A011775 Numbers k such that k divides phi(k) * sigma(k).

1, 6, 18, 24, 28, 40, 54, 72, 84, 96, 117, 120, 135, 162, 196, 200, 216, 224, 234, 252, 270, 288, 360, 384, 468, 486, 496, 540, 588, 600, 640, 648, 672, 756, 775, 819, 864, 891, 936, 1000, 1080, 1152, 1350, 1372, 1458, 1488, 1521, 1536, 1550, 1568, 1638, 1701, 1764

Offset

1,2

Comments

Comments from Farideh Firoozbakht, Dec 01 2005: (Start)

I. All numbers of the form 2^(4m-1)*5^n where m & n are natural numbers are in the sequence. Because if s=2^(4m-1)*5^n then phi(s)=2^(4m-2)*4*5^(n-1); sigma(s)=(2^(4m)-1)*(5^(n+1)-1)/4 so phi(s)*sigma(s)=6*((16^m-1)/15)*((5^(n+1)-1)/4)*(2^(4m-1)*5^n)= 6*((16^m-1)/15)*((5^(n+1)-1)/4)*s, note that (16^m-1)/15 and (5^(n+1)-1)/4 are integers, hence s divides phi(s)*sigma(s).

II. All numbers of the form 2^(2m-1)*3^n where m & n are natural numbers (A228104) are in the sequence. Because if s=2^(2m-1)*3^n then phi(s)=2^(2m-2)*2*3^(n-1); sigma(s)=(2^(2m)-1)*(3^(n+1)-1)/2 so phi(s)*sigma(s)=((3^(n+1)-1)/2)*((4^m-1)/3)*(2^(2m-1)*3^n) =((3^(n+1)-1)/2)*((4^m-1)/3)*s, note that ((3^(n+1)-1)/2 and (4^m-1)/3 are integers, hence s divides phi(s)*sigma(s).

So this sequence is infinite. Also it is obvious that perfect numbers (A000396) and multiply-perfect numbers(A007691) are subsequences of this sequence. (End)

Links

Donovan Johnson, Table of n, a(n) for n = 1..10000

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

Mathematica

Select[Range[1770], IntegerQ[DivisorSigma[1, # ]*EulerPhi[ # ]/# ] &] (* Farideh Firoozbakht, Dec 01 2005 *)

Prog

(PARI) is(n)=sigma(n)*eulerphi(n)%n==0 \\ Charles R Greathouse IV, Nov 27 2013

Crossrefs

Cf. A062354, A000396, A007691.

Sequence in context: A081318 A275168 A236359 * A015707 A236864 A101527

Adjacent sequences: A011772 A011773 A011774 * A011776 A011777 A011778

Keyword

nonn

Author

R. K. Guy

Extensions

Corrected and extended by David W. Wilson

Status

approved