1,1

If (sigma(m)-phi(m))/(4*m-sigma(m)-phi(m)) is a prime integer p not dividing m, then p*m is in the sequence. 135230346701100 is in the sequence and not divisible by 24. [From Jens Kruse Andersen (jens.k.a(AT)get2net.dk), Feb 17 2009]

Contribution from Farideh Firoozbakht (mymontain(AT)yahoo.com), Mar 30 2009: (Start)

If n=80*m is in the sequence and gcd(m,10)=1 then 200*m is also in the sequence.

proof: phi(200*m)+sigma(200*m)=phi(200)*phi(m)+sigma(200)*sigma(m)=80*phi(m)+

465*sigma(n)=5/2*(32*phi(m)+186*sigma(m))=5/2*(phi(80)*phi(m)+sigma(80)*sigma(m))=

5/2*(phi(80*m)+sigma(80*m))=5/2*(phi(n)+sigma(n))=5/2*(4*n)=5/2*(4*80*m)=4*(200*m)

so 200*m is in the sequence. (End)

R. K. Guy, Divisors and desires, Amer. Math. Monthly, 104 (1997), 359-360.

Zhang Ming-Zhi (typescript submitted to Unsolved Problems section of Monthly, Oct 01 1996.

phi(23760)+sigma(23760)=5760+89280=4*23760, so 23760 is in the sequence.

Cf. A000010, A000203, A011251, A011774, A015704.

Sequence in context: A031847 A069334 A118061 this_sequence A066234 A084691 A084692

Adjacent sequences: A011251 A011252 A011253 this_sequence A011255 A011256 A011257

nonn,nice

N. J. A. Sloane (njas(AT)research.att.com).

More terms from Jud McCranie (j.mccranie(AT)comcast.net)

1178491680 from Farideh Firoozbakht, Jan 31 2006

2 more terms from Jud McCranie (j.mccranie(AT)comcast.net), Jan 31 2006

24 divides all known terms of the sequence. If this is true for the next five terms then they are 6429564000, 14924714400, 25310621952, 26998616736 and 53138687040. - Farideh Firoozbakht, Mar 11 2006

More terms from Jens Kruse Andersen (jens.k.a(AT)get2net.dk), Feb 17 2009