The Lehmer-Comtet sequence

Question

The Lehmer-Comtet sequence is a sequence such that a(n) is the nth derivative of f(x) = x^x with respect to x as evaluated at x = 1.

Task

Take a non-negative integer as input and output the nth term of the Lehmer-Comtet sequence.

This is code-golf so you should minimize the file size of your source code.

Test Cases

OEIS 5727

Here are the first couple terms in order (copied from the OEIS)

1, 1, 2, 3, 8, 10, 54, -42, 944, -5112, 47160, -419760, 4297512, -47607144, 575023344, -7500202920, 105180931200, -1578296510400, 25238664189504, -428528786243904, 7700297625889920, -146004847062359040, 2913398154375730560, -61031188196889482880

Answer 1

Haskell, (削除) 77 (削除ここまで) 75 bytes, no differentiation builtins

x@(a:b)&y@(c:d)=a*c:zipWith(+)(b&y)(x&d)
s=1:s&(1:scanl(*)1[-1,-2..])
(s!!)

Try it online!

How it works

We represent a function as its infinite list of Taylor series coefficients about \$x = 1\$:

$$ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!} (x - 1)^n $$

is represented by \$[f(1), f'(1), f''(1), ...]\$.

The & operator multiplies two such functions using the product rule. This lets us recursively define the function \$s(x) = x^x\$ in terms of itself using the differential equation

$$ s(1) = 1, \\ s'(x) = s(x) \cdot (1 + \ln x), $$

where

$$ \ln x = \sum_{n=1}^\infty \frac{(-1)^{n-1}(n - 1)!}{n!}(x - 1)^n. $$

Answer 2

Mathematica, 19 bytes

D[x^x,{x,#-1}]/.x->1&

-18 bytes from @Not a tree

• Not a tree: Unless I'm missing something, you can get this a lot shorter: D[x^x,{x,#}]/.x->1&, 19 bytes.
• ZaMoC: actually 21 bytes.. but yes! a lot shorter!
• Not a tree: I don't think you need the -1 — the sequence from OEIS starts at n = 0.
• ZaMoC: ok then! 19 bytes it is

Answer 3

Octave with Symbolic Package, (削除) 36 (削除ここまで) 32 bytes

syms x
@(n)subs(diff(x^x,n),x,1)

The code defines an anonymous function which outputs a symbolic variable with the result.

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Answer 4

Haskell, 57 bytes

f 0=1
f n=f(n-1)-foldl(\a k->f(k-1)/(1-n/k)-a*k)0[1..n-1]

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No built-ins for differentiating or algebra. Outputs floats.

Answer 5

Python with SymPy, (削除) 77 (削除ここまで) (削除) 75 (削除ここまで) (削除) 58 (削除ここまで) 57 bytes

1 byte saved thanks to @notjagan

17 bytes saved thanks to @AndersKaseorg

from sympy import*
lambda n:diff('x^x','x',n).subs('x',1)

• Anders Kaseorg: lambda n:diff('x**x','x',10).subs('x',1) doesn’t require sympy.abc.
• Adalynn: Ummm ... where do you use n?
• Uriel: @ZacharyT thanks! coincidentally I tested anders' proposal right with n=10, so it gave the same result :) fixed now
• notjagan: -1 byte by replacing x**x with x^x.

Answer 6

SageMath, (削除) 33 (削除ここまで) 32 bytes

lambda n:diff(x^x,x,n).subs(x=1)

Try it on SageMathCell

Answer 7

Python 3, 150 bytes

lambda n:0**n or sum(L(n-1,r)for r in range(n))
L=lambda n,r:0<=r<=n and(0**n or n*L(n-2,r-1)+L(~-n,r-1)+(r-~-n)*L(~-n,r)if r else n<2or-~-n*L(n-1,0))

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Exponential runtime complexity. Uses the formula given in the OEIS page.

• Wheat Wizard: n>=r>=0 saves a byte.
• Wheat Wizard: You can also save a byte by putting 0**n after sum(...).
• Wheat Wizard: You can rearrange your addition to save a third byte.
• Wheat Wizard: Save two bytes by using bitwise or
• Wheat Wizard: Save two more bytes by using n<1 instead of 0**n

Answer 8

Python 3, (削除) 288 (削除ここまで) 261 bytes

Differentiation without differentiation built-in.

p=lambda a,n:lambda v:v and p(a*n,n-1)or a
l=lambda v:v and p(1,-1)
e=lambda v:v and m(e,a(p(1,0),l))or 1
a=lambda f,g:lambda v:v and a(f(1),g(1))or f(0)+g(0)
m=lambda f,g:lambda v:v and a(m(f(1),g),m(g(1),f))or f(0)*g(0)
L=lambda n,f=e:n and L(n-1,f(1))or f(0)

Try it online!

How it works

Each of the first five lines define functions and their derivatives and their results when evaluated at 1. Their derivatives are also functions.

• p is power i.e. a*x^n
• l is logarithm i.e. ln(x)
• e is exponential i.e. exp(x)
• a is addition i.e. f(x)+g(x)
• m is multiplication i.e. f(x)*g(x)

Usage: for example, exp(ln(x)+3x^2) would be represented as e(l()+p(3,2)). Let x=e(l()+p(3,2)). To find its derivative, call x(1). To find its result when evaluated at 1, call x(0).

Bonus: symbolic differentiation

• Wheat Wizard: You can save a lot of bytes by using exec compression. Try it online!

Answer 9

Python3+mpmath 52 bytes

from mpmath import*
lambda n:diff(lambda x:x**x,1,n)

-3 bytes, Thanks @Zachary T

• Wheat Wizard: You should change the language to python3+mpmath, since mpmath is not a standard library.
• Adalynn: You can change your first line to from mpmath import*, and the second to diff(lambda x:x**x,1,n). (just removing unnecessary spaces)

Answer 10

Pari/GP, 32 bytes

n->n!*Vec((y=1+x+O(x^n++))^y)[n]

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