n' = #div + sumdiv + 1

– Start the sequence S with an integer n (like 2023)

– compute the number A of divisors of n (2023 has A = 6 divisors, which are 1, 7, 17, 119, 289 and 2023),

– make the sum B of the said divisors (B = 2456 here)

– replace n by n' = A + B + 1 (n' = 2463 = 6 + 2456 + 1 here) 

– iterate until a prime is reached (2463 —> 3293 —> 3425 —> 4285 —> 5153 is prime)

– restart the iteration at that point with the smallest unused integer so far (which would be 1 here).

Question #1

How does the graph of S look if we start S with a(1) = 1?

S = 1,3,2,4,11,5,6,17,7,8,20,49,61,9,17,10,23,11,12,35,43,13,14,29,15,29,16,37,18,46,77,101,19,21,37,22,...

Explanation

1 —> 3 (prime, extend now S with 2)

2 (prime, extend now S with 4)

4 —> 11 (prime, extend now S with 5)

5 (prime, extend now S with 6)

6 —> 17 (prime, extend now S with 7)

7 (prime, extend now S with 8)

8 —> 20 —> 49 —> 61 (prime, etc.)

S is not a permutation of the positive integers as some terms may appear in different ways (6 and 9 are both followed by 17, for instance; 14 and 15 are both followed by 29, etc.)

Question #2

Could S enter into a loop at some point?

(many thanks to Dario Alpern for this online resource)

____________________

Update, a couple of hours later:

Giorgos Kalogeropoulos was quick to correct, extend S and deliver the corresponding graph:

> S = 1, 3, 2, 4, 11, 5, 6, 17, 7, 8, 20, 49, 61, 9, 17, 10, 23, 12, 35, 53, 13, 14, 29, 15, 29, 16, 37, 18, 46, 77, 101, 19, 21, 37, 22, 41, 24, 69, 101, 25, 35, 53, 26, 47, 27, 45, 85, 113, 28, 63, 111, 157, 30, 81, 127, 31, 32, 70, 153, 241, 33, 53, 34, 59, 36, 101, 38, 65, 89, 39, 61, 40, 99, 163, 42, 105, 201, 277, 43, 44, 91, 117, 189, 329, 389, 48, 135, 249, 341, 389, 50, 100, 227, 51, 77, 101, 52, 105, 201, 277, ...

> I don't think that this will enter a loop. I will examine further tomorrow.

> Best,

> GK.

Merci Giorgos !