[seqfan] Re: patterns continue for larger numbers? generalised question

[Neil Fernandez]

Hi Ali,

In message <1792432217.4274278.1570337524190 at mail.yahoo.com>, Ali Sada
via SeqFan <seqfan at list.seqfan.eu> writes

>If we add a positiveinteger n to and we apply the following algorithm:
>
>“If n iseven, we divide it by two;
>
>If n is odd,we add it to m; where m is the smallest square > n.”
>
>When Icontinued with this algorithm it reached either 1, or 11 as its lowest 
>point.

In message <3b07abda-fa2c-fdd0-7368-1d2652af49e1 at matcos.nl>, Matthijs
Coster via SeqFan <seqfan at list.seqfan.eu> writes

>I run my program till 100000.
>
>Here the records:
>
>1 8
>9 27
>18 28
>36 29
>49 31
>51 48
>81 58
>87 67
>174 68
>289 83
>313 85
>529 108
>973 112
>1033 113
>1873 123
>1897 125
>2115 136
>4230 137
>7245 139
>7249 149
>8309 150
>9029 253
>17789 262
>35435 265
>70269 292

I checked up to n=1.2*10^6 and always reached either the 1-cycle
(1,5,14,7,16,8,4,2,1)  or the 11-cycle (11,27,63,127,271,560,
280,140,70,35,71,152,76,38,19,44,22,11).

After 70269 (292 steps), the records for the number of steps taken to
reach 1 or 11 are

139929  299
279858  300
309553  301 
279895  314
280197  318
431913  332
558161  337
1113437 348
1156001 356

A generalised question that comes to mind is this:

is there any (n,t,u) (t>1,u>1) for which the sequence

a(1)=n;
a(i+1) = a(i)/t if n|t, otherwise a(i+1)= a(i) + smallest uth power > i

does not reach a cycle?

Given that nobody knows whether the 3n+1 sequence (numbers of steps
taken to reach 1 are in A006577) always reaches a cycle, let alone
whether it always reaches the 1-cycle, this may not be an easy question.
But it would be nice if we could find an (n,t,u) which *might* not reach
a cycle.

Neil

-- 
Neil Fernandez