[seqfan] Re: Another problem sequence that needs your help.

[RGWv]

Et al,

    I added the Mmca coding to the sequence and here are the first 15 terms. 
Mathematica calculates this very easily.

Bob.

2
6
12
630
2312310
4204200
212219517510
19148021098476270,
42528534634267283364442680
786914424185453199970788610804266056488050
3829646054051942938502659979286757965071707218349053660908889569730
102511850312034144559513273641764705839779523522691131104624175656908503117870251740580120
43447615978705154018443582201901172958097181487514161220973902400977841750139109089036870428621053570850601326666130203433335409541850
457155125222737012060234799010169791745364240080750784446339223714269930448615436395855152249061313299411119039918138348490220633665265525095898945258304347809625122148286022469953255816555023215500190
530818583760627400560567445546718267229825469640455916182951878107207926901523914246626035207685091467377316945780381144751351042290813415031284607198174493822881390970887871476305889740233398466625991787537723261321800041981574212204752063434768386493959831400592562283240


-----Original Message----- 
From: Ed Jeffery
Sent: Wednesday, June 29, 2011 10:55 PM
To: seqfan at list.seqfan.eu
Subject: [seqfan] Another problem sequence that needs your help.

Dear friends:


I recently submitted A192321={2, 6, 12, 630, 2312310, 4204200, 
212219517510,...} to OEIS with only the seven terms shown here because they 
are so tough for me to generate with the limited resources I have. I thought 
afterward that the sequence was rather lame and expected it to be rejected, 
but, to my surprise, I was asked to extend it and give a program for 
generating the terms. The problem is that I am not a programmer, so I wonder 
if one of you out there might be interested in writing something in Pari, 
Mathematica or Maple and provide more terms.

A general term a(n) of the sequence has prime factorization


a(n)=2^d_(n,1)*3^d_(n,2)*5^d_(n,3)*...*p_(k_n)^d_(n,k_n),


where p_(k_n) is the (k_n)-th prime, and the sequence of exponents 
d_(n,1),...,d_(n,k_n) (of length k_n) is given by the sequence of k_n digits 
from the n-th term of J. H. Conway's Look and Say
sequence A005150(defined with initial value 1).


For example, from Conway we have 
A005150={1,11,21,1211,111221,312211,13112221,...}, so, for n=4, 
A005150(4)=1211 -> {1,2,1,1}, which implies k_4=4, with d_(4,1)=1, 
d_(4,2)=2, d_(4,3)=d_(4,4)=1; hence a(4) = 2*3^2*5*7 = 630.

I know of no relation between terms which is why I was/am skeptical. I also 
didn't make the above representation of a(n) clear enough in my submission, 
so I need to edit it.

Thanks in advance,

Ed Jeffery

_______________________________________________

Seqfan Mailing list - http://list.seqfan.eu/