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

%I A300793 #47 Apr 06 2018 15:46:46
%S A300793 1,3,13,75,561,5355,63405,894915,14511105,263544435,5284255725,
%T A300793 116065424475,2778006733425,72093290744475,2017526711525325,
%U A300793 60547198550713875,1938662110170410625,65941564342927147875,2374177441960545346125,90211614359319635056875
%N A300793 a(n) is the n-th derivative of arcsinh(1/x) at x=1 times (-2)^n/sqrt(2) for n >= 1.
%H A300793 Frederik vom Ende, <a href="/A300793/a300793_2.pdf">Proof of the recursive formula for the a(n)</a>
%H A300793 Marcel Jacobse, <a href="/A300793/a300793.txt">Python Program to compute the sequence A300793</a>
%H A300793 Mathoverflow, <a href="https://mathoverflow.net/q/295019">Closed, sum-free form for the n-th derivative of arcsinh(1/x) in x=1</a>
%F A300793 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.
%F A300793 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.
%F A300793 a(n) = n!*[x^n]((log(sqrt((1-2*x)^2 + 1) + 1) - log(1 - 2*x))/sqrt(2)). - _Peter Luschny_, Apr 06 2018
%p A300793 a := n -> subs(x=1, (-2)^n/sqrt(2)*diff(arcsinh(1/x), x$n)):
%p A300793 seq(a(n), n=1..20); # _Peter Luschny_, Mar 14 2018
%p A300793 A300793_list := proc(len) local egf, ser, coef;
%p A300793 egf := (log(sqrt((1-2*x)^2+1)+1)-log(1-2*x))/sqrt(2):
%p A300793 ser := series(egf,x,len+1): coef := n -> round(n!*coeff(ser,x,n)):
%p A300793 seq(coef(n), n=1..len) end: A300793_list(20); # _Peter Luschny_, Apr 06 2018
%t A300793 (* Mathematica program from Bálint Koczor, TU Munich *)
%t A300793 alist[max_] := Module[{prevRow, buf, makeNewRow, ind},
%t A300793    (*definitions*)
%t A300793    ind[j_] := j + 1; (*to shift the index*)
%t A300793    makeNewRow[prevRow_, k_] := Table[
%t A300793      If[ind[j] > k, 0, prevRow[[ind[j]]]*(2 j - k)] +
%t A300793       If[j == 0, 0, prevRow[[ind[j] - 1]]*(2 j - 3 k - 1)]
%t A300793      , {j, 0, k}]; (*this is the recursion formula*)
%t A300793    prevRow = {-1}; (*initialize*)
%t A300793    buf = Table[
%t A300793      If[k == 0, -1, 0], {k, 0, max}];(*this will hold the resulting integers*)
%t A300793    Do[
%t A300793     prevRow = makeNewRow[prevRow, k];
%t A300793     buf[[k + 1]] = Total@prevRow;,(*sums up the previous row*)
%t A300793     {k, 1, max}];
%t A300793    Return@(buf*Table[(-1)^n, {n, 1, max + 1}]);
%t A300793    ];
%t A300793 alist[19]
%K A300793 nonn
%O A300793 1,2
%A A300793 _Frederik vom Ende_, Mar 13 2018