[seqfan] Re: Rank of a prime (Collatz/Syracuse)

[Jacques Tramu]

> Eric Angelini wrote:
> the idea behind my question was this one -- I'm trying to produce a
> little "number game" like the Collatz-one you all know:
> http://mathworld.wolfram.com/CollatzProblem.html
>
> a) start with an integer (here, "6")
> b) decompose it in prime factors (small --> big): "6" = 2.3
> c) write under each prime factor it's rank in the primes succession:
>          2.3
>          1 2 (because 2 is the 1st prime, and 3 is the 2nd prime)
> d) concatenate the ranks to produce a new integer: "12"
> e) start again the procedure from there
>
> REMARK: it may happen that the number obtained via concatenation
> is itself prime; simply replace it by it's own rank in the primes
> succession -- and go on with the procedure from there.
>


Let's call ea(n) this sequence - n= starter

I found the following :

->ea(4)
4       11      5       3       2       1

-> ea(5)
5       3       2       1

-> ea(6)
6       12      112     11114   1733    270     12223   7128    11122225 
33991010        13913661        2107998   12222775        33910130 
131212367       56113213        6837229 4201627 266366  112430  131359  7981 
969     278     134     119     47      15      23      9       22      15

-> ea(7)
7       4       11      5       3       2       1

-> ea(8)
8       111     212     1116    112211  52626   124441  28192   11111152 
111165448       1117261018      1910112963        252163429       42205629 
2914219 454002  127605  231542  110938  15631   44510   13605   23155 
3582    12246   12637   1509    296     11112   111290  131172  1127117 
76613   9470    13161   21328   11
111114  14142115        3625334 1125035 348169  78151   11369   1373    220 
1135    349     70      134     119       47      15      23      9       22 
15

-> ea(9)
9       22      15      23      9

-> ea(10)
10      13      6       12      112     11114   1733    270     12223   7128 
11122225        33991010        13
913661  2107998 12222775        33910130        131212367       56113213 
6837229 4201627 266366  112430  131359    7981    969     278     134 
119     47      15      23      9       22      15

-> ea(1000)
1000    111333  271217  23744   111111416       111668035       38123416 
111333214       111134771       52
041017  4504142 1193600 11111113374     1291293111      210659941 
11635952        111158582       112114318
13341506        1455641 512354  182534  151042  17438   11087   1344 
11111124        11285010        122351376
11112431073     272684284       117283810       134146302       121044136 
1118132200      111331051208    11
1654243638      14366699733     2225493779      81368636        112001924 
112921719       25911084        11
2331838 128221488       111122262170    13566196845     222320108177 
8863086101      405548474       116871062
145241063       5289904 111182002       14821986        12181067 
902891  522175  332349  234139  20756   11
691     22284   1122114 1242932 1126858 1141559 95098   17407   6627 
21515   3667    844     1147    1112    11
134     1862    1448    11142   122114  16152   1112122 155186  111368 
1111646 145756  116409  24089   6729    23
34      1277    206     127     31      11      5       3       2       1

-> ea(23)
23      9       22      15      23

-> ea(57233) - can somebody compute it ?
->ea(2^100) - can somebody compute it ?

So  the question is :
does any sequence terminates in "1"  or with the" 15 23 9 22 15 loop" ?
or 25 - ea(25)  is 25 33 25


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GBnums code:
void ea (n)
{
mpz u[] ; // factors
mpz tr[]; // sequence
print(n);
while(n > 1)
 {
     lfactors(u,n); // factorize into u
     vmap(u,pi);  // replace factors by rank
     n = catv(u); // concatenate
     print(n);
     if(vsearch(tr,n) > 0) break; // loop found
     vpush(tr,n); // remember n
     }
println('');
}

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http://www.echolalie.com/gbnums
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