A083264 - OEIS

%I #26 Jul 15 2018 12:23:34

%S 198,396,510,594,792,966,990,1020,1188,1386,1530,1566,1584,1782,1932,

%T 1980,2040,2178,2376,2550,2590,2772,2898,2970,3060,3132,3168,3198,

%U 3564,3570,3864,3960,4080,4158,4230,4356,4590,4698,4752,4830,4950,5100,5180

%N Numbers k such that the difference d of the largest and smallest prime factors of k is a composite divisor of k.

%C From _David A. Corneth_, Jul 14 2018: (Start)

%C No term k is a perfect power (or 1). If k is a perfect power then it's divisible by 0, a contradiction. Hence a term k has at least two prime factors.

%C All terms are even. Suppose a term k is odd. Then the smallest prime factor is > 2. Since k has at least two prime factors which are odd, the difference between the largest and smallest prime factor is even hence k is even. A contradiction, hence all terms are even.

%C All terms are of the form 2 * (p - 2) * p * m where p - 2 is composite, p is prime and m has all, if any, of its prime factors between 2 and p (inclusive). (End)

%H Harvey P. Dale, <a href="/A083264/b083264.txt">Table of n, a(n) for n = 1..1000</a>

%F Solutions to x mod d = 0 where d = A006530(x) - A020639(x) is composite.

%e 198 = 2*3*3*11 = 2*9*11 is in the sequence where d = 11 - 2 = 9 is composite.

%t ffi[x_] := Flatten[FactorInteger[x]]; ba[x_] := Table[Part[ffi[x], 2*w-1], {w, 1, lf[x]}]; lf[x_] := Length[FactorInteger[x]]; ma[x_] := Max[ba[x]]; mi[x_] := Min[ba[x]] Do[s=ma[ba[n]]-mi[ba[n]]; If[Mod[n, s]==0&&Greater[s, 2]&&!PrimeQ[s], Print[n]], {n, 1, 20000}]

%t dllpfQ[n_]:=Module[{c=Transpose[FactorInteger[n]][[1]],d},d=Last[c]-First[ c];If[d==0,d=1];Divisible[n,d]&&d>2&&CompositeQ[d]]; Select[ Range[ 6000],dllpfQ] (* _Harvey P. Dale_, Sep 26 2014 *)

%o (PARI) isok(n) = if (n>1, my(f=factor(n)[,1], d = vecmax(f) - vecmin(f)); (d > 1) && !isprime(d) && !(n % d)); \\ _Michel Marcus_, Jul 09 2018

%Y Cf. A025584, A033845, A071141, A006530, A020639, A083263.

%K nonn,easy

%O 1,1

%A _Labos Elemer_, May 12 2003

%E Name, Formula, and Example simplified by _Jon E. Schoenfield_, Jul 14 2018