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Let us define generalized Franel numbers $f^{(m)}_n$ through recurrence relations: $f^{(1)}_n=1$ for all $n$, and $$f^{(m)}_n=\sum\limits_{k=0}^n\binom{n}{k}^3f^{(m-1)}_k.$$ In fact $$f^{(m)}_n=\sum\limits_{k_1,\ldots,k_m}\left(\frac{n!}{k_1!\cdots k_m!}\right)^3, \quad \quad \quad \quad \quad \quad \quad \quad (1)$$ where the sum is over all nonnegative $k_1,\ldots,k_m$, such that $k_1+\cdots+k_m=n$. $f^{(2)}_n$ are ordinary Franel numbers: $$f^{(2)}_n=\sum\limits_{k=0}^n\binom{n}{k}^3.$$ Numerical evidence suggests the following congruences (I have checked only for smal values of $n$, $m$ and $p$) $$f^{(m)}_n\equiv 0 \quad (\mathrm{mod}\;m)$$ and $$ \quad \quad \quad \quad \quad \quad f^{(m)}_p\equiv m \quad (\mathrm{mod}\;p^3)\quad \quad \quad \quad (2)$$ if $p$ is prime. How these congruences can be proved?

Note that (2) is a generalization of the Gary Detlefs conjecture (see $$ \quad \quad \quad \quad \quad \quad f^{(2)}_p\equiv 2 \quad (\mathrm{mod}\;p^3)\quad \quad \quad \quad (3)$$ if $p$ is prime. Was (3) ever proved?

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$f^{(2)}_0 = 1\not\equiv 0 (\textstyle{mod}) 2$. Z.W. Sun works on congruences for Franel numbers. At some stage he always invokes Zeilbergers algorithm. –  Uwe Stroinski Aug 7 '13 at 12:23
As $f_0^{(m)}=1$ for all $m$, it is assumed of course that $n\ge 1$. One of Sun's paper is here –  Zurab Silagadze Aug 7 '13 at 12:43

1 Answer 1

Let $p$ be a prime number. Then $p$ divides $\binom{p}{k}$ whenever $1\le k\le p-1$. It follows that $$f_p^{(m)}=\sum_{k=0}^p\binom{p}{k}^3 f_k^{(m-1)} \equiv 1+f_p^{(m-1)}\pmod{p^3},$$ and by induction $p^3$ divides $f_p^{(m)}-m$. Note that in the Detlefs' conjecture, the difficult part is the converse: "it happens only if $p$ is a prime."

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