Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 http://oeis.org/A000172) $$ \quad \quad \quad \quad \quad \quad f^{(2)}_p\equiv 2 \quad (\mathrm{mod}\;p^3)\quad \quad \quad \quad (3)$$ if $p$ is prime. Was (3) ever proved?

share|improve this question
1  
$f^{(2)}_0 = 1\not\equiv 0 (\textstyle{mod}) 2$. 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 arxiv.org/abs/1112.1034 –  Zurab Silagadze Aug 7 '13 at 12:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.