A267510 Integers in A267509 such that (B0 + B1 + ... + Bm) is congruent to 0 mod m.

20, 22, 24, 26, 28, 30, 33, 36, 39, 40, 42, 44, 46, 48, 50, 55, 60, 62, 63, 64, 66, 68, 69, 70, 77, 80, 82, 84, 86, 88, 90, 93, 96, 99, 110, 121, 130, 132, 143, 150, 154, 156, 165, 169, 170, 176, 187, 190, 198, 200, 202, 204, 206, 208, 220, 222, 224, 226, 228, 231, 240

Offset: 1

Author

Abdul Gaffar Khan, Jan 16 2016

Keywords

• nonn
• base

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.

Examples

			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.
		

Crossrefs

Cf. A267509.

Programs

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 *)