OFFSET
1,1
COMMENTS
Each factorization is necessarily composed of multipliers that use only the single missing digit.
The single missing digit cannot be 0, 1, 5, or 6. Terms missing 2, 3, 4, 7, and 8 appear within a(1)-a(6). 52612606387341 = 9^6 * 99 * 999999 is an example of a term missing 9. - Michael S. Branicky, Apr 18 2024
Some terms are equal to the sum of two distinct smaller terms:
a(741) = a(635) + a(673)
a(1202) = a(1081) + a(1144)
a(1273) = a(1110) + a(1169)
a(1493) = a(1335) + a(1374)
a(2753) = a(2478) + a(2528)
a(2793) = a(2512) + a(2583)
a(3581) = a(3234) + a(3317)
a(4199) = a(3808) + a(3921)
a(4803) = a(4510) + a(4607) = a(4557) + a(4568)
a(5756) = a(5256) + a(5362)
a(6083) = a(5718) + a(5847)
a(7262) = a(6761) + a(6779)
a(7331) = a(6786) + a(6904)
a(9204) = a(8723) + a(8886)
a(9364) = a(8858) + a(8982)
a(9453) = a(8972) + a(8983) - Hans Havermann, Apr 21 2024
LINKS
Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1016 from Hans Havermann)
Michael S. Branicky and Hans Havermann, Table of n, a(n) for n = 1..10000, fully factored
EXAMPLE
10973607685048 = 22222*22222*22222 is in the sequence because it has nine distinct digits and may be factored using only its missing digit.
PROG
(Python)
import heapq
from itertools import islice
def agen(): # generator of terms
allowed = [2, 3, 4, 7, 8, 9]
v, oldt, h, repunits, bigr = 1, 0, list((d, d) for d in allowed), [1], 1
while True:
v, d = heapq.heappop(h)
if (v, d) != oldt:
s = set(str(v))
if len(s) == 9 and str(d) not in s:
yield v
oldt = (v, d)
while v > bigr:
bigr = 10*bigr + 1
repunits.append(bigr)
for c in allowed:
heapq.heappush(h, (bigr*c, c))
for r in repunits:
heapq.heappush(h, (v*d*r, d))
print(list(islice(agen(), 100))) # Michael S. Branicky, Apr 19 2024
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Hans Havermann, Apr 18 2024
STATUS
approved