

A182129


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


0



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
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OFFSET

1,2


COMMENTS

a(n) is the number of times you form the npower 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.


LINKS



EXAMPLE

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.


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:


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



