4), (5, 6), or (6, 5). Confirmed through 1 millionth term.

The conjecture is true. Proof: The conjecture holds for n <= 6. Suppose n >= 7 and the conjecture holds for lower values of n. If n is composite, then A063655(n) <= n/2+2. If n is prime, then A063655(n) = n+1 is even and A063655(A063655(n)) <= (n+1)/2+2. In both cases, n reaches a lower number and the conjecture holds for n. - Jason Yuen, Mar 30 2024

Examples

			n = 1 can be factored as 1*1 with minimum sum 2 (similarly, A063655(1) = 2). Then 2 = 1*2, so minimum sum is 3 = A063655(2). 3 = 1*3 which means the next number in the recursion is 4 = A063655(3). 4 = 2*2 which gives the same number 4 = A063655(4), hence this recursion will create a cycle at this point. Starting from n = 1 (including 1), we generated these numbers: (1, 2, 3, 4, 4, 4, ...). Therefore, a(1) = 4. a(2), a(3), and a(4) are trivially deduced from this example.
		

Links

• Jason Yuen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A063655, A056737.

Programs

Python

from sympy import divisors
def A369110(n):
    c = {n}
    while n<4 or n>5:
        c.add(n:=(d:=divisors(n))[((l:=len(d))-1)>>1]+d[l>>1])
    if n==5:
        c.add(6)
    return len(c) # Chai Wah Wu, Apr 25 2024