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#1 |
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Jan 2005
Transdniestr
503 Posts |
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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#2 |
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Jan 2005
Transdniestr
503 Posts |
5,22,102,510,3210,17490,?
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#3 |
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Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10,753 Posts |
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#4 |
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"Nancy"
Aug 2002
Alexandria
2,467 Posts |
%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) Alex |
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#5 |
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Jan 2005
Transdniestr
503 Posts |
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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