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A136539
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Numbers n such that n=6*phi(n)-sigma(n).
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6
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OFFSET
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1,1
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COMMENTS
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If 5*2^n-1 is prime (that is, n is in A001770) then m = 2^n*(5*2^n-1) is in the sequence. Proof: 6*phi(m)-sigma(m) = 6*2^(n-1)*(5*2^n-2) -(2^(n+1)-1)*5*2^n = 30*2^(2n-1)-6*2^n-5*2^(2n+1)+5*2^n = 5*2^(2n)-2^n = 2^n(5*2^n-1) = m.
The first seven terms of the sequence are of such form, with n=2, 4, 8, 10, 12, 14, 18. Are all terms of the sequence of this form?
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LINKS
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FORMULA
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EXAMPLE
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6*phi(76)-sigma(76)=6*36-140=76 so 76 is in the sequence.
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MATHEMATICA
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Do[If[n==6*EulerPhi[n]-DivisorSigma[1, n], Print[n]], {n, 85000000}]
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CROSSREFS
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KEYWORD
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more,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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