Haskell, 34 bytes
data T=E|N T T
f=E:scanl N(N E E)f
f is the infinite list of Fibonacci trees, represented as a custom type defined in the first line (E is the empty tree, N l r is the tree with left subtree l and right subtree r).
How?
Haskell is usually very well-suited for Fibonacci-related tasks. Assume we have a recurrence of the form $$ \begin{cases} \texttt{f}_0=\texttt{a}\\ \texttt{f}_1=\texttt{b}\\ \texttt{f}_{n+2}=\texttt{g}(\texttt{f}_{n+1},\texttt{f}_n), \end{cases} $$ where \$\texttt{g}\$ is a binary operator. Then the corresponding Haskell definition is simply
f=a:scanl g b f
Unfortunately, Haskell is not very well-suited for problems involving trees, since it can't (natively) handle nested lists or tuples. Defining my own data type for trees was the shortest way I could find to solve this issue.
Haskell, 34 bytes
data T=E|N T T
f=E:scanl N(N E E)f
f is the infinite list of Fibonacci trees, represented as a custom type defined in the first line (E is the empty tree, N l r is the tree with left subtree l and right subtree r).
How?
Haskell is usually very well-suited for Fibonacci-related tasks. Assume we have a recurrence of the form $$ \begin{cases} \texttt{f}_0=\texttt{a}\\ \texttt{f}_1=\texttt{b}\\ \texttt{f}_{n+2}=\texttt{g}(\texttt{f}_{n+1},\texttt{f}_n), \end{cases} $$ where \$\texttt{g}\$ is a binary operator. Then the corresponding Haskell definition is simply
f=a:scanl g b f
Unfortunately, Haskell is not very well-suited for problems involving trees, since it can't (natively) handle nested lists or tuples. Defining my own data type for trees was the shortest way I could find to solve this issue.
Haskell, 2534 bytes
[n|n<-[1..],moddata T=E|N T T
f=E:scanl N(60^nN E E)n>0]f
The infinite list of non-Hamming numbers
How?
(Hopefully) a positive integer \$n\$ is Hamming if and only if \$n\$ does not divide \60ドル^n\$.Try it online!
Haskell, 25 bytes
[n|n<-[1..],mod(60^n)n>0]
The infinite list of non-Hamming numbers
How?
(Hopefully) a positive integer \$n\$ is Hamming if and only if \$n\$ does not divide \60ドル^n\$.