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%I #26 Mar 21 2024 12:38:52
%S 1,3,25,222,3166,47016,951544,19827408,520029520,13952218560,
%T 449559799360,14756761434240,563961412362880,21893890640563200,
%U 968019931702297600,43385863589508249600,2178487766250586470400,110704921777161066700800,6222745685273069016064000
%N a(n) = ( Product {k = 1..n} 3*k - 2 ) * ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 2) ).
%C Original name was: s(1)*s(2)*...*s(n)(1/s(1) - 1/s(2) + ... + c/s(n)), where c=(-1)^(n+1) and s(k) = 3k-2 for k = 1,2,3,...
%H L. Euler, <a href="http://eulerarchive.maa.org">De fractionibus continuis observationes</a>, The Euler Archive, Index Number 123, Section 5
%F From _Peter Bala_, Feb 20 2015: (Start)
%F a(n) = A007559(n) * Sum {k = 1..n} (-1)^(k+1)/(3*k - 2).
%F Recurrence: a(n+1) = 3*a(n) + (3*n - 2)^2*a(n-1) with a(1) = 1 and a(2) = 3.
%F The triple factorial numbers A007559 also satisfy this second-order recurrence equation. This leads to the continued fraction representation a(n)/A007559(n) = 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + ... + (3*n - 2)^2/(3 ))))).
%F Taking the limit as n -> infinity gives the generalized continued fraction: Sum {k >= 1} (-1)^(k+1)/(3*k - 2) = 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + 10^2/(3 + ... ))))) due to Euler.
%F The alternating sum has the value 1/3*( log(2) + Pi/sqrt(3) ) = A113476. Cf. A024396. (End)
%F a(n) ~ sqrt(2*Pi) * (sqrt(3)*Pi + 3*log(2)) * 3^(n-2) * n^(n-1/6) / (GAMMA(1/3) * exp(n)). - _Vaclav Kotesovec_, Feb 21 2015
%p a[1] := 1: a[2] := 3: for n from 3 to 20 do a[n] := 3*a[n-1]+(3*n-5)^2*a[n-2] end do: seq(a[n], n = 1 .. 20); # _Peter Bala_, Feb 20 2015
%t Table[Product[3*k-2,{k,1,n}] * Sum[(-1)^(k+1)/(3*k-2),{k,1,n}],{n,1,20}] (* _Vaclav Kotesovec_, Feb 21 2015 *)
%o (Magma) I:=[1,3]; [n le 2 select I[n] else 3*Self(n-1)+(3*n-5)^2*Self(n-2): n in [1..20]]; // _Vincenzo Librandi_, Feb 21 2015
%Y Cf. A007559, A024199, A024383, A024396, A113476.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_
%E New name from _Peter Bala_, Feb 20 2015
%E a(18)-a(19) from _Vincenzo Librandi_, Feb 21 2015
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