A342142 Take a(n), reverse it, divide the larger of the two numbers by the smaller and keep only the remainder: this remainder is present in a(n) as a substring of digits.
10, 20, 25, 30, 40, 50, 52, 60, 70, 80, 89, 90, 98, 100, 101, 110, 138, 180, 200, 202, 220, 295, 300, 303, 330, 400, 404, 410, 440, 500, 505, 510, 511, 520, 521, 530, 540, 550, 592, 600, 606, 660, 700, 707, 770, 800, 808, 810, 820, 831, 880, 890, 899, 900, 909, 940, 990, 998, 1000, 1001, 1010, 1089
Offset: 1
Examples
a(1) = 10, which reversed is 1 (leading zeros are erased); 10/1 leaves a remainder 0, which is present in a(1); a(2) = 20, which reversed is 2 (leading zeros are erased); 20/2 leaves a remainder 0, which is present in a(2); a(3) = 25, which reversed is 52; 52/25 leaves a remainder 2, which is present in a(3); ... a(50) = 831, which reversed is 138; 831/138 leaves a remainder 3, which is present in a(50); etc.
Crossrefs
Cf. A342114 (where the terms of this sequence are used).
Programs
-
Mathematica
lst={};k=1;Do[While[!StringContainsQ[ToString@k,ToString@Mod[#2,#]&@@(Sort@{k,IntegerReverse@k})],k++];AppendTo[lst,k];k++,{n,61}];lst (* Giorgos Kalogeropoulos, May 08 2022 *)
Comments