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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)

a(n) can be also computed by replacing all consecutive runs of zeros in the binary expansion of n with * (multiplication sign), and then performing that multiplication, still in binary, after which the result is converted into decimal. See the example below.

Each subgroup {0,a,b,...} of nimber addition can be assigned an integer 1+2^a+2^b+...

These integers ordered by size give this sequence.

Without nimbers the sequence may be defined as follows:

The powerset af a set {0,...,n-1} with the symmetric difference as group operation forms the elementary abelian group (Z_2)^n.

The elements of the group can be numbered lexicographically from 0 to 2^n-1, with 0 representing the neutral element:

{}-->0 , {0}-->2^0=1 , {1}-->2^1=2 , {0,1}-->2^0+2^1=3 , ... , {0,...,n-1}-->2^n-1

So the subgroups of (Z_2)^n can be represented by subsets of {0,...,2^n-1}.

So each subgroup {0,a,b,...} of (Z_2)^n can be assigned an integer 1+2^a+2^b+...

For each (Z_2)^n there is a finite sequence of these numbers ordered by size, and it is the beginning of the finite sequence for (Z_2)^(n+1).

This leads to the infinite sequence:

* 1, (1 until here for (Z_2)^0)

* 3, (2 until here for (Z_2)^1)

* 5, 9, 15, (5 until here for (Z_2)^2)

* 17, 33, 51, 65, 85, 105, 129, 153, 165, 195, 255, (16 until here for (Z_2)^3)

* 257, 513, 771, 1025, 1285, 1545, 2049, 2313, 2565, 3075, 3855, 4097, 4369, 4641, 5185, 6273, 8193, 8481, 8721, 9345, 10305, 12291, 13107, 15555, 16385, 16705, 17025, 17425, 18465, 20485, 21845, 23205, 24585, 26265, 26985, 32769, 33153, 33345, 33825, 34833, 36873, 38505, 39321, 40965, 42405, 43605, 49155, 50115, 52275, 61455, 65535, (67 until here for (Z_2)^4)

* 65537, ...

The number of subgroups of (Z_2)^n is 1, 2, 5, 16, 67, 374, 2825, ... (A006116)

Comment from Tilman Piesk, Aug 27 2013: (Start)

Boolean functions correspond to integers, and belong to small equivalence classes (sec). So a sec can be seen as an infinite set of integers (represented in A227722 by the smallest one). Some secs contain only one odd integer. These unique odd integers, ordered by size, are shown in this sequence. (While the smallest integers from these secs are shown in A227963.)

(End)

Trivial fixed points are the powers of 2. How many nontrivial cases exist like 3, 15, 255, 65535: the first 5 terms of A051179. More?

83623935 is the next such term (see also A050474 and A203966). - Michel Marcus, Nov 13 2015

Conjecture: For n>=3, each term a(n), when considered as a GF(2)[X]-polynomial, is divisible by GF(2)[X] -polynomial (x + 1) ^ A000225(n-1) (= A051179(n-2)). If this holds, then for n>=3, a(n) = A048720bi(A136386(n),A048723bi(3,A000225(n-1))) = A048720bi(A136386(n),A051179(n-2)).

More generally, reading in base B >= 2: a(n) = (B^2^n - 1)/(B-1).

Recurrence: a(n) = a(n-1)*(B^K + 1) and a(0)=1 where K = floor(log_B a(n-1)) + 1.

B = 2 gives A051179; B = 3 gives A059918.

Two Boolean functions belong to the same small equivalence class (sec) when they can be expressed by each other by negating arguments. E.g., when f(p,~q,r) = g(p,q,r), then f and g belong to the same sec. Geometrically this means that the functions correspond to hypercubes with 2-colored vertices that are equivalent up to reflection (i.e., exchanging opposite hyperfaces).

Boolean functions correspond to integers, so each sec can be denoted by the smallest integer corresponding to one of its functions. There are A000231(n) small equivalence classes of n-ary Boolean functions. Ordered by size they form the finite sequence A_n. It is the beginning of A_(n+1) which leads to this infinite sequence A.

Two Boolean functions belong to the same big equivalence class (bec) when they can be expressed by each other by negating and permuting arguments. E.g., when f(~p,r,q) = g(p,q,r), then f and g belong to the same bec. Geometrically this means that the functions correspond to hypercubes with binarily colored vertices that are equivalent up to rotation and reflection.

Boolean functions correspond to integers, so each bec can be denoted by the smallest integer corresponding to one of its functions. There are A000616(n) big equivalence classes of n-ary Boolean functions. Ordered by size they form the finite sequence A_n. It is the beginning of A_(n+1), which leads to this infinite sequence A.

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

This sequence is obtained by applying Run Length Transform to the right-shifted version of the sequence A001317: 1, 3, 5, 15, 17, 51, 85, 255, 257, ...