OFFSET
1,1
COMMENTS
From David A. Corneth, Jul 14 2018: (Start)
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.
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.
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)
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
EXAMPLE
198 = 2*3*3*11 = 2*9*11 is in the sequence where d = 11 - 2 = 9 is composite.
MATHEMATICA
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}]
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Labos Elemer, May 12 2003
EXTENSIONS
Name, Formula, and Example simplified by Jon E. Schoenfield, Jul 14 2018
STATUS
approved