A182129 Number of iterations of the orbit n -> (sum of the decimal digits of n)^n starting with n, needed to stabilize.

0, 5, 3, 3, 5, 3, 3, 5, 5, 1, 6, 6, 2, 6, 9, 5, 2, 6, 2, 7, 5, 5, 6, 6, 6, 3, 5, 2, 9, 7, 6, 13, 12, 9, 5, 9, 2, 10, 9, 7, 15, 9, 7, 4, 7, 2, 6, 3, 7, 12, 6, 9, 9, 5, 2, 10, 12, 10, 14, 7, 8, 8, 11, 2, 13, 10, 5, 9, 8, 15, 9, 6, 2, 17, 13, 8, 9, 5, 15, 12

Offset: 1

Author

Michel Lagneau, Apr 13 2012

Keywords

• nonn
• base

Comments

a(n) is the number of times you form the n-power of the sum of the digits before reaching the last number of the cycle.

Generalization and conjecture: Let k be a positive integer. The number of iterations of the orbit k -> (sum of the decimal digits of k)^n is finite for any exponent n and any starting value k.

Example with n = 17; start with k = 3.

3^17 = 129140163, sum of the decimal digits = 27,

27^17 = 2153693963075557766310747, sum of the decimal digits = 117,

117^17 = 144264558065210807467328187211661877, sum of the decimal digits = 153,

153^17 = 13796036156758195415808856807283698713, sum of the decimal digits = 189,

189^17 = 501014933601411817143935347829544613629, sum of the decimal digits = 153 is already in the trajectory.

Examples

			0 is in the sequence 1^1 -> 1;
For the power 2, a(2) = 5:
    2 ->       2^2 =   4;
    4 ->       4^2 =  16;
   16 ->   (1+6)^2 =  49;
   49 ->   (4+9)^2 = 169;
  169 -> (1+6+9)^2 = 256 is the end of the cycle because 256 -> (2+5+6)^2 = 169 is already in the trajectory. Hence we obtain the map: 2 -> 4 -> 16 -> 49 -> 169 -> 256 with 5 iterations.
		

Crossrefs

Cf. A177148, A182128.

Programs

Maple

with(numtheory) : T :=array(1..500) :W:=array(1..500):for n from 1 to 80 do : k:=0:nn:=n:for it from 1 to 50 do:T :=convert(nn,base,10) :l:=nops(T):s1:=sum(T[i],i=1..l):s:=s1^n:k:=k+1:W[k]:=s:nn:=s:od: z:= [seq(W[i],i=1..k)]:V:=convert(z,set):n1:=nops(V): printf(`%d, `,n1):od: