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First acknowledging the helpful response to the earlier version of this sequence, I have found a complete expression for the sequence in closed form, and would be interested in improvements to the base 2 notation of the sum. Is there a standard form for such notation? Does it depend on the context? The sequence is:

$$S_n = \left(\frac{1}{n!}\right) \sum_{i_1i_2...i_{n-1} =\ 00...0}^{11...1} \frac{(n-1)!}{(n-1)^{i_1}(n-2)^{i_2},..., 2^{i_{n-2}}1^{i_{n-1}}}{\big[\ n^{i_1},(n-1)^{i_2},...,2^{i_{n-1}}\big]}$$

The $i_n$ to the right of the summation sign are digits of a base-$2$ number, and those same digits $(0,1)$ appear as exponents in the expression. The numeral $1$ in the denominator is an awkward convenience, and I'm sure there's a more elegant way of expressing this.

On the other hand (up to typos) it seems to work well. What I would really like is to put this in recursive form, because the list in brackets, multiplied together term by term, gives (half) the coefficients for $S_{n+1}$. So I think it lends itself to a form like $S_{n+1}= f(S_n)$.

The list in brackets, $\big[n^{1_1},...,\text{ etc.}\big]$, is a somewhat arbitrary property of each term $-$ its salient feature is providing a new coefficient for $S_{n+1}$. In terms of the earlier post, I have excluded from the lists the negations, ($n'$, etc.), which do not help with coefficients for $S_{n+1}$ in any event..

If someone knows the general family to which such a distribution belongs that would interest me. Also, I'd be interested in seeing a good Mathematica routine for this. Thanks.

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Is $[n^{i_1},\dots,2^{i_{n-2}}]$ a number? A tuple? Please define the notation in this question. – Greg Martin Nov 5 '11 at 23:31
It's a tuple, yes. – daniel Nov 5 '11 at 23:41
Example: n=3: S = (1/6)(2[1,1]+2[1,2]+1[3,1]+1[2,3]). – daniel Nov 5 '11 at 23:49
Here is my naive Mathematica. It generates a list of coefficients, and then a list of the objects (tuples) that go with each coefficient. The possible objects are the numbers 2,3,...,n, and so in this program, in the list of tuples, each "1" indicates the absence of one of the numbers. Each list of tuples, when multiplied together, give a coefficient in S_n+1. – daniel Nov 6 '11 at 3:56
Just a tiny tip on your Mathematica: there are things in there like While[] and Break[]; those would make for slightly neater code than what you currently have. :) – J. M. Nov 6 '11 at 4:34

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