The problem:

Count the number of permutations of n distinct objects that leave none of them fixed.

This is actually a well-studied problem with pretty well-established terminology (derangements). Mathematica even has a built-in function for it, Subfactorial[n], which it is perfectly capable of producing if you feed it a simple enough recurrence:

RSolve[{a[n] == (n - 1) (a[n - 1] + a[n - 2]),
        a[0] == 1, a[1] == 0}, a[n], n] // FullSimplify
(* output: {{a[n] -> Subfactorial[n]}} *)

But when I attacked the problem, I was unable to come up with this simple recurrence, and only produced recurrences that RSolve couldn't handle.

Your mission:

Demonstrate how one could successfully use the analytical features (simplification, solving, reduction, etc.) of Mathematica to obtain the closed-form solution Subfactorial[n], beginning from one of my recurrences below.

(note: the intent of the last restriction above is to weed out responses that merely derive the simpler formula directly from the problem. It is okay for a solution to not meet this restriction, so long as it doesn't require me to have a Jimmy Neutron-style "brain blast!" to see it)

I do not mind if a bit of guessing, trial and error, or manipulation-by-hand is required. I am merely seeking techniques applicable to this problem which may also be applicable to others.

My first attempt: Count ways to partition into cycles of size $> 1$.

Suppose that you have $N$ objects whose fates have yet to be decided. From these, pick elements to form $m$ cycles ($m\ge1$) of size $b>1$, dividing by $m!$ to account for the order in which the cycles can be chosen. To ensure the same $b$ is never considered twice, a second parameter $b_\text{max}$ is used to ensure that $b$ strictly decreases between recursive evalutaions.

$$ \text{RSolve}\left[\left\{a[n,~b_\text{max}]=\sum _{b=2}^{b_\text{max}} \sum _{m=1}^{\left\lfloor \frac{n}{b}\right\rfloor } \frac{n! a[n-b m,~b-1]}{m! b^m (n-b m)!},~a[0,~b]=1\right\},~a[j,~k],~\{j,~k\}\right] $$

(* % // InputForm *)
RSolve[{a[n, bmax] == Sum[(a[-(b*m) + n, -1 + b]*n!)/(b^m*m!*(-(b*m) + n)!), 
{b, 2, bmax}, {m, 1, Floor[n/b]}], a[0, b] == 1}, a[j, k], {j, k}]

Unfortunately, this is not even a valid way to invoke RSolve (it will merely spit back out what you typed); RSolve doesn't like multivariable recurrences.

My second attempt: A single valued recurrence.

I tried another approach using the inclusion-exclusion principle, which eventually turned into this:

To count the ways to arrange $N$ objects with no fixed points, subtract from $N!$ the number of arrangements with at least one fixed point. For the latter, simply pick $m\ge1$ fixed points, and then the remaining $N-m$ have no fixed points (giving a recurrence):

$$ \text{RSolve}\left[\left\{f[n]=n!-\sum _{m=1}^n \binom{n}{m} f[n-m],~f[0]=1\right\},~f[n],~n\right] $$

RSolve[{f[n] == n! - Sum[Binomial[n, m]*f[-m + n], {m, 1, n}], f[0] == 1}, f[n], n]

But again, Mathematica just spits the same form back out.

Too long, didn't read, what's the question?: Ctrl+f this page for "your mission".

  • $\begingroup$ Its strange on my computer I have no problem to obtain a solution in terms of BesselK functions a[n] -> (BesselI[n, -2] BesselK[1, 2] + BesselI[1, 2] BesselK[n, 2])/( BesselI[1, 2] BesselK[0, 2] + BesselI[0, 2] BesselK[1, 2]). What is your Mathematica version ? $\endgroup$ Sep 19, 2016 at 4:06
  • $\begingroup$ @cyrille.piatecki my version is 10.4 $\endgroup$
    – Exp HP
    Sep 19, 2016 at 14:44

1 Answer 1


Not precisely solving the recurrences from the question, but a different approach to getting the formula from the very definition.

Make a condition such that elem is on a kth position; will be used to exclude such a situation:

cond[k_, elem_] := Table[_, k - 1]~Join~{elem}~Join~{___}
a[n_] := Alternatives @@ Table[cond[i, i], {i, 1, n}] /; n > 1
a[1] := {_}

Let's take for example lists of length up to max = 10:

seq = Length /@ Table[DeleteCases[Permutations[Range[n]], a[n]], {n, 1, max}]

{0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961}

This is the same as Subfactorial:

Subfactorial /@ Range[max]

{0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961}

One can use FindSequenceFunction to guess what's the origin of sequence seq:



In fact, max = 4 is sufficient for FindSequenceFunction to return the correct solution.

EDIT: I'd like to emphasize that FindSequenceFunction is a guess on what the underlying formula might be. As an illustrative example, consider a polynomial

f[x_] := -7 + 12 x - 6 x^2 + x^3

Its first 4 values are

f /@ Range[4]

{0, 1, 2, 9}

which happen to coincide with Subfactorial. On the other hand, f /@ Range[5] is {0, 1, 2, 9, 28}, and applying FindSequenceFunction yields (correctly) f[x]. Hence, FindSequenceFunction, and a somewhat related FindFormula, should be used with caution and treated more as hints rather than a definite solution.

  • $\begingroup$ Huh, never would've expected there to be a feature that can guess analytical solutions from brute force values! I have clarified the "from my recurrences" criterion, because this does meet the rule in spirit. $\endgroup$
    – Exp HP
    Sep 19, 2016 at 15:08

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