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, 77 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!!)

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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

Answer 3

Octave with Symbolic Package, 36 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, 77 75 58 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)

Answer 6

SageMath, 33 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.

Answer 8

Python 3, 288 261 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)

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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

Answer 9

Python3+mpmath 52 bytes

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

-3 bytes, Thanks @Zachary T

Answer 10

Pari/GP, 32 bytes

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

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