$p$-adic-valuation of an expression involving Bernoulli numbers












4












$begingroup$


Let $p = 43, 67$ or $163$ (three primes such that $h(mathbb{Q}(sqrt{-p})) = 1$) and consider $k = (p+1)/2$. I'm interested in computing the $p$-adic valuation of the expression
begin{equation}
1+frac{l}{B_l},
end{equation}

where $l = k + m(p-1),~mgeq 0$. Note that, by the Kummer congruences and the fact that $h(-p) equiv -2B_k~(mathrm{mod}~p)$, we have
begin{equation}
v_p(1+l/B_l)geq 1.
end{equation}

In fact, I have observed by numerical computations that this inequality is an equality for many value of $m$. I have also observed that if $m$ is such that $2m+1 = dp$ for an integer $d$, then the valuation of $1+l/B_l$ seems to be equal to $2$.



These observations seem very mysterious for me, so I was wondering if anybody had an idea to explain this? Thanks !










share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Let $p = 43, 67$ or $163$ (three primes such that $h(mathbb{Q}(sqrt{-p})) = 1$) and consider $k = (p+1)/2$. I'm interested in computing the $p$-adic valuation of the expression
    begin{equation}
    1+frac{l}{B_l},
    end{equation}

    where $l = k + m(p-1),~mgeq 0$. Note that, by the Kummer congruences and the fact that $h(-p) equiv -2B_k~(mathrm{mod}~p)$, we have
    begin{equation}
    v_p(1+l/B_l)geq 1.
    end{equation}

    In fact, I have observed by numerical computations that this inequality is an equality for many value of $m$. I have also observed that if $m$ is such that $2m+1 = dp$ for an integer $d$, then the valuation of $1+l/B_l$ seems to be equal to $2$.



    These observations seem very mysterious for me, so I was wondering if anybody had an idea to explain this? Thanks !










    share|cite|improve this question











    $endgroup$















      4












      4








      4


      4



      $begingroup$


      Let $p = 43, 67$ or $163$ (three primes such that $h(mathbb{Q}(sqrt{-p})) = 1$) and consider $k = (p+1)/2$. I'm interested in computing the $p$-adic valuation of the expression
      begin{equation}
      1+frac{l}{B_l},
      end{equation}

      where $l = k + m(p-1),~mgeq 0$. Note that, by the Kummer congruences and the fact that $h(-p) equiv -2B_k~(mathrm{mod}~p)$, we have
      begin{equation}
      v_p(1+l/B_l)geq 1.
      end{equation}

      In fact, I have observed by numerical computations that this inequality is an equality for many value of $m$. I have also observed that if $m$ is such that $2m+1 = dp$ for an integer $d$, then the valuation of $1+l/B_l$ seems to be equal to $2$.



      These observations seem very mysterious for me, so I was wondering if anybody had an idea to explain this? Thanks !










      share|cite|improve this question











      $endgroup$




      Let $p = 43, 67$ or $163$ (three primes such that $h(mathbb{Q}(sqrt{-p})) = 1$) and consider $k = (p+1)/2$. I'm interested in computing the $p$-adic valuation of the expression
      begin{equation}
      1+frac{l}{B_l},
      end{equation}

      where $l = k + m(p-1),~mgeq 0$. Note that, by the Kummer congruences and the fact that $h(-p) equiv -2B_k~(mathrm{mod}~p)$, we have
      begin{equation}
      v_p(1+l/B_l)geq 1.
      end{equation}

      In fact, I have observed by numerical computations that this inequality is an equality for many value of $m$. I have also observed that if $m$ is such that $2m+1 = dp$ for an integer $d$, then the valuation of $1+l/B_l$ seems to be equal to $2$.



      These observations seem very mysterious for me, so I was wondering if anybody had an idea to explain this? Thanks !







      number-theory p-adic-number-theory bernoulli-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 30 at 17:16







      David Ayotte

















      asked Jan 28 at 16:51









      David AyotteDavid Ayotte

      213




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