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Expansion of 1/(1 - x/(1 - 1^2*x/(1 - 2^2*x/(1 - 3^2*x/(1 - 4^2*x/(1 - ...)))))), a continued fraction.
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%I #30 Oct 02 2023 06:56:27

%S 1,1,2,8,76,1540,53684,2812148,205054036,19805016628,2444724910292,

%T 375282530128052,70102075181928148,15655136160745164340,

%U 4118456236678107528404,1260512820941791994429876,444069171743010266366969044,178408825363590577961830752052

%N Expansion of 1/(1 - x/(1 - 1^2*x/(1 - 2^2*x/(1 - 3^2*x/(1 - 4^2*x/(1 - ...)))))), a continued fraction.

%C Invert transform of Euler (or secant) numbers (A000364), shifted right one place.

%H Alois P. Heinz, <a href="/A303943/b303943.txt">Table of n, a(n) for n = 0..243</a>

%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>

%F a(n) ~ 2^(4*n - 1) * n^(2*n - 3/2) / (Pi^(2*n - 3/2) * exp(2*n)). - _Vaclav Kotesovec_, Jun 08 2019

%p b:= proc(u, o) option remember;

%p `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))

%p end:

%p a:= proc(n) option remember; `if`(n<1, 1,

%p add(a(n-i)*b((i-1)*2, 0), i=1..n))

%p end:

%p seq(a(n), n=0..20); # _Alois P. Heinz_, Jun 13 2018

%p # Alternative:

%p T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1)

%p else (n - k)^2 * T(n, k - 1) + T(n - 1, k) fi fi end:

%p a := n -> T(n, n): seq(a(n), n = 0..17); # _Peter Luschny_, Oct 02 2023

%t nmax = 17; CoefficientList[Series[1/(1 - x/(1 + ContinuedFractionK[-k^2 x, 1, {k, 1, nmax}])), {x, 0, nmax}], x]

%t nmax = 17; CoefficientList[Series[1/(1 - x Sum[Abs[EulerE[2 k]] x^k, {k, 0, nmax}]), {x, 0, nmax}], x]

%t a[0] = 1; a[n_] := a[n] = Sum[Abs[EulerE[2 (k - 1)]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 17}]

%Y Cf. A000290, A000364, A112934, A305532.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Jun 04 2018