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A011254
Numbers k such that phi(k) + sigma(k) = 4*k.
4
23760, 59400, 153720, 4563000, 45326160, 113315400, 402831360, 731601000, 803685120, 865950624, 919501200, 1178491680, 3504597120, 3786686400, 6429564000, 14924714400, 25310621952, 26998616736, 53138687040, 86955675840, 513969369984, 1054373308800, 1868445408960
OFFSET
1,1
COMMENTS
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. - Jens Kruse Andersen, Feb 17 2009
If k=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(k) = (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(k) + sigma(k)) = (5/2)*(4*k) = 5/2*(4*80*m) = 4*(200*m) so 200*m is in the sequence. - Farideh Firoozbakht, Mar 30 2009
REFERENCES
David Wells, Prime Numbers: The Most Mysterious Figures in Math, Hoboken, New Jersey, John Wiley & Sons (2005), p. 75.
Zhang Ming-Zhi (typescript submitted to Unsolved Problems section of Monthly, Oct 01 1996.
LINKS
Donovan Johnson, Table of n, a(n) for n = 1..25 (terms < 5*10^12)
Richard K. Guy, Divisors and desires, Amer. Math. Monthly, 104 (1997), 359-360.
Kelley Harris, On the classification of integers n that divide phi(n)+sigma(n), J. Num. Theory 129 (2009) 2093-2110
EXAMPLE
phi(23760) + sigma(23760) = 5760 + 89280 = 4*23760, so 23760 is in the sequence.
MATHEMATICA
Select[Range[1000000], DivisorSigma[1, #] + EulerPhi[#] == 4 # &] (* David Nacin, Feb 28 2012 *)
PROG
(PARI) is(n)=eulerphi(n)+sigma(n)==4*n \\ Charles R Greathouse IV, Nov 27 2013
CROSSREFS
KEYWORD
nonn,nice
EXTENSIONS
More terms from Jud McCranie
1178491680 from Farideh Firoozbakht, Jan 31 2006
2 more terms from Jud McCranie, 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, Feb 17 2009
a(21) from Donovan Johnson, Feb 28 2012
a(22)-a(23) from Donovan Johnson, Apr 04 2012
STATUS
approved