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A099650
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Solutions to x+phi[x]=sigma[x]/2.
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0
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OFFSET
| 1,1
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COMMENTS
| If 5*2^n-1 is prime then m=3*2^(n+1)*(5*2^n-1) is in the sequence because m+phi(m)=2^(n+1)*3*(5*2^n-1)+2^(n+1)*(5*2^n-2)=2^(n+1) *(20*2^n-5)=2^(n+1)*5*(2^(n+2)-1)=1/2*4*(2^(n+2)-1)*(5*2^n)= 1/2*sigma(3)*sigma(2^(n+1))*sigma(5*2^n-1)=1/2*sigma(3*2^(n+1) *(5*2^n-1))=1/2*sigma(m). So 3*2^(A001770+1)*(5*2^A001770-1) is a subsequence of this sequence. A110084 is this subsequence. Next term is greater than 10^8. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Aug 04 2005
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EXAMPLE
| n=456: phi[456]=144, sigma[456]=1200.
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MATHEMATICA
| Do[If[DivisorSigma[1, m] == 2m + 2 EulerPhi[m], Print[m]], {m, 100000000}] (Firoozbakht)
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CROSSREFS
| Cf. A000010, A000203.
Cf. A001770, A110084.
Sequence in context: A145528 A203058 A116331 * A200784 A077578 A048110
Adjacent sequences: A099647 A099648 A099649 * A099651 A099652 A099653
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KEYWORD
| nonn
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AUTHOR
| Labos E. (labos(AT)ana.sote.hu), Nov 05 2004
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EXTENSIONS
| Two more terms from Farideh Firoozbakht (mymontain(AT)yahoo.com), Aug 04 2005
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