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%I #20 Nov 25 2015 23:15:14
%S 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,
%T 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,
%U 2,3,3,3,31,2,2,2,2,2,3,31
%N 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.
%H <a href="/index/Ge#GF2X">Index entries for sequences operating on GF(2)[X]-polynomials</a>
%e 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.
%e The triangle starts:
%e [empty row for n=1]
%e 2
%e 3
%e 2, 2,
%e 3, 3
%e 2, 3
%e 7
%e 2, 2, 2
%e 3, 7
%e 2, 3, 3
%e 11
%e 2, 2, 3
%e 13
%e 2, 7
%e 3, 3, 3
%p f:= proc(n)
%p local L,P,R;
%p L:= convert(n,base,2);
%p P:= add(L[i]*X^(i-1),i=1..nops(L));
%p R:= Factors(P) mod 2;
%p op(sort([seq(eval(r[1],X=2)$r[2], r=R[2])]));
%p end proc:
%p seq(f(n), n=1..50); # _Robert Israel_, Jun 07 2015
%o (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
%Y Cf. A014580, A027746, A091222 (row lengths).
%K nonn,tabf
%O 2,1
%A _Franklin T. Adams-Watters_, Jun 07 2015
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