

A305669


Take the sum of the digits of a number, put it at the left side and delete the same number of digits at the right side. Repeat the process. Sequence lists numbers that reach themselves after some steps.


0



0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 14, 17, 21, 32, 41, 51, 53, 54, 65, 81, 85, 95, 98, 101, 108, 109, 116, 171, 179, 210, 321, 632, 811, 910, 917, 1013, 1071, 1112, 1113, 1114, 1116, 1271, 1291, 1312, 1313, 1315, 1316, 1323, 1375, 1381, 1415, 1516, 1517, 1585
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OFFSET

0,3


COMMENTS

Fixed points of the process (numbers that reach themselves in a single step) are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1818, 17171, 272727, 1313131, 2626262, 3939393, 12121212, 24242424, 36363636, 48484848, etc.
The number of steps to return to the original number or to enter another cycle depends on the number of digits. Here below all possible steps against the number of digits from 1 to 8:
Digits Steps
1 1
2 3, 12
3 3, 9
4 1, 3, 10, 31
5 1, 9
6 1, 4, 13, 21, 39
7 1, 2, 4, 5, 6, 10, 12, 20, 23, 30, 60
8 1, 3, 26, 78


LINKS



MAPLE

P:=proc(q) local a, b, c, d, k, n, x;
for n from 1 to q do a:=convert(n, base, 10); d:=a; x:=0;
while x<10^(ilog10(n)+1) do x:=x+1; b:=convert(a, `+`);
c:=ilog10(b)+1; b:=convert(b, base, 10);
for k from 1 to nops(a)c do a[k]:=a[k+c]; od;
for k from 1 to c do a[nops(a)c+k]:=b[k]; od;
if a=d then print(n); break; fi; od; od; end: P(10^6);


CROSSREFS



KEYWORD

nonn,easy,base


AUTHOR



STATUS

approved



