OFFSET
1,1
COMMENTS
If Bi is congruent to 0 mod m for all i=1,2,...,m then the integer n = (Bm,...,B1,B0) is a member of this sequence if and only if n is a member of A267509.
EXAMPLE
22 is a term, as f(x)=B0+B1x=2+2x=2*(1+x)=g(x)*h(x) with g(x)=2, h(x)=1+x, and neither g(x) nor h(x) is a unit in the ring of integers implies that f(x) is reducible over the ring of integers and 2+2=4=0 mod 1.
121 is a term, as f(x)=B0+B1x+B2x^2=1+2x+1x^2=1+2x+x^2=(1+x)*(1+x)=g(x)*h(x) with g(x)=1+x=h(x) and neither g(x) nor h(x) is a unit in the ring of integers implies that f(x) is reducible over the ring of integers and 1+2+1=4=0 mod 2.
MATHEMATICA
okQ[n_] := MatchQ[Factor[(id = IntegerDigits[n]).x^Range[lg = Length[id] - 1, 0, -1]][[0]], Times | Power] && Divisible[Total[id], lg]; Select[ Range[240], okQ] (* Jean-François Alcover, Feb 01 2016 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Abdul Gaffar Khan, Jan 16 2016
STATUS
approved