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The Bell numbers $B(n)$ can be given as a sum of the (signed) Stirling numbers of the second kind $S(n,k)$ as $B(n)=\sum_{k=0}^nS(n,k)$. There are also the so-called complementary Bell numbers defined by $$B_1(n):=\sum_{k=0}^n(-1)^kS(n,k).$$

QUESTION. Given any prime $p$ and a positive integer $k$, is it true that $$B_1(n+p^k)\equiv B_1(n+1)-kB_1(n) \mod p?$$

EDIT. In the definition of $B_1(n)$ there was a typo: $(-1)^n$ was meant to be $(-1)^k$.

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    $\begingroup$ You have tested this for some range of values of $p$ and $k$? or you have some other reason for believing it? $\endgroup$
    – Gerry Myerson
    Commented Dec 14, 2016 at 2:25
  • $\begingroup$ Thanks for asking. This has been well-tested for many values of $p$ and $k$, so I'm a "believer" my evidence. :-) $\endgroup$
    – T. Amdeberhan
    Commented Dec 14, 2016 at 2:26
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    $\begingroup$ (You might also edit the title to make it more informative, e.g. "Conjecture: congruence for complementary Bell numbers".) $\endgroup$
    – Noam D. Elkies
    Commented Dec 14, 2016 at 3:34
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    $\begingroup$ Thanks, I altered the "catchy" title to a more straight-forward one. $\endgroup$
    – T. Amdeberhan
    Commented Dec 14, 2016 at 4:05

2 Answers 2

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The definition of the complementary Bell numbers should be $$B_1(n):=\sum_{k=0}^n(-1)^kS(n,k).$$

Define the polynomials $B_n(x)$ by $$B_n(x)=\sum_{k=0}^nx^kS(n,k),$$ so $B_1(n) = B_n(-1)$. These polynomials are called Bell polynomials or exponential polynomials. Christian Radoux proved the congruence $$ B_{n+p^k}(x) \equiv B_{n+1}(x) + (x^p+x^{p^2}+\cdots+x^{p^k})B_n(x) \pmod p,$$ where congruence is coefficientwise as polynomials in $x$. (See Anne Gertsch and Alain M. Robert, Some congruences concerning the Bell numbers. Bull. Belg. Math. Soc. Simon Stevin 3 (1996), no. 4, 467–475, for references and another proof.) The congruence for the complementary Bell numbers is the case $x=-1$. (Touchard's congruence is the case $x=1$.)

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  • $\begingroup$ Thank you for the reference. The results are interesting because I see that both Bell numbers can be treated uniformly. Up-voted. $\endgroup$
    – T. Amdeberhan
    Commented Dec 14, 2016 at 20:12
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Looks like you've rediscovered "Touchard's congruence" (the case $k=1$) and the generalization attributed by Wikipedia to Hurst and Schultz (arXiv:0906.0696).

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  • $\begingroup$ Looking more carefully, I see that Touchard's congruence is for the ordinary Bell numbers $B(n)$, not the complementary Bell numbers $B_1(n)$. But there's a good chance that an elementary proof of generalized Touchard for $B(n)$ can be adapted to give the corresponding congruence for $B_1(n)$. $\endgroup$
    – Noam D. Elkies
    Commented Dec 14, 2016 at 3:37
  • $\begingroup$ (1) Yes, the numbers and the results have close semblance but different; (2) I wish to see a more elegant proof, perhaps one that puts both numbers under the same umbrella. $\endgroup$
    – T. Amdeberhan
    Commented Dec 14, 2016 at 4:06
  • $\begingroup$ Now that the question has been answered properly I should probably make this into a comment and then delete my "answer" which turned out to answer a different question. $\endgroup$
    – Noam D. Elkies
    Commented Dec 15, 2016 at 16:07
  • $\begingroup$ Please do not delete. It's a good reference for people who wish to explore. Instead, edit your first line only, saying "a similar or complementary result can be found ..." $\endgroup$
    – T. Amdeberhan
    Commented Dec 15, 2016 at 16:48
  • $\begingroup$ The references (to Wikipedia and Hurst-Schultz) can still appear in the comment. I thought this would be better because the comment would be seen before Ira Gessel's answer (which refers to Touchard's congruence without explanation). $\endgroup$
    – Noam D. Elkies
    Commented Dec 15, 2016 at 18:15

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