

A256170


Irregular triangle where the nth row contains the binary representations of the factors of the member of GF(2)[x] whose binary representation is 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, 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
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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^(i1), 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. AdamsWatters, Jun 07 2015


STATUS

approved



