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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. 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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified December 12 22:06 EST 2019. Contains 329963 sequences. (Running on oeis4.)