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%I #65 Nov 11 2023 13:55:05
%S 1,3,5,15,17,51,85,255,257,641,771,1285,1923,3205,3855,4369,9615,
%T 10897,13107,21845,32691,54485,65535,65537,114689,163455,164737,
%U 196611,274177,319489,327685,344067,494211,573445,822531,823685,958467,974849,983055
%N Squarefree products of factors of Fermat numbers (A023394).
%C 641 is the first member not in sequences A001317, A004729, etc.
%C Conjectured (by Munafo, see link) to be the same as: numbers n such that 2^^n == 1 mod n, where 2^^n is A014221(n).
%C It is clear from the observations by _Max Alekseyev_ in A023394 and the Chinese remainder theorem that any squarefree product b of divisors of Fermat numbers satisfies 2^(2^b) == 1 (mod b), hence satisfies Munafo's congruence above. The converse is true iff all Fermat numbers are squarefree. However, if nonsquarefree Fermat numbers exist, the criterion that is equivalent with Munafo's property would be "numbers b such that each prime power that divides b also divides some Fermat number". - _Jeppe Stig Nielsen_, Mar 05 2014
%C Also numbers b such that b is (squarefree and) a divisor of A051179(m) for some m. Or odd (squarefree) b where the multiplicative order of 2 mod b is a power of two. - _Jeppe Stig Nielsen_, Mar 07 2014
%C From _Jianing Song_, Nov 11 2023: (Start)
%C Also squarefree numbers k such that there exists i >= 1 such that k divides 2^^i - 1, where 2^^i = 2^2^...^2 (i times) = A014221(i): 2^^i == 1 (mod k) if and only if ord(2,k) divides 2^^(i-1) (ord(a,k) is the multiplicative order of a modulo k), so such i exists if and only if ord(2,k) is a power of 2. For such k, k divides 2^^i - 1 if and only if 2^^(i-2) >= log_2(ord(2,k)).
%C Note that 2^^(i-1) divides 2^^i implies that 2^^i - 1 divides 2^^(i+1) - 1, so this sequence is also squarefree numbers k such that k divides 2^^i - 1 for all sufficiently large i. (End)
%H Robert G. Wilson v, T. D. Noe and Ray Chandler, <a href="/A094358/b094358.txt">Table of n, a(n) for n = 1..3393</a> (Original 55 terms from Robert G. Wilson, extended to 1314 terms from T. D. Noe)
%H Sourangshu Ghosh and Pranjal Jain, <a href="https://www.researchgate.net/publication/350605122_On_Fermat_Numbers_and_Munafo%27s_Conjecture">On Fermat Numbers and Munafo's Conjecture</a>, (2021).
%H R. Mestrovic, <a href="http://arxiv.org/abs/1202.3670">Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof</a>, arXiv preprint arXiv:1202.3670 [math.HO], 2012-2018. - From _N. J. A. Sloane_, Jun 13 2012
%H Robert Munafo, <a href="http://mrob.com/pub/seq/a094358.html">Sequence A094358, 2^^A(N) = 1 mod N</a>
%e 3 is a term because it is in A023394.
%e 51 is a term because it is 3*17 and 17 is also in A023394.
%e 153 = 3*3*17 is not a term because its factorization includes two 3's.
%e See the Munafo link for examples of the (conjectured) 2^^n == 1 (mod n) property.
%t kmax = 10^6;
%t A023394 = Select[Prime[Range[kmax]], IntegerQ[Log[2, MultiplicativeOrder[2, #] ] ]&];
%t Reap[For[k = 1, k <= kmax, k++, ff = FactorInteger[k]; If[k == 1 || AllTrue[ff, MemberQ[A023394, #[[1]]] && #[[2]] == 1 &], Print[k]; Sow[k]]]][[2, 1]] (* _Jean-François Alcover_, Nov 03 2018 *)
%o (PARI) ( isOK1(n) = n%2==1 && hammingweight(znorder(Mod(2,n)))==1 ) ; ( isOK2(n) = issquarefree(n) && isOK1(n) ) \\ isOK1 and isOK2 differ only if n contains a squared prime that divides a Fermat number (none are known) \\ _Jeppe Stig Nielsen_, Apr 02 2014
%Y Cf. A023394, A014221, A092188, A001317, A004729.
%K nonn
%O 1,2
%A _Robert Munafo_, Apr 26 2004
%E Edited by _T. D. Noe_, Feb 02 2009
%E Example brought in line with name/description by _Robert Munafo_, May 18 2011
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