A397699 Associated primes of A397698.

3, 2, 2, 2, 3, 2, 3, 3, 2, 2, 2, 2, 3, 2, 5, 2, 2, 2, 2, 2, 3, 2, 3, 3, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 5, 2, 2, 2, 2, 2, 3, 2, 3, 3, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 5, 2, 5, 2

Offset

1,1

Comments

If A397698(n) is divisible by p^k (with k maximized) for more than one prime p, a(n) is the largest such p.

Conjecture: All primes appear in this sequence.

From Michael S. Branicky, Jul 05 2026: (Start)

The conjecture is likely true (based on an argument by Allan Wechsler; see Links).

Let ok(p, i) denote the following predicate: whether any a(s) = p over all s obtained by deleting a digit from p^i (discarding any resulting s with leading zeros). Empirically, one finds that all primes p < 10^4 satisfy ok(p, 4). (End)

Links

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Allan Wechsler, Prime Powers Sequence, SeqFan mailing list, Jul 05 2026.

Formula

a(n) = A356840(A397698(n)). - Pontus von Brömssen, Jul 05 2026

Example

A397698(1) = 81 = 3^4; the associated prime is 3, so a(1) = 3.
A397698(21) = 216 = 2^3*3^3; since 3 > 2, a(21) = 3.

Prog

(Python)
from sympy import factorint
def f(n): return max((k, p) for p, k in factorint(n).items())
def cands(s): yield from map(int, set(s[:i]+d+s[i:] for i in range(len(s)+1) for d in "0123456789"[int(i==0):]))
def a(n): return sorted(f(t)+(-t, ) for t in cands(str(n)))[-1][1]
print([a(n) for n in range(1, 89)]) # Michael S. Branicky, Jul 05 2026

Crossrefs

Cf. A356840, A397698.

Sequence in context: A207384 A085034 A119323 * A102299 A302481 A306542

Adjacent sequences: A397695 A397697 A397698 * A397700 A397701 A397702

Keyword

nonn,base,new

Author

Ali Sada, Jul 05 2026

Extensions

More terms from Michael S. Branicky, Jul 05 2026

Status

approved