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A112730
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Numbers n such that the equation sigma(x)-phi(x)=n has at least one solution.
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1
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2, 5, 7, 10, 11, 14, 15, 16, 18, 20, 22, 23, 24, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 47, 48, 50, 52, 54, 56, 59, 60, 62, 63, 64, 66, 67, 68, 72, 73, 74, 75, 76, 78, 79, 80, 83, 84, 86, 87, 88, 90, 92, 94, 95, 96, 98, 100, 102, 104, 106, 107, 108, 110, 112
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OFFSET
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1,1
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COMMENTS
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It is conjectured that if n>2 then all solutions of the equation sigma(x)-phi(x)=n (*) are less than or equal to (n-1)^2/4 and 2 is the only number n such that the equation (*) has infinitely many solutions. In fact in the case n=2 prime numbers are all solutions of (*). All numbers of the form 2p+1 where p is prime are in the sequence because p^2 is a solution for the equation sigma(x)-phi(x)=2p+1. All numbers of the form 3*2^n-1 are in the sequence because 2^(n+1) is a solution for the equation sigma(x)-phi(x)=3*2^n-1 and etc.
The conjecture in the previous comment was established by Luke Pebody, see the Rivera link.
Theorem (Luke Pebody): If integers n>2 and m satisfy sigma(m)-phi(m)=n then m<=(n-1)^2/4.
Proof: Case I: m=1. Then n=sigma(m)-phi(m)=0 is not more than 2.
Case II: m is prime. Then n=sigma(m)-phi(m)=(m+1)-(m-1)=2 is not more than 2.
Case III: m has at least one nontrivial divisor. Let m=pq where 1<p<m. Then sigma(m) is the sum of all of the distinct factors of m. Thus sigma(m)>=m+p+1. Phi(m) is certainly no greater than the number of integers smaller than or equal to m that are not divisible by p. Thus phi(m)<=m-q. Thus n=sigma(m)-phi(m)>=p+q+1. Finally, the arithmetic mean of two numbers is always greater than their geometric mean, so Sqrt(m)=Sqrt(pq)<=(p+q)/2<=(n-1)/2. Squaring both sides, m<=(n-1)^2/4.
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LINKS
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EXAMPLE
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5 is in the sequence because 4 is a solution to the equation sigma(x)-phi(x)=5.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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