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O.g.f.: 1/(1 - x/(1 - 2^4*x/(1 - 3^4*x/(1 - 4^4*x/(1 - 5^4*x/(1 - 6^4*x/(1 -...))))))), a continued fraction.
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%I #28 Oct 02 2023 16:23:48

%S 1,1,17,1585,485729,372281761,601378506737,1820943071778385,

%T 9489456505643743169,79759396929125826861121,

%U 1027412704023984825792488657,19464301715272748317827942755185,524230105465412991467916306841439009,19509134827116013764271741468197795034081

%N O.g.f.: 1/(1 - x/(1 - 2^4*x/(1 - 3^4*x/(1 - 4^4*x/(1 - 5^4*x/(1 - 6^4*x/(1 -...))))))), a continued fraction.

%C Compare to the continued fraction for the Euler numbers (A000364):

%C 1/(1-x/(1-2^2*x/(1-3^2*x/(1-4^2*x/(1-5^2*x/(1-6^2*x/(1-...))))))).

%C From _Vaclav Kotesovec_, Sep 24 2020: (Start)

%C In general, if s>0 and g.f. = 1/(1 - x/(1 - 2^s*x/(1 - 3^s*x/(1 - 4^s*x/(1 - 5^s*x/(1 - 6^s*x/(1 -...))))))), a continued fraction, then

%C a(n,s) ~ c(s) * d(s)^n * (n!)^s / sqrt(n), where

%C d(s) = (2*s*Gamma(2/s) / Gamma(1/s)^2)^s

%C c(s) = sqrt(s*d(s)/(2*Pi)). (End)

%F a(n) ~ c * d^n * (n!)^4 / sqrt(n), where d = 4096 * Pi^2 / Gamma(1/4)^8 = 1.353976395034780345656335026823167975194... and c = sqrt(2*d/Pi) = 64 * sqrt(2*Pi) / Gamma(1/4)^4 = 0.9284223954634658948993105287957575... - _Vaclav Kotesovec_, Aug 25 2017, updated Sep 23 2020

%e G.f.: A(x) = 1 + x + 17*x^2 + 1585*x^3 + 485729*x^4 + 372281761*x^5 +...

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

%p else (k - n - 1)^4 * T(n, k - 1) + T(n - 1, k) fi fi end:

%p a := n -> T(n, n): seq(a(n), n = 0..13); # _Peter Luschny_, Oct 02 2023

%t nmax = 20; CoefficientList[Series[1/Fold[(1 - #2/#1) &, 1, Reverse[Range[nmax + 1]^4*x]], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Aug 25 2017 *)

%o (PARI) {a(n)=local(CF=1+x*O(x^n)); for(k=1, n, CF=1/(1-(n-k+1)^4*x*CF)); polcoeff(CF, n)}

%o for(n=0, 20, print1(a(n), ", "))

%Y Cf. A000364, A216966, A337807, A337808, A337809.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 26 2013