A293871 Numbers having 11 as substring of their digits.
11, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 211, 311, 411, 511, 611, 711, 811, 911, 1011, 1100, 1101, 1102, 1103, 1104, 1105, 1106, 1107, 1108, 1109, 1110, 1111, 1112, 1113, 1114, 1115, 1116, 1117, 1118, 1119, 1120, 1121, 1122, 1123, 1124, 1125, 1126, 1127, 1128, 1129, 1130, 1131
Offset: 1
Links
Crossrefs
Cf. A011540, A011531, A011532, A011533, A011534, A011535, A011536, A011537, A011538, A011539: analog for substrings '0' through '9'.
Cf. A293870, A293872, A293873, A293874, A293875, A293876, A293877, A293878, A293879, A293880: same for substrings '10' - '20'.
Cf. A121031: subsequence of terms divisible by 11.
Numbers divisible by k and having k as a substring: A121022 (2), A121023 (3), A121024 (4), A121025 (5), A121026 (6), A121027 (7), A121028 (8), A121029 (9), A121030 (10), A121031 (11), A121032 (12), A121033 (13), A121034 (14), A121035 (15), A121036 (16), A121037 (17), A121038 (18), A121039 (19), A121040 (20).
Cf. A121041.
Programs
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Mathematica
Select[Range[2000], StringContainsQ[IntegerString[#], "11"] &] (* Paolo Xausa, Feb 25 2024 *)
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PARI
is_A293871 = has(n,p=11,m=10^#Str(p))=until(p>n\=10,n%m==p&&return(1))
Formula
a(n) ~ n. - Charles R Greathouse IV, Nov 02 2022