A182129 - OEIS

%I #18 Jan 28 2022 07:47:59

%S 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,

%T 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,

%U 2,13,10,5,9,8,15,9,6,2,17,13,8,9,5,15,12

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

%C 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.

%C 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.

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

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

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

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

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

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

%e 0 is in the sequence 1^1 -> 1;

%e For the power 2, a(2) = 5:

%e     2 ->       2^2 =   4;

%e     4 ->       4^2 =  16;

%e    16 ->   (1+6)^2 =  49;

%e    49 ->   (4+9)^2 = 169;

%e   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.

%p 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:

%Y Cf. A177148, A182128.

%K nonn,base

%O 1,2

%A _Michel Lagneau_, Apr 13 2012