Very small constants from poly-bernoulli

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If we define
$$a_n(m)=sumlimits_k=0^nk!nbrace k(k+1)^m(-1)^n-k$$
$$prodlimits_k=2^n+11-kx=sumlimits_k=0^nt(n,k)x^k$$
for $n>0$, $mgeqslant0$, so
$$sumlimits_k=0^nt(n,k)a_n(m-k)=0$$
for $n>0$, $|m|geqslant0$, ex.
$$a_1(m)-2a_1(m-1)=0$$
$$a_2(m)-5a_2(m-1)+6a_2(m-2)=0$$
Another words
$$a_1(m)=fraca_1(m+1)2$$
$$a_2(m)=frac5a_2(m+1)-a_2(m+2)6$$
Then if
$$f(n)=sumlimits_k=1^inftya_n(-k)$$
so $f(1)=1$, $f(2)=0$.



For $n>2$ we have very small constants with rational ratio, ex.
$$g(n)=fracf(n)f(3)$$
$$g(4)=frac454, g(5)=frac254, g(6)=-frac67938, g(7)=-frac43398, g(8)=frac6691994,$$
$$g(9)=13430, g(10)=-frac11492150669232, g(11)=frac25700873197041316$$
Also, surprisingly, it looks like $f(3)=2^-55$.



Is there a way to identify them without precisely calculation?










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    up vote
    0
    down vote

    favorite












    If we define
    $$a_n(m)=sumlimits_k=0^nk!nbrace k(k+1)^m(-1)^n-k$$
    $$prodlimits_k=2^n+11-kx=sumlimits_k=0^nt(n,k)x^k$$
    for $n>0$, $mgeqslant0$, so
    $$sumlimits_k=0^nt(n,k)a_n(m-k)=0$$
    for $n>0$, $|m|geqslant0$, ex.
    $$a_1(m)-2a_1(m-1)=0$$
    $$a_2(m)-5a_2(m-1)+6a_2(m-2)=0$$
    Another words
    $$a_1(m)=fraca_1(m+1)2$$
    $$a_2(m)=frac5a_2(m+1)-a_2(m+2)6$$
    Then if
    $$f(n)=sumlimits_k=1^inftya_n(-k)$$
    so $f(1)=1$, $f(2)=0$.



    For $n>2$ we have very small constants with rational ratio, ex.
    $$g(n)=fracf(n)f(3)$$
    $$g(4)=frac454, g(5)=frac254, g(6)=-frac67938, g(7)=-frac43398, g(8)=frac6691994,$$
    $$g(9)=13430, g(10)=-frac11492150669232, g(11)=frac25700873197041316$$
    Also, surprisingly, it looks like $f(3)=2^-55$.



    Is there a way to identify them without precisely calculation?










    share|cite|improve this question

























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      If we define
      $$a_n(m)=sumlimits_k=0^nk!nbrace k(k+1)^m(-1)^n-k$$
      $$prodlimits_k=2^n+11-kx=sumlimits_k=0^nt(n,k)x^k$$
      for $n>0$, $mgeqslant0$, so
      $$sumlimits_k=0^nt(n,k)a_n(m-k)=0$$
      for $n>0$, $|m|geqslant0$, ex.
      $$a_1(m)-2a_1(m-1)=0$$
      $$a_2(m)-5a_2(m-1)+6a_2(m-2)=0$$
      Another words
      $$a_1(m)=fraca_1(m+1)2$$
      $$a_2(m)=frac5a_2(m+1)-a_2(m+2)6$$
      Then if
      $$f(n)=sumlimits_k=1^inftya_n(-k)$$
      so $f(1)=1$, $f(2)=0$.



      For $n>2$ we have very small constants with rational ratio, ex.
      $$g(n)=fracf(n)f(3)$$
      $$g(4)=frac454, g(5)=frac254, g(6)=-frac67938, g(7)=-frac43398, g(8)=frac6691994,$$
      $$g(9)=13430, g(10)=-frac11492150669232, g(11)=frac25700873197041316$$
      Also, surprisingly, it looks like $f(3)=2^-55$.



      Is there a way to identify them without precisely calculation?










      share|cite|improve this question















      If we define
      $$a_n(m)=sumlimits_k=0^nk!nbrace k(k+1)^m(-1)^n-k$$
      $$prodlimits_k=2^n+11-kx=sumlimits_k=0^nt(n,k)x^k$$
      for $n>0$, $mgeqslant0$, so
      $$sumlimits_k=0^nt(n,k)a_n(m-k)=0$$
      for $n>0$, $|m|geqslant0$, ex.
      $$a_1(m)-2a_1(m-1)=0$$
      $$a_2(m)-5a_2(m-1)+6a_2(m-2)=0$$
      Another words
      $$a_1(m)=fraca_1(m+1)2$$
      $$a_2(m)=frac5a_2(m+1)-a_2(m+2)6$$
      Then if
      $$f(n)=sumlimits_k=1^inftya_n(-k)$$
      so $f(1)=1$, $f(2)=0$.



      For $n>2$ we have very small constants with rational ratio, ex.
      $$g(n)=fracf(n)f(3)$$
      $$g(4)=frac454, g(5)=frac254, g(6)=-frac67938, g(7)=-frac43398, g(8)=frac6691994,$$
      $$g(9)=13430, g(10)=-frac11492150669232, g(11)=frac25700873197041316$$
      Also, surprisingly, it looks like $f(3)=2^-55$.



      Is there a way to identify them without precisely calculation?







      calculus recurrence-relations interpolation constants bernoulli-numbers






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      edited Sep 4 at 10:41

























      asked Sep 4 at 9:09









      user565184

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