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

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

Offset

1,2

Links

Table of n, a(n) for n=1..20.

Frederik vom Ende, Proof of the recursive formula for the a(n)

Marcel Jacobse, Python Program to compute the sequence A300793

Mathoverflow, Closed, sum-free form for the n-th derivative of arcsinh(1/x) in x=1

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

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]

Crossrefs

Sequence in context: A000670 A032036 A305535 * A222427 A026072 A063646

Adjacent sequences: A300790 A300791 A300792 * A300794 A300795 A300796

Keyword

nonn

Author

Frederik vom Ende, Mar 13 2018

Status

approved