A256170 Irregular triangle where the n-th row contains the binary representations of the factors of the member of GF(2)[x] whose binary representation is n.

2, 3, 2, 2, 3, 3, 2, 3, 7, 2, 2, 2, 3, 7, 2, 3, 3, 11, 2, 2, 3, 13, 2, 7, 3, 3, 3, 2, 2, 2, 2, 3, 3, 3, 3, 2, 3, 7, 19, 2, 2, 3, 3, 7, 7, 2, 11, 3, 13, 2, 2, 2, 3, 25, 2, 13, 3, 3, 7, 2, 2, 7, 3, 11, 2, 3, 3, 3, 31, 2, 2, 2, 2, 2, 3, 31

Offset: 2

Author

Franklin T. Adams-Watters, Jun 07 2015

Keywords

• nonn
• tabf

Examples

			9 is 1001 in binary, so it corresponds to x^3 + 1 in GF(2)[x]. This factors as (x+1) * (x^2+x+1), which have binary representations 3 and 7; so row 9 is 3, 7.
The triangle starts:
[empty row for n=1]
2
3
2, 2,
3, 3
2, 3
7
2, 2, 2
3, 7
2, 3, 3
11
2, 2, 3
13
2, 7
3, 3, 3
		

Links

• Index entries for sequences operating on GF(2)[X]-polynomials

Crossrefs

Cf. A014580, A027746, A091222 (row lengths).

Programs

Maple

f:= proc(n)
  local L,P,R;
  L:= convert(n,base,2);
  P:= add(L[i]*X^(i-1),i=1..nops(L));
  R:= Factors(P) mod 2;
  op(sort([seq(eval(r[1],X=2)$r[2], r=R[2])]));
end proc:
seq(f(n), n=1..50); # Robert Israel, Jun 07 2015

PARI

arow(n)=my(fm=factor(Pol(binary(n))*Mod(1,2)),x=2,np,r,k);for(k=1,(np=#fm~),fm[k,1]=eval(lift(fm[k,1])));r=vector(sum(j=1,np,fm[j,2]));k=0;for(j=1,np,for(i=1,fm[j,2],r[k++]=fm[j,1]));r