A372880 a(1) = 1; a(2) = 3; for n > 2, a(n) is the smallest proper multiple of a(n-1) that contains a(n-2) as subsequence of its digits.

1, 3, 12, 36, 612, 1836, 168912, 10810368, 16366897152, 51703028103168, 1563447866811697152, 23520172003575940628103168, 1155558163424267804668132116971520, 12369352104691609178206055357839959406281031680

Offset

1,2

Comments

It is unknown whether a(n+1)/a(n) -> oo as n -> oo.

The same rule starting from terms 1, 2 gives A004643 and its multiples are as easy as A004643(n+1)/A004643(n) = 2 or 5 alternately.

Links

Table of n, a(n) for n=1..14.

Kevin Ryde, C Code

Formula

a(n) <= a(n-2)*10^k + (a(n-1) - (a(n-2)*10^k mod a(n-1))), where k is the number of decimal digits in a(n-1). - Michael S. Branicky, May 17 2024

Example

a(7) = 168912; 16812 = 92*1836 = 92*a(6) and "16812" contains a(5) = 612 as a subsequence.

Prog

(Python)
def subseq(x, y):
    i = 0
    j = 0
    while i != len(x) and j != len(y):
        if x[i] == y[j]:
           i += 1
        j += 1
    return i == len(x)
def a(n):
    if n == 1:
        return 1
    A = 1
    B = 3
    for _ in range(n-2):
        s = str(A)
        i = 1
        while not subseq(s, str(B*i)):
            i += 1
        A, B = B, B*i
    return B
(Python)
from itertools import count, islice
def is_subseq(s, p):
    while s and p:
        if p%10 == s%10: s //= 10
        p //= 10
    return s == 0
def agen(): # generator of terms
    an2, an1 = [1, 3]
    yield from [an2, an1]
    while True:
        an = next(i*an1 for i in count(1) if is_subseq(an2, i*an1))
        an2, an1 = an1, an
        yield an
print(list(islice(agen(), 11))) # Michael S. Branicky, May 15 2024
(C) /* See links. */

Crossrefs

Cf. A004643.

Sequence in context: A064028 A326241 A110950 * A102744 A145951 A083215

Adjacent sequences: A372877 A372878 A372879 * A372881 A372882 A372883

Keyword

nonn,base,more

Author

Bryle Morga, May 15 2024

Extensions

a(12)-a(13) from Michael S. Branicky, May 15 2024

a(14) from Kevin Ryde, Jun 23 2024

Status

approved