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A126029
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Perfect m-th roots: a(n) = minimum number N such that SOPF(N)*ND(N))^n =SOD(N) where SOPF(N) =sum of prime factors of N, ND(N) = num of divisors of N, SOD(N) = sum of divisors of N.
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1
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OFFSET
| 1,1
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COMMENTS
| 35 is actually the only solution for m=1. 14844221560107739 is most likely minimal but it hasn't been proved. No solutions have been found (minimal or otherwise) where the number was not squarefree.
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LINKS
| Mersenne Forum, Mersenne forum thread
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EXAMPLE
| 22446139 factors as: 31*67*101*107, SOPF(N) =sum of prime factors of N = 31+67+101+107 = 306, ND(N) = num of divisors of N = 2^4 = 16, SOD(N) = sum of divisors of N = (31+1)*(67+1)*(101+1)*(107+1) = 23970816, SOPFN(N)*ND(N))^2 = (306*16)^2 = 23970816 = SOD(N). As this number turns out to be minimal, it would be the 2nd term in the sequence.
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PROG
| (PARI) By picking primes p such that p+1 is very smooth, one has a much better chance of solving the relation
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CROSSREFS
| Cf. A126028.
Sequence in context: A196542 A139473 A110596 * A107736 A139474 A023929
Adjacent sequences: A126026 A126027 A126028 * A126030 A126031 A126032
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KEYWORD
| hard,nonn,uned,bref
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AUTHOR
| Fred Schneider (frederick.william.schneider(AT)gmail.com), Dec 14 2006
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