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%I #17 Jan 20 2020 21:42:41
%S 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,17,
%T 26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,51,44,45,46,47,48,
%U 49,34,51,52,53,54,39,56,57,58,59,60,61,62,63,64,65,66,67
%N Remultiply n first "upward", from GF(2)[X] to N, and then remultiply that result back "downward", from N to GF(2)[X]: a(n) = A234741(A234742(n)).
%H Antti Karttunen, <a href="/A236851/b236851.txt">Table of n, a(n) for n = 0..8192</a>
%H <a href="/index/Ge#GF2X">Index entries for sequences operating on polynomials over GF(2)</a>
%F a(n) = A234741(A234742(n)).
%F To compute a(n): factor n as a polynomial over GF(2) (where n is mapped to such polynomials via the binary representation of n), that is, find first a unique multiset of terms i, j, ..., k from A014580 for which i x j x ... x k = n, where x stands for the carryless multiplication (A048720). Then divide from those i, j, ..., k the ones that are in A091214 (composite integers in N) to their constituent prime factors (in N), and multiply all back together (including the factors that are in A091206 and thus not changed) with the carryless multiplication (A048720).
%F Compare this to how primes are "broken" in a similar way in A235027 (cf. also A235145).
%e 5 ('101' in binary) = 3 x 3 (3 = '11' in binary). 3 is in A091206, so it stays intact, and 3 x 3 = 5, thus a(5)=5.
%e 25 ('11001' in binary) = 25 (25 is irreducible in GF(2)[X]). However, it divides as 5*5 in Z, so the result is 5 x 5 = 17, thus a(25)=17, 25 being the least n which is not fixed by this function.
%e 43 ('101011' in binary) = 3 x 25, of which the latter factor divides to 5*5, thus the result is 3 x 5 x 5 = 3 x 17 = 15 x 5 = 51.
%o (Scheme) (define (A236851 n) (A234741 (A234742 n)))
%Y A236850 gives the fixed points.
%Y Cf. A234741, A234742, A236380, A236852, A236836-A236837, A236846-A236847.
%K nonn
%O 0,3
%A _Antti Karttunen_, Feb 02 2014
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