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There is a remarkable and unexplained agreement: if 3 and 7 are replaced by 11 and 14619833 is replaced by 14920303, the result is sequence A056637 (least prime of class n-, according to the Erdős-Selfridge classification of primes).

From David A. Corneth, Oct 18 2016: (Start):

If a(n) * k + 1 is prime then a(n + 1) <= a(n) * k + 1.

a(18), a(19), ..., a(23) <= 309554053859, 619108107719, 19811459447009, 433142367554861, 866284735109723, 22523403112852799 respectively. (End)

Conjecture: a(n) is the smallest prime p such that b(p) = n, where f(2) = 0 and for an odd prime p, f(p) = 1 + max{q|(p-1), q prime} f(q). In other words, a(n) is the smallest prime p such that A364332(primepi(p)) = n. Verified for n <= 13. - Jianing Song, Apr 28 2024

This sequence is related to a problem about googology (see Zhihu).

There are three numbers named A, B and C, and initially A>1, and B=C=0.

Now take the following steps:

Step 1. Let B=B+1, C=C+1.

Step 2. Check if A is divisible by (B+1). If yes, go to Step 3; if no, go back to Step 1.

Step 3. Let A=A/(B+1)*B^C.

Step 4. Let B=0.

Step 5. If A=1, then stop, otherwise go back to Step 1.

This program will always stop. When the program stops, C will be a very large number.

When Step 3 reached, B+1 will be the smallest prime factor of A, and this factor is eliminated while some smaller factors; i.e., factors of B, are added.

The properties of this problem rely on the prime factors of A. A can be described by the ordinal: O(A) = n_(k-1)*omega^(l_(k-1)) + ... + n_2*omega^(l_2) + n_1*omega^(l_1) + n_0 where A = 2^(n_0) * 3^(n_1) * 5^(n_2) * p(k)^(n_(k-1)). Every operation will monotonically decrease the ordinal.

Note that the exponential of omega in each term is l_k instead of k, and this is because the correspondence between prime factors and powers of omega is more complicated than "p(k+1) to omega^k".

When B+1=2, the increase of C is only 1, then every prime factor 2 corresponds ordinal 1, or omega^0. This agrees with a(p)=0 when p=2 in this sequence.

When B+1=3, any number of exponential of 2 can be increased, then every prime factor 3 corresponds ordinal omega.

When B+1=5, only the power of 2 can be increased as well, then prime factor 5 also corresponds omega.

When B+1=7, any number of exponentials of 2 and 3 can be both increased, then the corresponding ordinal is (1+omega)*omega=omega*omega=omega^2.

The rule is, the power of omega for a given prime number p equals the maximum among those for prime factors of (p-1), plus one.

This sequence is similar to but different from A083647, the difference begins at the 52nd term; i.e. for p=239. That is because 238=2*7*17 and a(7)=3>2=a(17) although 7<17.

From Jianing Song, Apr 28 2024: (Start)

a(n) is one less than the maximum length of a sequence d_0 = prime(n), d_1, ..., d_m = 2 such that d_i is a prime factor of d_{i-1} - 1. For example, for n = 52 (then prime(n) = 239), we have a(n) = 3 since that a sequence with maximum length is 239, 7, 3, 2. It is clear that a(n) >= A083647(n), as the latter takes d_i to be largest prime factor of d_{i-1} - 1.

Conjecture: the smallest prime p such that f(p) = m (namely a(primepi(p)) = m) is p = A082449(m). Verified for m <= 13. (End)

The primality tree starts at an odd prime at the root node. The branches of each node are the odd primes that divide the value of (the node minus 1). The length of the longest branch is related to A126805. Following the branch with the largest label gives A083647.