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A215012 Composite numbers n such that sigma(n)/n leaves a remainder which divides n.

%I A215012 #28 Oct 28 2024 02:01:01
%S A215012 12,18,20,24,40,56,88,104,180,196,224,234,240,360,368,420,464,540,600,
%T A215012 650,780,992,1080,1344,1504,1872,1888,1890,1952,2016,2184,2352,2376,
%U A215012 2688,3192,3276,3724,3744,4284,4320,4680
%N A215012 Composite numbers n such that sigma(n)/n leaves a remainder which divides n.
%C A215012 The numbers and the program were provided by _Charles R Greathouse IV_.
%C A215012 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 A215012 Donovan Johnson, <a href="/A215012/b215012.txt">Table of n, a(n) for n = 1..1000</a>
%e A215012 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 A215012 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 *)
%t A215012 Select[Range[5000],CompositeQ[#]&&Mod[#,Mod[DivisorSigma[1,#],#]]==0&] // Quiet (* _Harvey P. Dale_, May 24 2019 *)
%o A215012 (PARI) is(n)=my(t=sigma(n)%n);t && n%t==0 && !isprime(n)
%Y A215012 Cf. A000203, A071862.
%K A215012 nonn
%O A215012 1,1
%A A215012 _J. M. Bergot_, Jul 31 2012
%E A215012 Terms a(24)-a(41) from _John W. Layman_, Jul 31 2012