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A267510
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Integers in A267509 such that (B0 + B1 + ... + Bm) is congruent to 0 mod m.
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0
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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