A381242 Lexicographically earliest sequence of distinct terms > 1 such that no term is a substring of the product of any two terms.

2, 3, 20, 22, 28, 30, 200, 202, 220, 248, 280, 300, 2000, 2002, 2020, 2022, 2200, 2480, 2800, 3000, 3252, 3272, 20000, 20002, 20020, 20022, 20200, 20220, 22000, 23252, 24800, 28000, 30000, 32520, 32720, 200000, 200002, 200020, 200022, 200200, 200202, 200220

Offset: 1

Author

Dominic McCarty, Mar 24 2025

Keywords

• nonn
• base

Comments

The "two terms" mentioned in the name are not necessarily distinct, so no term is the substring of the square of any term.

It can be shown that the digits 1 and 6 never appear, and that all terms start with either a 2 or a 3.

If k is in the sequence, then so is 10*k.

Conjecture: The digit 9 never appears.

Examples

			For calculating a(3):
If 4 was in the sequence, 3*4 = 12 would have 2 as a substring, so it is disallowed.
If 5 was in the sequence, 3*5 = 15 would have 5, itself, as a substring, so it is disallowed.
The first term that is allowed is 20, since 20 is the first term not to have 2, 3, or itself as a substring of any of the following: 2*20 = 40, 3*20 = 60, 20*20 = 400.
So, a(3) = 20.
		

Crossrefs

Cf. A382453

Programs

Python

a = [2]
while len(a) < 20:
    a.append(a[-1]+1)
    while any(any(str(k) in str(a[i]*a[j]) for k in a) for i in range(len(a)) for j in range(i,len(a))): a[-1] += 1
print(a)

Python

from itertools import count, islice
def agen():    # generator of terms
    slst, alst, an = [], [], 2
    S = {"4"}  # strings of products of two terms (including self products)
    while True:
        alst.append(an)
        slst.append(str(an))
        yield an
        for k in count(an+1):
            sk = str(k)
            if any(sk in s for s in S): continue
            Pk = [str(ai*k) for ai in alst] + [str(k*k)]
            if any(si in s for s in Pk for si in slst+[sk]): continue
            an = k
            S |= set(Pk)
            break
print(list(islice(agen(), 42))) # Michael S. Branicky, Mar 26 2025