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After 0 and 1, numbers n all of whose prime divisors encode an irreducible polynomial over GF(2) (are terms of A091206).
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%I #19 Feb 27 2020 20:51:30

%S 0,1,2,3,4,6,7,8,9,11,12,13,14,16,18,19,21,22,24,26,27,28,31,32,33,36,

%T 37,38,39,41,42,44,47,48,49,52,54,56,57,59,61,62,63,64,66,67,72,73,74,

%U 76,77,78,81,82,84,88,91,93,94,96,97,98,99,103,104,108,109

%N After 0 and 1, numbers n all of whose prime divisors encode an irreducible polynomial over GF(2) (are terms of A091206).

%C After 0 and 1, positive integers which are products of p * q * ... * r, where p, q, ..., r are terms of A091206.

%C Also fixed points of A236852(n). Proof: if k is a term of this sequence, the operation described in A236852 reduces to an identity operation. On the other hand, if k is not a term of this sequence, then it has at least one prime divisor which is reducible in polynomial ring GF(2)[X], which is thus "broken" by A236852 (A234742) to two or more separate factors (either prime or not), and because the original factor was prime, and N is a unique factorization domain, the new product computed over the new set of factors (with one or more "broken" pieces) cannot be equal to the original k. (Compare this to how primes are "broken" in a similar way in A235027, also A235145.)

%C Note: This sequence is not equal to all n for which A234741(n) = A236846(n). The first counterexample occurs at a(325) = 741 (= 3*13*19) for which we have: A236846(741) = 281 (= 3 x 247 = 3 x (13*19)) while A234741(741) = 329 (= 3 x 13 x 19). Contrast this with the behavior of the "dual sequence" A236850, where the corresponding property holds.

%H Antti Karttunen, <a href="/A236860/b236860.txt">Table of n, a(n) for n = 1..12696</a>

%o (Scheme, with _Antti Karttunen_'s IntSeq-library, two alternative implementations)

%o (define A236860 (MATCHING-POS 1 0 (lambda (n) (or (zero? n) (every (lambda (p) (= 1 (A091225 p))) (ifactor n))))))

%o (define A236860 (FIXED-POINTS 1 0 A236852))

%o (PARI) isp(p) = polisirreducible( Mod(1, 2) * Pol(binary(p))); \\ A091206

%o isok(n) = if ((n==0), 1 , my(f=factor(n)); prod(k=1, #f~, isp(f[k,1])) != 0); \\ _Michel Marcus_, Dec 22 2018

%Y Complement: A236848.

%Y Subsequence of A236842.

%Y Fixed points of A236852.

%Y Cf. A091206, A236850.

%K nonn

%O 1,3

%A _Antti Karttunen_, Mar 08 2014