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By the multinomial theorem we have that $$(a + b + c)^n = \sum_{0 \leq i \leq j \leq n} \frac{n!}{i! (j-i)! (n-j)!} a^i b^{j-i} c^{n-j},$$ and so $$\sum_{0 \leq i \leq j \leq n} \frac{1}{i! (j-i)! (n-j)!} a^i b^{j-i} c^{n-j} = \frac{1}{n!} (a+b+c)^n.$$ I am interested, however, in simplifying the summation given by $$\sum_{0 \leq i \leq j \leq n} \left(\frac{1}{i! (j-i)! (n-j)!}\right)^2 a^i b^{j-i} c^{n-j}. \quad \quad (\dagger)$$ Due to the squared coefficients it now becomes a bit complicated to write the summation in terms of $(a+b+c)^n$... Is there a way to simplify the summation $(\dagger)$ in terms of powers of $n$? This summation also reminds me of a "tensorised version" of the terms in the power expansion of a Bessel function. Do you see any possible simplifications? By any chance is there a name for this "multinomial like" expression with squared coefficients?

OtherQuestions
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  • Do you only want the sum of coefficients, or an expression for the whole (including the varibles)? – D S Mar 04 '24 at 14:47
  • @DS I'm looking for a simplification of the whole, including the variables – OtherQuestions Mar 04 '24 at 14:57
  • Any reason to believe this has a closed form? – D S Mar 04 '24 at 14:57
  • No reason :( I just want to see whether I can simplify that summation further. Actually, let $S_n$ denote the sum $(\dagger)$ above. Ultimately I am interested in the series $S:= \sum_{n=0}^\infty S_n$ and whether I can write $S$ in a simpler form, e.g. as a summation over $n\in \mathbb{N}$ (eliminating the summations over $i$ and $j$). – OtherQuestions Mar 04 '24 at 15:03
  • Your $S$ will contain terms like $a^2bc$. – D S Mar 04 '24 at 15:19
  • Indeed, $S_4$ has that term – OtherQuestions Mar 04 '24 at 15:22
  • First try finding the sum for $a=b=c=1$. Then you can move to arbitrary values of $a,b,c$. – D S Mar 04 '24 at 15:23
  • That will be the sum of the coefficients for all $n$... I'm not trying to compute the sum. I'm trying to rewrite it as a sum over $n$ with the summations over $i$ and $j$ suppressed. Surely at the expense of using some other functions, like Bessel functions. Thanks for having a look though. – OtherQuestions Mar 04 '24 at 15:26

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