Query on sequence of dynamics of iterating A113172 Scrabble value of English word for the number n.

[Jonathan Post]

Thank you, Maximilian Hasler.

I wrestled with the same question of index (0 or 1 step to get to orbit).

I DON'T know there aren't any other cycles than the orbit of 9. I
don't even know that A113172(n) <= n for all sufficiently large n.
Hence I don't know that there are not infinite orbits.

So I continue (by hand: someone coding this might be nice) and, for
that matter, coding would allow someone to do an extension of A113172
to answer some of these questions:

n   trajectory of iterated A113172(n)
31	 	31,15,13,11,9
32		32,18, 12
33		33,20,12
34		34,19,8,9
35		35,22,18,12
36		36,22,18,12
37		37,20,12
38		38,21,15,13,11,9
39		39,16,14,11,9
40		40,11,9
41		41,14,11,9
42		42,17,12
43		43,19,8,9
44		44,18,12
45		45,21,15,13,11,9
46		46,21,15,13,11,9
47		47,19,8,9
48		48,20,12
49		49,15,13,11,9
50		50,14,11,9
51		51,17,12
52		52,20,12
53		53,22,18,12
54		54,21,15,13,11,9
55		55,24,19,8,9
56		56,24,19,8,9
57		57,22
58		58,23,20,12
59		59,18,12
60		60,15,13,11,9
61		61,18,12
62		62,21,15,13,11,9
63		63,23,20,12
64		64,22,18,12
65		65,5,22,18,12
66		66,25,22,18,12
67		67,23,20,12
68		68,24,19,8,9
69		69,19,8,9
70		70,13,11,9
71		71,16,14,11,9
72		71,19,8,9

Length to orbit or fixed point is a record:
0,2,38, what next?

You're right, though, that I should cease psychoanalysis of seqfans,
as I am unqualified, and it's rude.

On 7/1/08, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> On Tue, Jul 1, 2008 at 15:53, Jonathan Post <jvospost3 at gmail.com> wrote:
>  > There must be a better way than this artifice:
>  > a(n) = -k if k is the least number iterations of A113172(n) to reach a
>  > fixed point
>  > a(n) = 0 if n is a fixed point under A113172
>  > a(n) = +j if j is the least number of iterations before entering a cycle.
>  >
>  > n  a(n)  comment
>  > 0   3   because 0->ZERO->13-> THIRTEEN ->11->
>  > ELEVEN->9->NINE->4->FOUR->7->SEVEN->8->EIGHT->9 looping
>
> > ...
>
> > 8   2  because  (8,9)
>  > 9   1  because 9 begins the (9,4,7,8,9,4,7,8,9,4,7,8,9,...) loop
>
>
> Q: why a(9)=1 ?
>  Zero iterations are needed to get into the loop!
>
>
>  > 17 -2 because (17,12) gets to a fixed point
>
>
> again : ONE iteration takes to the f.p.
>  else it is not reasonable to say that 12 is at distance 0 of a fixed point.
>
>  Maybe a better convention would be:
>  a(n)>0 is the number of elements in orbit(n) ending at a fixed point,
>  a(n)<=0 is - the number of steps needed to reach a point of a cycle
>  (I suppose there aren't any others than the orbit of 9.)
>
>  Then:
>  a(n)=1  <=>  n is a fixed point
>  a(n)=0  <=>  n is element of a cycle
>
>  Alternative:
>
> > I'm getting a sense of who's in the equivalence class who hates "word"
>  > sequences and who is the in equivalence class who finds them
>  > interesting.
>
>
> I don't think these are equivalence classes.
>  I even think that each of the two "belongs to" relations are ill
>  defined. At least for me, "hating"(?) or "finding interesting" is not
>  a function of the "word" kw alone.
>
>  Regards,
>
> Maximilian
>