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Find the number of arrangements of $k \mbox{ }1'$s, $k \mbox{ }2'$s, $\cdots, k \mbox{ }n'$s - total $kn$ cards - so that no same numbers appear consecutively. For $k=2$ we can compute it by using the PIE, and it is $$\frac{1}{2^n} \sum_{i=0}^n (-1)^i \binom{n}{i} (2n-i)! 2^i$$

Pedro
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hkju
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  • We have seen similar problems here before: http://math.stackexchange.com/questions/76213/how-many-arrangements-of-a-2b-3c-4d-5e-have-no-identical-consecutive-lett My impression is that there is no general formula for the solution, but I would love to be proven wrong! –  Apr 09 '12 at 02:30
  • quite different problem! – hkju Apr 09 '12 at 03:57
  • Not so different. –  Apr 09 '12 at 04:00
  • the problem you mentioned deals with the different number of cards, but here we consider the same number of cards. – hkju Apr 09 '12 at 06:41
  • The cases $k=2$ and $k=3$ are tabulated here: https://oeis.org/A114938 and https://oeis.org/A193638 . For $k=3$ there is no simple formula given. –  Apr 09 '12 at 14:17
  • @Byron Thanks for the compliments. I'm not sure I understand your bounty - what would merit the award? – Jair Taylor Apr 11 '12 at 02:30
  • Hmm, I have not previously worked with Laguerre polynomials, but it seems that they have a simple expansion in terms of my $q_k$ polynomials: $L_k(x) = \sum_{i=0}^k (-1)^i q_i(x)$! That's rather nice. – Jair Taylor Apr 11 '12 at 02:39
  • @Jair Sorry, I should have explained. I will simply award you the bounty, but I have to wait 24 hours. I will give it to you sometime tonight. –  Apr 11 '12 at 12:11
  • Ah, yes - I figured that out after posting. Thanks! – Jair Taylor Apr 11 '12 at 14:43

1 Answers1

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I believe the answer is given by $$\int_0^\infty e^{-x} q_k(x)^n \, dx$$ where $q_k(x) = \sum_{i=1}^k \frac{(-1)^{i-k}}{i!} {k-1 \choose i-1}x^i$ for $k\geq 1$, and $q_0(x) = 1$. In general if we allow $k_i$ of the $i$th number the answer should be $$\int_0^\infty e^{-x} \prod_i q_{k_i}(x) \, dx$$

You can check that this agrees with the sequences oeis.org/A114938 and oeis.org/A193638 above. I do not (quite) have a proof of this, although I'm very close. The method is my own, and has not been published anywhere as far as I know. I'd be happy to give you more information in private, but I'm not sure I want to expose it publicly until it's proven. Please let me know if you think this is noteworthy and any potential applications.

Edit: Following some information given to me by Byron, I found that this formula is already known and that in fact $q_n(x) = (-1)^{n}L_n^{(-1)}(x)$ where $L_n^{(\alpha)} (x) $ denotes the generalized Laguerre polynomial. See Section 6 here for a labelled version. I should have mentioned this sooner; thanks Byron!

Alex Riley
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Jair Taylor
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  • I must be missing something. When $k=2$ and $n=1$, the correct answer is zero, but that's not what your formula gives. –  Apr 09 '12 at 22:36
  • Ahhh, sorry! I forgot a factor of $(-1)^{i-k}$ in the expression for $q_k$. I will fix it. – Jair Taylor Apr 10 '12 at 01:46
  • You can check that $q_2(x) = -x + \frac{x^2}{2}$ and then $\int_0^\infty e^{-x} ( -x + \frac{x^2}{2}) , dx = 0$ as desired. – Jair Taylor Apr 10 '12 at 01:48
  • Here is some numerical evidence in the form of a Sage worksheet - it reproduces oeis.org/A114938. See http://sagenb.org/home/pub/4677/ – Jair Taylor Apr 10 '12 at 01:56
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    Well, I'm convinced! That is a wonderful formula. –  Apr 10 '12 at 01:59
  • Thanks! Now I just need to figure out something I can do with it. :) – Jair Taylor Apr 10 '12 at 18:11
  • @jair Thanks for the answer. By the way, is there any recursion formula for $q_k (x)$'s? – hkju Apr 11 '12 at 00:21
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    Yes - I don't know if this helps, but $q_k(x) = \int_0^x q_{k-1}(t), dt - q_{k-1}(x)$. Also, $q_k(x)$ is the inverse Laplace of $(1-x)^{k-1}/x^{k+1}$. Can I ask why this question came up? – Jair Taylor Apr 11 '12 at 01:14
  • One of my students asked me several days ago. No other reasons! Your previous comments are greatly helpful. – hkju Apr 11 '12 at 01:23