%I #24 Jul 27 2024 09:29:16
%S 1,3,7,5,21,9,11,15,49,63,13,27,19,29,107,17,273,83,25,65,69,23,121,
%T 45,31,53,151,39,35,189,37,51,251,819,173,245,41,43,233,195,47,207,93,
%U 57,997,139,55,119,127,33,1911,95,79,441,59,105,367,101,61,455,67,111,475,85,1281,269,73,1365,81,503,457,287,87,123,1549,125,179,315
%N GF(2)[X] factorization prime shift towards larger terms.
%C Permutation of the odd numbers, A005408.
%C Let a x b stand for the carryless binary multiplication of positive integers a and b, that is, the result of operation A048720(a,b). With n having a unique factorization as A014580(i) x A014580(j) x ... x A014580(k), 1 <= i <= j <= ... <= k, a(n) = A014580(1+i) x A014580(1+j) x ... x A014580(1+k).
%H Antti Karttunen, <a href="/A305421/b305421.txt">Table of n, a(n) for n = 1..21845</a>
%H <a href="/index/Ge#GF2X">Index entries for sequences related to polynomials in ring GF(2)[X]</a>
%F For all n >= 1:
%F A305422(a(n)) = n.
%F A268389(a(n)) = A007814(n).
%F a(A000079(n)) = A001317(n).
%e For n = 12, which by its binary representation '1100' corresponds with (0,1)-polynomial x^3 + x^2, which over GF(2)[X] is factored as (x)(x)(x+1), i.e., 12 = A048720(2,A048720(2,3)) = A048720(A014580(1), A048720(A014580(1),A014580(2))), we then form a(12) as A048720(A014580(2), A048720(A014580(2),A014580(3))) = A048720(3,A048720(3,7)) = 27. Note that x, x+1 and x^2 + x + 1 are the three smallest irreducible (0,1)-polynomials when factored over GF(2)[X], and their binary representations 2, 3 and 7 are the three initial terms of A014580.
%o (PARI)
%o A091225(n) = polisirreducible(Pol(binary(n))*Mod(1, 2));
%o A305420(n) = { my(k=1+n); while(!A091225(k),k++); (k); };
%o A305421(n) = { my(f = subst(lift(factor(Pol(binary(n))*Mod(1, 2))),x,2)); for(i=1,#f~,f[i,1] = Pol(binary(A305420(f[i,1])))); fromdigits(Vec(factorback(f))%2,2); };
%Y Cf. A305422 (a left inverse).
%Y Cf. A001317, A014580, A091225, A193231, A268389, A305420, A305423.
%Y Cf. also A003961, A300841.
%K nonn
%O 1,2
%A _Antti Karttunen_, Jun 07 2018