A241181 -id:A241181 - cp's OEIS frontend

A241183 Start with n; add to it any of its digits; repeat; a(n) = minimal number of steps needed to reach 2n, or -1 if 2n cannot be reached.

1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 4, 4, 3, 3, 5, 4, 4, 3, 7, 6, 5, 5, 5, 4, 5, 5, 5, 4, 5, 7, 6, 6, 7, 6, 6, 7, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, -1, 8, 8, 8, 7, 8, 8, 8, 8, 8, 10, 9, 9, 9, 10, 11, 10, 10, 10, 11, 12, 11, 12, 12, 11, 12, 12, 14, 12, 13, 15, 12, 13, 14, 14, 14, 14, 14, 15, 14, 14, 16, 16, 16, 15, 15, 17, 16, 15, 15, 18

Offset: 1

N. J. A. Sloane, Apr 23 2014

Is it a theorem that a(n) always exists?
Does any number take two steps to reach 2n by the shortest path? If the answer is yes, we should add the sequence of the smallest numbers that take n steps to go from k to 2k.
If the answer is no, then the sequence of the smallest numbers that take at least n steps to go from k to 2k.
There cannot be a 2 in this list.  The maximum possible increase in two steps is 18 (2 nines) so no number larger than 18 can be doubled in 2 steps by digit addition. Since the minimum number of steps have been found through 18, and since none of them required exactly 2 steps, then there can be no 2s in this sequence. - David Consiglio, Jr., May 12 2014
a(50) does not exist. - Hiroaki Yamanouchi, Sep 05 2014
Except for n=50, 2n can be reached from all n<=10000. - Robert Price, Mar 20 2019

			Examples (in condensed notation):
1+1=2
2+2=4
...
9+9=18
10+1=11+1=12+2=14+1=15+5=20
11+1=12+1=13+3=16+6=22
12+1=13+3=16+1=17+7=24
13+1=14+4=18+8=26
14+4=18+8=26+2=28
15+5=20+2=22+2=24+4=28+2=30
15+1=16+6=22+2=24...
15+1=16+1=17+7=24...
16+6=22+2=24+2=26+6=32
17+1=18+8=26+6=32+2=34
18+8=26+2=28+8=36
19+1=20+2=22+2=24+2=26+6=32+2=34+4=38
20+2=22+2=24+2=26+6=32+3=35+5=40
...

Eric Angelini, Posting to Sequence Fans Mailing List, Apr 20 2014.

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 49 terms from Hiroaki Yamanouchi)

Related sequences: A241173, A241174, A241175, A241176, A241177, A241178, A241179, A241180, A241181, A241182, A241183.

A241183[n_] := Module[{c=1, nx=n},
   While[ ! AnyTrue[nx =
       Union[Flatten[nx + IntegerDigits[nx]]], # == 2 n &], c++];
   Return[c]];
Join[Table[A241183[i], {i, 49}], -1, Table[A241183[i], {i, 51, 100}]] (* Robert Price, Mar 18 2019 *)
(* The following is a program to detect any n that cannot reach 2n *)
f[n_] := Module[{n2 = 2 n, totest = {2 n}, i},
   While[Length[totest] > 0,
    x = First[totest]; totest = Rest[totest];
    For[i = 1, i <= 9, i++,
     If[MemberQ[IntegerDigits[x - i], i],
      If[! MemberQ[totest, x - i], AppendTo[totest, x - i]]] ];
    If[MemberQ[totest, n], Return[False]]]; Return[True]];
Select[Range[100], f[#] &] (* Robert Price, Mar 20 2019 *)

a(11) corrected (was 5 should be 4) by David Consiglio, Jr., May 12 2014
a(11) corrected in example by David Consiglio, Jr., May 20 2014
a(13) corrected (including the example) and a(23)-a(49) from Hiroaki Yamanouchi, Sep 05 2014
Escape clause added to definition at the suggestion of Robert Price. - N. J. A. Sloane, Mar 18 2019
a(50)-a(100) from Robert Price, Mar 18 2019

cp's OEIS Frontend

A241183 Start with n; add to it any of its digits; repeat; a(n) = minimal number of steps needed to reach 2n, or -1 if 2n cannot be reached.

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