A099646 Function f(n) = 1 + Sum(digit^2 of n) is iterated and a(n) is the length of terminal cycle at initial value n.
9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 1, 1, 9, 9, 9, 9, 9, 9, 9, 9, 9, 1, 9, 9, 9, 9, 9, 9, 1, 9, 9, 9, 1, 9, 9, 9, 9, 9, 1, 1, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 1, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9
Offset: 1
Examples
For n = 1: iteration-list= {1,2,5,26,41,18,66,73,59,107,51,[27,54,42,21,6,37,59,107,51],27... with t = 11 transient and c = a(1) = 9, the cycle-length; For n = 35: list={36,46,53,[35],35,...} with transient t = 3, c = a(35) = 1 the cycle-length.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
ed[x_] :=IntegerDigits[x]; f[x_] :=Apply[Plus, ed[x]^2]+1; itef[x_, ho_] :=NestList[f, x, ho]; tmc=Table[Length[Union[itef[w, 100]]], {w, 1, 256}]; c1=Table[Min[Flatten[Position[itef[w, Length[Union[itef[w, 100]]]] -Last[itef[w, Length[Union[itef[w, 100]]]]], 0]]], {w, 1, 256}]; (* transient-length= *) c1-1; (* cycle-length= *) c=tmc-(c1-1); (* ho=iteration number is chosen by trial and error *) (* program provides t, t+c and c lengths[=unknown-in-advance] for any similar iterations if f modified *) (* Second program: *) With[{nn = 10^3}, Table[Function[s, Length@ KeySelect[s, Length@ Lookup[s, #] > 1 &]]@ PositionIndex@ NestList[1 + Total[ IntegerDigits[#]^2] &, n, nn], {n, 105}]] (* Michael De Vlieger, Jul 24 2017 *)
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