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A027687
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4-perfect (quadruply-perfect or sous-triple) numbers: sum of divisors of n is 4n.
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42
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30240, 32760, 2178540, 23569920, 45532800, 142990848, 1379454720, 43861478400, 66433720320, 153003540480, 403031236608, 704575228896, 181742883469056, 6088728021160320, 14942123276641920, 20158185857531904, 275502900594021408
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OFFSET
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1,1
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COMMENTS
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It is conjectured that there are only finitely many terms. - N. J. A. Sloane, Jul 22 2012
Odd perfect number (unlikely to exist) and infinitely many Mersenne primes will make the sequence infinite - take the product of the OPN and coprime EPNs.
Theorem: If k>1 and p=a(n)/2^(k-2)+1 is prime then for each positive integer m, 2^(k-1)*p^m is a solution to the equation sigma(phi(x))=2*x-2^k, which implies the equation has infinitely many solutions.
Proof: sigma(phi(2^(k-1)*p^m)) = sigma(2^(k-2)*(p-1)*p^(m-1)) = sigma(2^(k-2)*(p-1))*sigma(p^(m-1)) = sigma(a(n))*(p^m-1)/(p-1) = 4*a(n)*(p^m-1)/(p-1) = 2^k*(p^m-1) = 2*(2^(k-1)*p^m)-2^k.
It seems that for all such equations there exist such an infinite set of solutions. So I conjecture that the sequence is infinite! (End)
If 3 were prepended to this sequence, then it would be the sequence of integers k such that numerator(sigma(k)/k)=4. - Michel Marcus, Nov 22 2015
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REFERENCES
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R. K. Guy, Unsolved Problems in Number Theory, B2.
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LINKS
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EXAMPLE
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30240 = 2^5*3^3*5*7
sigma(30240) = (2^6-1)/1*(3^4-1)/2*(5^2-1)/4*(7^2-1)/6
= (63)*(40)*(6)*(8)
= (7*3^2)*(2^3*5)*(2*3)*(2^3)
= 2^7*3^3*5*7
= (2^2) * (2^5*3^3*5*7)
= 4 * 30240 (End)
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MATHEMATICA
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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Jean-Yves Perrier (nperrj(AT)ascom.ch)
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EXTENSIONS
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STATUS
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approved
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