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A115747
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Numbers n such that phi(n) + sigma(n) = 5/2*n.
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1
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18, 20, 88, 368, 1504, 24448, 98048, 5238976, 25161728, 2730992944, 33995232256, 412316336128, 1391737114624, 7732492570624
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OFFSET
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1,1
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COMMENTS
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If p = 3*2^(m-1)-1 is an odd prime then 2^m*p is in the sequence because phi(2^m*p) = 2^(m-1)*(3*2^(m-1)-2), sigma(2^m*p) = (2^(m+1)-1)*(3*2^(m-1)) so phi(2^m*p)+sigma(2^m*p) = 2^(m-1)*(3* 2^(m-1)-2)+(2^(m+1)-1)*(3*2^(m-1)) = 3*2^(2m-2)-2^m+3*2^(2m)-3*2^ (m-1) = 2^(m-1)*(3*2^(m-1)-2+3*2^(m+1)-3) = 2^(m-1)*(3*5*2^(m-1)-5) = 5/2*2^m*(3*2^(m-1)-1) = 5/2*(2^m*p). Except 18 & 5238976 all known terms of the sequence are of the form 2^m*(3*2^(m-1)-1), where (3*2^(m-1)-1) is prime.
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LINKS
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EXAMPLE
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25161728 is in the sequence because phi(25161728) + sigma(25161728) = 12578816 + 50325504 = 5/2*25161728.
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MATHEMATICA
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Do[If[DivisorSigma[1, n]+EulerPhi[n]==5/2*n, Print[n]], {n, 200000000}]
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PROG
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(PARI) isok(n) = eulerphi(n) + sigma(n) == 5*n/2; \\ Michel Marcus, Jul 14 2015
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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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