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a(0) = 2, a(n) = (-1)^n + (-2)^n + (1/2) * Sum_{j=1..n} (1-(-1)^j-(-2)^j) * binomial(n,j) * a(n-j) for n > 0.
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%I #20 Feb 28 2024 20:40:14

%S 2,1,5,25,161,1321,13025,149605,1963841,29004721,475975745,8591917885,

%T 169193833121,3609452038921,82924458549665,2041207822721365,

%U 53594538159184001,1495143168658285921,44164021453758342785,1377005070100813288045,45193800193226286112481

%N a(0) = 2, a(n) = (-1)^n + (-2)^n + (1/2) * Sum_{j=1..n} (1-(-1)^j-(-2)^j) * binomial(n,j) * a(n-j) for n > 0.

%C Inverse binomial transform of A370092 + A370456.

%F E.g.f.: 2*(1 + exp(x))/(1 + exp(x) + exp(2*x) - exp(3*x)).

%o (SageMath)

%o def a(m):

%o if m==0:

%o return 2

%o else:

%o return (-1)^m+(-2)^m+1/2*sum([(1-(-2)^j-(-1)^j)*binomial(m,j)*a(m-j) for j in [1,..,m]])

%o list(a(m) for m in [0,..,20])

%o (SageMath)

%o f=2*(1+e^x)/(1+e^x+e^(2*x)-e^(3*x))

%o print([(diff(f,x,i)).subs(x=0) for i in [0,..,20]])

%o (PARI) seq(n)={my(p=exp(x + O(x*x^n))); Vec(serlaplace(2*(1 + p)/(1 + p + p^2 - p^3)))} \\ _Andrew Howroyd_, Feb 26 2024

%Y Cf. A370092, A370456.

%K nonn

%O 0,1

%A _Prabha Sivaramannair_, Feb 26 2024