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Odd values of n for which a regular n-gon can be constructed by compass and straightedge.
21

%I #43 Aug 09 2024 01:56:20

%S 3,5,15,17,51,85,255,257,771,1285,3855,4369,13107,21845,65535,65537,

%T 196611,327685,983055,1114129,3342387,5570645,16711935,16843009,

%U 50529027,84215045,252645135,286331153,858993459,1431655765,4294967295

%N Odd values of n for which a regular n-gon can be constructed by compass and straightedge.

%C If there are no more Fermat primes, then 4294967295 is the last term in the sequence.

%C From _Daniel Forgues_, Jun 17 2011: (Start)

%C The 31 = 2^5 - 1 terms of this sequence are the nonempty products of distinct Fermat primes. The 5 known Fermat primes are in A019434.

%C Prepending the empty product, i.e., 1, to this sequence gives A004729.

%C The initial term for this sequence is thus a(1) (offset=1), since a(0) should correspond to the omitted empty product, term a(0) of A004729.

%C Rows 1 to 31 of SierpiƄski's triangle, if interpreted as a binary number converted to decimal (A001317), give a(1) to a(31). (End)

%D Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 73 at pp. 181-182.

%H Wilfrid Keller, <a href="http://www.prothsearch.com/fermat.html">Prime factors k.2^n + 1 of Fermat numbers F_m</a>.

%H OEIS Wiki, <a href="/wiki/Constructible_odd-sided_polygons">Constructible odd-sided polygons</a>.

%H OEIS Wiki, <a href="/wiki/Sierpinski&#39;s_triangle">Sierpinski's triangle</a>.

%F Each term is the product of distinct odd Fermat primes.

%F Sum_{n>=1} 1/a(n) = -1 + Product_{n>=1} (1+1/A019434(n)) = 0.7007354948... >= 1003212011/1431655765 = sigma(2^32-1)/(2^32-1) - 1, with equality if there are only five Fermat primes (A019434). - _Amiram Eldar_, Jan 22 2022

%t Union[Times@@@Rest[Subsets[{3,5,17,257,65537}]]] (* _Harvey P. Dale_, Sep 20 2011 *)

%Y Cf. A019434. Essentially same as A004729.

%Y Coincides with A001317 for the first 31 terms only. - _Robert G. Wilson v_, Dec 22 2001

%Y Cf. A053576.

%K hard,nonn,nice

%O 1,1

%A _Ken Takusagawa_