cp's OEIS Frontend

The terms 1 and 2 correspond to degenerate polygons.

These are also the numbers for which phi(n) is a power of 2: A209229(A000010(a(n))) = 1. - Olivier Gérard Feb 15 1999

From Stanislav Sykora, May 02 2016: (Start)

The sequence can be also defined as follows: (i) 1 is a member. (ii) Double of any member is also a member. (iii) If a member is not divisible by a Fermat prime F_k then its product with F_k is also a member. In particular, the powers of 2 (A000079) are a subset and so are the Fermat primes (A019434), which are the only odd prime members.

The definition is too restrictive (though correct): The Georg Mohr - Lorenzo Mascheroni theorem shows that constructibility using a straightedge and a compass is equivalent to using compass only. Moreover, Jean Victor Poncelet has shown that it is also equivalent to using straightedge and a fixed ('rusty') compass. With the work of Jakob Steiner, this became part of the Poncelet-Steiner theorem establishing the equivalence to using straightedge and a fixed circle (with a known center). A further extension by Francesco Severi replaced the availability of a circle with that of a fixed arc, no matter how small (but still with a known center).

Constructibility implies that when m is a member of this sequence, the edge length 2*sin(Pi/m) of an m-gon with circumradius 1 can be written as a finite expression involving only integer numbers, the four basic arithmetic operations, and the square root. (End)

If x,y are terms, and gcd(x,y) is a power of 2 then x*y is also a term. - David James Sycamore, Aug 24 2024

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

From Daniel Forgues, Jun 17 2011: (Start)

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.

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

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.

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

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)

Odd terms of A051913.

This sequence is infinite (unlike A004729), because it contains any A058383(n) times any power of 3.

A regular polygon of a(n) sides can be constructed if one also has an angle trisector.

If n is present, then 2n is present also, as shifting binary representation left never produces any carries.

Note: Start indexing from n=1 if you want just composite numbers. a(0)=1 is the only nonprime, noncomposite in this list.

The first term with three prime divisors is a(11) = 255 = 3*5*17.

The next terms with three prime divisors are

255, 3855, 13107, 21845, 24415, 28527, 30583, 31215, 31611, 31695, 32691, 48059, 56283, 56797, 61935, 65365, 87805, 98005, ...

Of these 24415 (= 5*19*257) is the first one with at least one prime factor that is not a Fermat prime (A019434).

The first term with four prime divisors is a(427) = 65535 = 3*5*17*257.

The first terms which are not multiples of any Fermat prime are: 511, 959, 3647, 4039, 4847, 5371, 7141, 7231, 7679, 7913, 8071, 9179, 12179, ... (511 = 7*73, 959 = 7*137, ...)

Similar to A004729, which allows each Fermat prime to occur 0 or 1 times. Applying Euler's phi function to these numbers produces numbers in A143513.

If the well-known conjecture that there are only five prime Fermat numbers F_k = 2^(2^k) + 1, k=0,1,2,3,4, is true, then we have exactly Sum_{n>=1} 1/a(n) = Product_{k=0..4} F_k/(F_k-1) = 4294967295/2147483648 = 1.9999999995343387126922607421875. - Vladimir Shevelev and T. D. Noe, Dec 01 2010

phi(x) is a power of 2 if and only if x is a power of 2 multiplied by a product of distinct Fermat primes. So if, as is conjectured, there are only 5 Fermat primes, then there are only 32 possibilities for the odd part of x, namely the divisors of 2^32-1, given in A004729.

The same numbers, in increasing order, are given in A003401.

The first entry in row n is the n-th divisor of 2^32-1 for 0 <= n <= 31 (A004729) and is 2^(n+1) for n >= 32. The last entry in row n is given in A058215.

The ratio of adjacent terms is 2 except for five terms (if there are exactly five Fermat primes). - T. D. Noe, Jun 21 2012