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Python, 40 bytes f=lambda x:x*(-1)**x g=lambda x:3*f(x)+1 Try it online! Some outputs are floats that equal integers because (-1)**(-3) gives a float for instance. Based off ideas from Peter Tay …

Python 2, 96 bytes Try it online! to test the bijection. f=lambda l:len(l)and f(l[0])*2+1<<f(l[1:]) Takes void lists to non-negative integers. 42 bytes. g=lambda n:n*[g]and[g(n/(n&-n)/2)]+g(len …

Python 2, 42 bytes lambda s:(s.split("1"+len(s)/2*"0")*2)[:2] Try it online! The inverse is: def F(a,b): return a+"1"+len(a+b)*"0"+b A tempting idea for the inverse is to encode the pair …

Haskell, 38 bytes f(' ':a:b)=(a:)<$>f b f s=(drop 1s,"") Try it online! Not using built-in parsing. 21 bytes (doesn't quite work) last.(("",""):).reads Try it online! (As pointed out by …

Python, 22 bytes lambda s:s[0]+s[:0:-1] Try it online! Based on Mitchell Spector's idea. Leaves the first digit in place and reverses the rest. Input and output are as strings.

Haskell, 29 bytes (+1)?(2?) z?f=(iterate f z!!) Try it online! An Ackermann-like function. Starting the base numerical value at 2 makes for a simple base case. An annoying number of bytes are sp …

Haskell, 17 bytes [[1..x]|x<-[1..]] Try it online! Since the challenge seems to allow non-flat output, we can simply generate a list of the lists [1],[1,2],[1,2,3,],..., as suggested by @AZTECCO. H …

Python 2, 31 bytes R=1, while 1:print R;R+=len(R), Try it online! Thanks to @Danis for saving a byte here over R+=R[-1]+1,. This Prints: (1,) (1, 1) (1, 1, 2) (1, 1, 2, 3) (1, 1, 2, 3, 4) (1, 1, 2, …

Haskell, 18 bytes f(h:t)=h:reverse t Try it online! Based on Mitchell Spector's idea. Leaves the first digit in place and reverses the rest. Input and output are as strings.

Python 2, 14 bytes print{`str`:1} Outputs: {"<type 'str'>": 1} The backticks get the string representation in Python 2. Usually, this outputs inside creates single quotes, which Python recognize …

Haskell, 35 bytes h 1=[] h n=mod n 2:h(div n 2) h.(+1) Haskell doesn't have a binary built-in, so the (reversed) conversion is done manually. To remove the initial 1, the base case has 1 transform t …

Haskell, 27 bytes foldr1(\a b->2^a*(2*b+1)-1) Try it online! Uses a different bijection from the Cantor pairing function. Every positive integer can be uniquely split into a power of 2 times an odd …

Python 2, 38 bytes f=lambda a,*l:l and(a-~a<<f(*l))-1or a Try it online! Takes input splatted like f(1,2,3). Uses the pairing function \$p(a,b)=(2a+1)2^b\$. We use bit-shift <<b to shorten *2**b, an …

Python, 43 The inverse of Sp3000's function, implemented recursively. f=lambda n,k=1:n>k and k+f(n-k,k+1)or n%k+1 The function is a one-cycle followed by a two-cycle followed by a three-cycle, . …

Python, 20 bytes lambda n:bin(~n)[4:] Test: >> [bin(~n)[4:] for n in range(16)] ['', '0', '1', '00', '01', '10', '11', '000', '001', '010', '011', '100', '101', '110', '111', '0000'] Doing lam …