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A011251
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Numbers n such that phi(n) + sigma(n) = 3n.
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7
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312, 560, 588, 1400, 85632, 147492, 556160, 569328, 1590816, 2013216, 3343776, 4695456, 9745728, 12558912, 22013952, 23336172, 30002960, 52021242, 75007400, 137617728, 153587720, 699117024, 904683264, 2468053248, 2834395104, 21669802880, 48444151296
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OFFSET
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1,1
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COMMENTS
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n is necessarily composite.
From the Math. Rev.: if both q=7*2^{r-2}+2^s-1 and p=49*2^{2r-s-4}+7*2^{r-2}-5*2^{r-s-2}-1 are prime for 1 <= s < r-2, then n=2^r*3*p*q is a solution to the equation phi(n)+sigma(n)=3n . [R. K. Guy]
If 7*2^n-1 is prime then m = 2^(n+2)*3*(7*2^n-1) is in the sequence. Because phi(m) = 2^(n+2)*(7*2^n-2); sigma(m) = 7*2^(n+2)*(2^(n+3)-1) so phi(m)+sigma(m) = 2^(n+2)*((7*2^n-2)+(7*2^(n+3)-7)) = 2^(n+2)*(63*2^n-9) = 3*(2^(n+2)*3*(7*2^n-1)) = 3*m. Hence A112729 = 2^(A001771+2)*3*(7*2^A001771-1) is a subsequence of this sequence. - Farideh Firoozbakht, Dec 01 2005
If both numbers p=7*2^n+2^k-1 & q=49*2^(2n-k)+2^(n-k)*(7*2^k-5)-1 are prime and m=2^(n+2)*3*p*q then phi(m)+sigma(m)=3*m. Namely m is in the sequence. - Farideh Firoozbakht, Jan 11 2007
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REFERENCES
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Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004. See Section B42, p. 151.
David Wells, Prime Numbers: The Most Mysterious Figures in Math, Hoboken, New Jersey, John Wiley & Sons (2005), 75.
Ming Zhi ZHANG, A note on the equation phi(n)+sigma(n)=3n, Sichuan Daxue Xuebao 37 (2000), no. 1, 39-40; MR1755990 (2001a:11009).
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LINKS
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MATHEMATICA
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Reap[ Do[ If[ EulerPhi[n] + DivisorSigma[1, n] == 3 n, Print[n]; Sow[n]], {n, 0, 10^8, 2}]][[2, 1]] (* Jean-François Alcover, Feb 16 2012 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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