A094830 Start with x = n, repeatedly replace x with x + sum of squares of digits of x until you reach a prime; sequence gives number of steps.

1, 0, 0, 6, 0, 4, 0, 11, 5, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 5, 13, 9, 0, 4, 11, 12, 17, 3, 0, 8, 0, 8, 7, 1, 7, 3, 0, 5, 7, 4, 0, 3, 0, 3, 7, 8, 0, 2, 2, 2, 6, 3, 0, 10, 2, 3, 1, 3, 0, 3, 0, 2, 2, 18, 2, 11, 0, 2, 6, 9, 0, 10, 0, 1, 1, 2, 5, 1, 0, 16, 2, 8, 0, 3, 11, 6, 10, 2, 0, 4, 1, 15, 2, 1, 9, 2, 0, 7, 7, 1

Offset: 1

Author

Jason Earls, Jun 13 2004

Keywords

• base
• easy
• nonn
• look

Comments

It is currently not known if this sequence is well defined, i.e. if for all n>=1 there exists an integer v such that a(n) = v. If the sequence is not well defined then the given programs are incorrect as they will get stuck in an infinite loop for some integers. - Peter Luschny, Oct 28 2016

Examples

			a(4)=6 because 4 -> 20 -> 24 -> 44 -> 76 -> 161 -> 199, takes 6 steps to reach a prime.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A003132, A094877.

Programs

Maple

f:= proc(n) local x,k;
x:= n;
for k from 0 do
  if isprime(x) then return k fi;
  x:= x + add(t^2, t = convert(x,base,10))
od;
end proc:
map(f, [$1..100]); # Robert Israel, Oct 27 2016

Mathematica

p[n_]:=Length[NestWhileList[#+Total[IntegerDigits[#]^2]&,n, !PrimeQ[ #]&]]-1; Array[p,100] (* Harvey P. Dale, Dec 03 2011 *)

PARI

a(n) = my(k=0); while(!isprime(n),d=digits(n); n+=vecsum(vector(#d,i,d[i]^2)); k++) ;k \\ David A. Corneth, Oct 27 2016