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

2,1

Links

Table of n, a(n) for n=2..77.

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

Example

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

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

Prog

(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

Crossrefs

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

Sequence in context: A300708 A240755 A309806 * A051686 A079294 A242789

Adjacent sequences: A256167 A256168 A256169 * A256171 A256172 A256173

Keyword

nonn,tabf

Author

Franklin T. Adams-Watters, Jun 07 2015

Status

approved