%I #20 Jul 09 2024 09:02:24

%S 0,1,3,0,-20,-45,21,308,540,-585,-4235,-5676,11232,54145,51975,

%T -182400,-654160,-380205,2680425,7516400,1320900,-36753255,-82175665,

%U 24032700,477852900,850446025,-749925189,-5944471092,-8220606800,14049061455,71102953305,71989187536,-220682377872

%N a(0)=0, a(1)=1, a(n)=((2*n-1)*a(n-1)-5*n*a(n-2))/(n-1).

%C n divides a(n) iff the binary representation of n ends with an even number of zeros (i.e. n is in A003159)

%H Harvey P. Dale, <a href="/A102840/b102840.txt">Table of n, a(n) for n = 0..1000</a>

%F log(abs(a(n))) is asymptotic to c*n where c=0.80... [c = log(5)/2 = 0.8047189562... - _Vaclav Kotesovec_, Feb 15 2019]

%F a(n) ~ sqrt(n) * 5^(n/2) / sqrt(8*Pi) * ((sqrt(2 + sqrt(5)) + sqrt(38 + 25*sqrt(5)) / (16*n)) * sin(n*arctan(2)) - (sqrt(-2 + sqrt(5)) - sqrt(-38 + 25*sqrt(5)) / (16*n)) * cos(n*arctan(2))). - _Vaclav Kotesovec_, Feb 15 2019

%F From _Seiichi Manyama_, Jul 09 2024: (Start)

%F G.f.: x/(1 - 2*x + 5*x^2)^(3/2).

%F a(n+1) = binomial(n+2,2) * A343773(n). (End)

%t RecurrenceTable[{-5 n a[n-2] + (2*n - 1) a[n-1] + (1 - n) a[n] ==

%t 0, a[0] == 0, a[1] == 1}, a, {n, 0, 30}] (* _Vaclav Kotesovec_, Feb 15 2019 *)

%t nxt[{n_,a_,b_}]:={n+1,b,(b(2n+1)-5a(n+1))/n}; NestList[nxt,{1,0,1},40][[;;,2]] (* _Harvey P. Dale_, Apr 22 2024 *)

%o (PARI) a(n)=if(n<2,if(n,1,0),1/(n-1)*((2*n-1)*a(n-1)-5*n*a(n-2)))

%Y Cf. A102839, A343773.

%K sign

%O 0,3

%A _Benoit Cloitre_, Feb 27 2005