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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
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.
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
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