I am interested in the efficient computability of sequences.


Is it possible some ``interesting sequences'' be computed via addition formulae/semigroup operation?


Here is an artificial example.


Suppose one finds an associative operation $f : \mathbb{Z}^2 \times \mathbb{Z}^2 \to \mathbb{Z}^2$ and a sequence $a\_n$ such that:


$f([a\_{2n},a\_{2n+1}],[a\_{2m},a\_{2m+1}]) = [ a\_{2n+2m},a\_{2n+2m+1} ]$


$f$ is associative and the result is in [$a\_{2k},a\_{2k+1}$] which resembles a semigroup


Let $a\_n = A000069(n)$ where [A000069 Odious numbers: numbers with an odd number of 1's in their binary expansion.](http://oeis.org/A000069)


From the OEIS comment:
$a\_{2n+1} + a\_{2n} = A017101(n) = 8n+3$


one can find $n,m$.


From:
$a\_n = \frac{1}{2} (4n + 1 + (-1)^{A000120(n)})$


[A000120 1's-counting sequence: number of 1's in binary expansion of n](http://oeis.org/A000120)


one can find 


[ $a\_{2n+2m},a\_{2n+2m+1}$ ]


and this seems to complete associativity and closure.


1. So is [A000069(2n),A000069(2n+1)] $ \in \mathbb{Z}^2 $a semigroup?
2. If yes what type of semigroup is it (the semigroup operation involves counting ones in the binary expansion)?
3. Are there other similar examples/constructions of sequences that are not rational functions or related to multiples of points on curves (e.g. Fibonacci, Somos4).



Sequences, semigroups, addition formulae.
Is your functional equation a non-vacuous constraint? Since summation is group operation, the pairs [2n,2n+1] form a group 2N, which map to your group [a(2n),a(2n+1)]. For example, for A000069



```
1, 2, 4, 7, 8, 11, 13, 14, 16, 19, 21, 22, 25, 26, 28, 31, 32, 35, 37, 38, 41, 42, ...
0  1  2  3  4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21  
   ^  ^     ^           ^   ^           ^       ^   ^       ^           ^       ^   
----  ----  -----   -----   -----  ------  ------   -----   -----  ------  ------

```

each pair [2n,2n+1] is a group element is, so that we have 



```
[a(6),a(7)] + [a(10),a(11)] = [a(16),a(17)]

```

and yet there seems to be no relation among a(6), a(10) and a(16).


