%I
%S 12,18,20,24,40,56,88,104,180,196,224,234,240,360,368,420,464,540,600,
%T 650,780,992,1080,1344,1504,1872,1888,1890,1952,2016,2184,2352,2376,
%U 2688,3192,3276,3724,3744,4284,4320,4680
%N Composite numbers n such that sigma(n)/n leaves a remainder which divides n.
%C The numbers and the program were provided by _Charles R Greathouse IV_.
%C If n belongs to the sequence, then sigma(n)= d*n + rem, so sigma(n)/n = d + rem/n. Since rem is a divisor of n, n = rem*r, thus rem/n = 1/r. Then sigma(n)/n = d + 1/r and contfrac(sigma(n)/n) = [d, r], and length(contfrac(sigma(n)/n)) = 2. That is, A071862(n) = 2. [_Michel Marcus_, Aug 29 2012]
%H Donovan Johnson, <a href="/A215012/b215012.txt">Table of n, a(n) for n = 1..1000</a>
%e 24 has the divisors 1,2,3,4,6,12,24, which sum to be 60. Divide 60 by 24 and the remainder is 12, which is a divisor of 24.
%t a={}; For[n=1, n<=5000, n++, If[!PrimeQ[n], {s=DivisorSigma[1, n]; If[Mod[n, Mod[s,n]] == 0, AppendTo[a,n]]; }]; ]; a (* _John W. Layman_, Jul 31 2012] *)
%o (PARI) is(n)=my(t=sigma(n)%n);t && n%t==0 && !isprime(n)
%Y Cf. A000203, A071862.
%K nonn
%O 1,1
%A _J. M. Bergot_, Jul 31 2012
%E Terms a(24)a(41) from _John W. Layman_, Jul 31 2012
