A241181 Start with n; add to it any of its digits; repeat; a(n) = minimal number of steps needed to reach a prime.

1, 0, 0, 3, 0, 2, 0, 2, 2, 1, 0, 1, 0, 2, 2, 1, 0, 1, 0, 6, 1, 5, 0, 4, 2, 3, 1, 5, 0, 6, 0, 2, 5, 1, 2, 4, 0, 1, 3, 4, 0, 4, 0, 3, 2, 3, 0, 2, 1, 5, 2, 2, 0, 1, 4, 1, 4, 3, 0, 3, 0, 3, 3, 3, 1, 2, 0, 2, 4, 4, 0, 1, 0, 2, 4, 1, 3, 3, 0, 4, 1, 3, 0, 2, 3, 2, 2

Offset: 1

Author

N. J. A. Sloane, Apr 23 2014

Keywords

• easy
• nonn
• base

Comments

a(n) = 0 iff n is a prime.

Is it a theorem that a(n) always exists?

Yes: the proof is similar to that of Robert Israel for A241180. - Rémy Sigrist, Jul 25 2020

Examples

			Examples, in condensed notation:
1+1=2
2
3
4+4=8+8=16+1=17
5
6+6=12+1=13
7
8+8=16+1=17
9+9=18+1=19
10+1=11
11
12+1=13
13
14+4=18+1=19
15+1=16+1=17
16+1=17
17
18+1=19
19
20+2=22+2=24+2=26+6=32+2=34+3=37
...
		

References

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

Links

• Hiroaki Yamanouchi, Table of n, a(n) for n = 1..100000

Crossrefs

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

Programs

Mathematica

A241181[n_] := Module[{c, nx},
   If[PrimeQ[n], Return[0]];
   c = 1; nx = n;
   While[ ! AnyTrue[nx = Flatten[nx + IntegerDigits[nx]], PrimeQ], c++];
   Return[c]];
Table[A241181[i], {i, 100}] (* Robert Price, Mar 17 2019 *)

Extensions

a(23)-a(87) from Hiroaki Yamanouchi, Sep 05 2014