A289351 Starting from one digit move right by x steps, x being the value of the digit. If the steps go beyond the least significant digits they continue from the left side. Then repeat the process from the reached digit. The sequence lists the numbers such that all the digits are touched just one time and the last run ends in the initial digit.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15, 17, 19, 31, 33, 35, 37, 39, 51, 53, 55, 57, 59, 71, 73, 75, 77, 79, 91, 93, 95, 97, 99, 111, 114, 117, 141, 144, 147, 171, 174, 177, 222, 225, 228, 252, 255, 258, 282, 285, 288, 411, 414, 417, 441, 444, 447, 471, 474, 477

Offset: 0

Author

Paolo P. Lava, Jul 03 2017

Keywords

• nonn
• base
• easy

Comments

Apart from a(0), only zeroless numbers.

If we move left instead of right, the sequence is the same up to a(103); here, a(103)=1223 while in the other sequence a(103) would be 1322.

Examples

			13894: for instance, let us start from 8. Moving eight steps right we are at 1. Then, moving one step right we are at 3. Then 3 steps right we are at 4. Again after 4 steps we are at 9. After an additional 9 steps we end at 8 again. All the digits have been touched and we are again at the digit we started from.
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A014261 (2 digits terms), A071073 (3 digits terms up to 588), A284515, A284591.

Programs

Maple

P:=proc(q) local a,b,d,k,n,t; print(0); for n from 1 to q do d:=ilog10(n)+1; a:=convert(n,base,10);
for k from 1 to trunc(d/2) do b:=a[k]; a[k]:=a[d-k+1]; a[d-k+1]:=b; od; b:=array(1..d);
for k from 1 to d do b[k]:=0; od; t:=1; for k from 1 to d do
if ((t+(a[t] mod d)) mod d)>0 then b[(t+(a[t] mod d)) mod d]:=1; t:=(t+(a[t] mod d)) mod d;
else b[d]:=1; t:=d; fi; od; if add(b[k],k=1..d)=d then print(n); fi; od; end: P(10^9);

Mathematica

Select[Range[0,477],(n=IntegerDigits@#;Last[m=Mod[Accumulate@Mod[n,s=Length@n],s]]==0&&Sort@m+1==Range@s)&] (* Giorgos Kalogeropoulos, Nov 21 2021 *)