OFFSET
1,4
COMMENTS
a(n) = 0 iff n is a prime.
Is it a theorem that a(n) always exists?
REFERENCES
Eric Angelini, Posting to Sequence Fans Mailing List, Apr 20 2014
LINKS
Hiroaki Yamanouchi, Table of n, a(n) for n = 1..100000
EXAMPLE
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
...
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 *)
CROSSREFS
KEYWORD
easy,nonn,base
AUTHOR
N. J. A. Sloane, Apr 23 2014
EXTENSIONS
a(23)-a(87) from Hiroaki Yamanouchi, Sep 05 2014
STATUS
approved