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Old 2007-01-11, 01:49   #1
grandpascorpion
 
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Default What's the next term in the sequence (part deux)?

5,22,102,510,82110,17490,?

The solution involves basic number theory.

39 and values derived from fitted polynomials aren't what I'm looking for :)

Last fiddled with by grandpascorpion on 2007-01-11 at 02:23 Reason: rounding errors gave bad values in the sequence
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Old 2007-01-11, 03:39   #2
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Default Ugh, one of the terms was wrong. Here's the correct sequence:

5,22,102,510,3210,17490,?
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Old 2007-01-11, 08:36   #3
xilman
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Quote:
Originally Posted by grandpascorpion View Post
5,22,102,510,82110,17490,?

The solution involves basic number theory.

39 and values derived from fitted polynomials aren't what I'm looking for :)
42 ;-)

Paul
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Old 2007-01-11, 09:52   #4
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%I A048251
%S A048251 5,22,102,510,3210,17490,112890,600270
%N A048251 Smallest number whose sum of divisors is 6^n.
%F A048251 a(n)=Min{x : A000203(x)=6^n}
%e A048251 Sigma[x]=1296=6.6.6.6 if x={510, 642, 710, 742, 782, 795, 862, 935, 1177, 1207, 1219}; the smallest is a(4)=510.
%Y A048251 A006532, A020477, A019422, A019423, A018427.
%Y A048251 Sequence in context: A083586 A087439 A033452 this_sequence A017971 A017972 A008485
%Y A048251 Adjacent sequences: A048248 A048249 A048250 this_sequence A048252 A048253 A048254
%K A048251 nonn
%O A048251 1,1
%A A048251 Labos E. (labos(AT)ana.sote.hu)

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Old 2007-01-11, 13:19   #5
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Yep, I came up with this and found the sequence later (once the minimum terms were correct).

I was thinking that the sigma would be the perfect power of a number n such that n had two or more factors. In this context, that would seem to mean n=2*3. That would be an interesting thing to prove I think.

Last fiddled with by grandpascorpion on 2007-01-11 at 13:22
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