cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Frederik vom Ende

Frederik vom Ende's wiki page.

Frederik vom Ende has authored 1 sequences.

A300793 a(n) is the n-th derivative of arcsinh(1/x) at x=1 times (-2)^n/sqrt(2) for n >= 1.

Original entry on oeis.org

1, 3, 13, 75, 561, 5355, 63405, 894915, 14511105, 263544435, 5284255725, 116065424475, 2778006733425, 72093290744475, 2017526711525325, 60547198550713875, 1938662110170410625, 65941564342927147875, 2374177441960545346125, 90211614359319635056875
Offset: 1

Author

Frederik vom Ende, Mar 13 2018

Keywords

Programs

  • Maple
    a := n -> subs(x=1, (-2)^n/sqrt(2)*diff(arcsinh(1/x), x$n)):
    seq(a(n), n=1..20); # Peter Luschny, Mar 14 2018
    A300793_list := proc(len) local egf, ser, coef;
    egf := (log(sqrt((1-2*x)^2+1)+1)-log(1-2*x))/sqrt(2):
    ser := series(egf,x,len+1): coef := n -> round(n!*coeff(ser,x,n)):
    seq(coef(n), n=1..len) end: A300793_list(20); # Peter Luschny, Apr 06 2018
  • Mathematica
    (* Mathematica program from Bálint Koczor, TU Munich *)
    alist[max_] := Module[{prevRow, buf, makeNewRow, ind},
       (*definitions*)
       ind[j_] := j + 1; (*to shift the index*)
       makeNewRow[prevRow_, k_] := Table[
         If[ind[j] > k, 0, prevRow[[ind[j]]]*(2 j - k)] +
          If[j == 0, 0, prevRow[[ind[j] - 1]]*(2 j - 3 k - 1)]
         , {j, 0, k}]; (*this is the recursion formula*)
       prevRow = {-1}; (*initialize*)
       buf = Table[
         If[k == 0, -1, 0], {k, 0, max}];(*this will hold the resulting integers*)
       Do[
        prevRow = makeNewRow[prevRow, k];
        buf[[k + 1]] = Total@prevRow;,(*sums up the previous row*)
        {k, 1, max}];
       Return@(buf*Table[(-1)^n, {n, 1, max + 1}]);
       ];
    alist[19]

Formula

Proved (see links): a(n) = (-1)^n*Sum_{j=0..n-1} b(j,n) for any n >= 1 where {b(j,n)} for n=1,2,... and j any integer is a recursive sequence given by b(0,1)=-1, b(j,n)=0 if j < 0 or j >= n and b(j,n+1) = b(j,n)*(2j-n) + b(j-1,n)*(2j-3n-1) for all n >= 1 and 0 <= j <= n.
Empirical (by Martin Rubey on mathoverflow, see links): a(1)=1, a(2)=3, a(3)=13, a(n) = 4(n-2)^2*(n-3)*a(n-3) - 2(3n-5)*(n-2)*a(n-2) + (4n-5)*a(n-1) for all n >= 4.
a(n) = n!*[x^n]((log(sqrt((1-2*x)^2 + 1) + 1) - log(1 - 2*x))/sqrt(2)). - Peter Luschny, Apr 06 2018