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A366492 Analog of A121805, but starting with 4.

%I A366492 #18 Nov 18 2023 13:13:12
%S A366492 4,48,129,221,233,265,318,402,426,490,494,539,635,691,708,795,853,891,
%T A366492 910,919,1010,1011,1022,1043,1074,1115,1166,1227,1298,1379,1470,1471,
%U A366492 1482,1503,1534,1575,1626,1687,1758,1839,1930,1931,1942,1963,1994,2036,2098,2180,2182
%N A366492 Analog of A121805, but starting with 4.
%C A366492 If instead we start with 3, the sequence is the two-term sequence [3, 36].
%C A366492 The present sequence is finite, with last term a(199900) = 9999945.
%H A366492 N. J. A. Sloane, <a href="/A366492/b366492.txt">Table of n, a(n) for n = 1..20000</a>
%o A366492 (Python)
%o A366492 from itertools import islice
%o A366492 def agen(start=4): # generator of terms
%o A366492     an, y = start, 1
%o A366492     while y < 10:
%o A366492         yield an
%o A366492         an, y = an + 10*(an%10), 1
%o A366492         while y < 10:
%o A366492             if str(an+y)[0] == str(y):
%o A366492                 an += y
%o A366492                 break
%o A366492             y += 1
%o A366492 print(list(islice(agen(), 50))) # _Michael S. Branicky_, Nov 18 2023
%Y A366492 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.
%Y A366492 Cf. A330128, A330129.
%K A366492 nonn,base,fini
%O A366492 1,1
%A A366492 _N. J. A. Sloane_, Nov 14 2023