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A366492
Analog of A121805, but starting with 4.
13
4, 48, 129, 221, 233, 265, 318, 402, 426, 490, 494, 539, 635, 691, 708, 795, 853, 891, 910, 919, 1010, 1011, 1022, 1043, 1074, 1115, 1166, 1227, 1298, 1379, 1470, 1471, 1482, 1503, 1534, 1575, 1626, 1687, 1758, 1839, 1930, 1931, 1942, 1963, 1994, 2036, 2098, 2180, 2182
OFFSET
1,1
COMMENTS
If instead we start with 3, the sequence is the two-term sequence [3, 36].
The present sequence is finite, with last term a(199900) = 9999945.
LINKS
PROG
(Python)
from itertools import islice
def agen(start=4): # generator of terms
an, y = start, 1
while y < 10:
yield an
an, y = an + 10*(an%10), 1
while y < 10:
if str(an+y)[0] == str(y):
an += y
break
y += 1
print(list(islice(agen(), 50))) # Michael S. Branicky, Nov 18 2023
CROSSREFS
Comma sequences in base 10, starting with 1, 2, 4, 5, 6, 7, 8, 9, 10 are A121805, A139284, A366492, A367337, A367350, A367351, A367352, A367353, A367354. Starting with 3 is trivial, and those starting with 11, 12, 13 are essentially duplicates.
Sequence in context: A362402 A373435 A048608 * A275033 A192418 A162673
KEYWORD
nonn,base,fini
AUTHOR
N. J. A. Sloane, Nov 14 2023
STATUS
approved