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A094358
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Squarefree products of factors of Fermat numbers (A023394).
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15
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1, 3, 5, 15, 17, 51, 85, 255, 257, 641, 771, 1285, 1923, 3205, 3855, 4369, 9615, 10897, 13107, 21845, 32691, 54485, 65535, 65537, 114689, 163455, 164737, 196611, 274177, 319489, 327685, 344067, 494211, 573445, 822531, 823685, 958467, 974849, 983055
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OFFSET
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1,2
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COMMENTS
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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).
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
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
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)).
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)
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LINKS
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Robert G. Wilson v, T. D. Noe and Ray Chandler, Table of n, a(n) for n = 1..3393 (Original 55 terms from Robert G. Wilson, extended to 1314 terms from T. D. Noe)
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EXAMPLE
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3 is a term because it is in A023394.
51 is a term because it is 3*17 and 17 is also in A023394.
153 = 3*3*17 is not a term because its factorization includes two 3's.
See the Munafo link for examples of the (conjectured) 2^^n == 1 (mod n) property.
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MATHEMATICA
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kmax = 10^6;
A023394 = Select[Prime[Range[kmax]], IntegerQ[Log[2, MultiplicativeOrder[2, #] ] ]&];
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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Example brought in line with name/description by Robert Munafo, May 18 2011
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STATUS
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approved
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