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A136416 Numbers n such that 1+(x+1)^k+(x+1)^n is a primitive polynomial over GF(2) for some k where 0

%I A136416 #12 Aug 30 2014 18:57:21
%S A136416 2,3,4,5,6,7,9,10,11,12,15,17,18,20,21,22,23,25,28,29,31,33,35,36,39,
%T A136416 41,42,47,49,52,54,55,57,58,60,62,63,65,66,68,71,73,74,76,79,81,84,86,
%U A136416 87,89,92,93,94,95,97,98,100,102,103,105,106,108,110,111,113,118,119,121
%N A136416 Numbers n such that 1+(x+1)^k+(x+1)^n is a primitive polynomial over GF(2) for some k where 0<k<n.
%C A136416 Iff the trinomial T(x)=1+x^k+x^n is irreducible (A073571) then the polynomial T(x+1)=1+(x+1)^k+(1+x)^n is irreducible.
%C A136416 The order of T(x+1) is in general different from the order of T(x).
%C A136416 So this sequence is different from A073726: for example, 1+(x+1)^7+(1+x)^10 is primitive but 1+(x+1)^3+(1+x)^10 is not (while 1+x^7+x^10 and 1+x^3+x^10 are mutual reciprocal and have the same order).
%H A136416 Joerg Arndt, Mar 31 2008, <a href="/A136416/b136416.txt">Table of n, a(n) for n = 1..211</a>
%H A136416 Joerg Arndt, <a href="http://www.jjj.de/fxt/#fxtbook">Matters Computational (The Fxtbook)</a>
%e A136416 10 is in the sequence because 1+(x+1)^7+(1+x)^10 is a primitive polynomial over GF(2).
%K A136416 nonn
%O A136416 1,1
%A A136416 _Joerg Arndt_, Mar 31 2008, May 02 2009