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a(n) = H_n(1,2) where H_n is the n-th hyperoperator.
3

%I #42 Aug 09 2024 10:08:26

%S 3,3,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,

%T 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,

%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1

%N a(n) = H_n(1,2) where H_n is the n-th hyperoperator.

%C See A054871 for definitions and key links.

%C Sequence is also the decimal expansion of 2989/9000.

%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (1).

%F From _Elmo R. Oliveira_, Jul 16 2024: (Start)

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

%F a(n) = 1 for n >= 3. (End)

%F E.g.f.: exp(x) + 2 + 2*x + x^2/2. - _Elmo R. Oliveira_, Aug 09 2024

%e a(0) = H_0(1,2) = 2+1 = 3;

%e a(1) = H_1(1,2) = 1+2 = 3;

%e a(2) = H_2(1,2) = 1*2 = 2;

%e a(3) = H_3(1,2) = 1^2 = 1;

%e a(4) = H_4(1,2) = 1^^2 = 1.

%Y Cf. A054871.

%K nonn,cons,easy

%O 0,1

%A _Natan Arie Consigli_, Aug 24 2015