A094358 Squarefree products of factors of Fermat numbers (A023394).

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

Offset: 1

Author

Robert Munafo, Apr 26 2004

Keywords

• nonn

Comments

641 is the first member not in sequences A001317, A004729, etc.

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

From Jianing Song, Nov 11 2023: (Start)

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)

Examples

			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.
		

Links

• 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)
• Sourangshu Ghosh and Pranjal Jain, On Fermat Numbers and Munafo's Conjecture, (2021).
• R. Mestrovic, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012-2018. - From _N. J. A. Sloane_, Jun 13 2012
• Robert Munafo, Sequence A094358, 2^^A(N) = 1 mod N

Crossrefs

Cf. A023394, A014221, A092188, A001317, A004729.

Programs

Mathematica

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 *)

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

Extensions

Edited by T. D. Noe, Feb 02 2009

Example brought in line with name/description by Robert Munafo, May 18 2011