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a(n) = 5 written in base n.
3

%I #16 Jun 16 2024 05:21:45

%S 11111,101,12,11,10,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,

%T 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,

%U 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5

%N a(n) = 5 written in base n.

%C The term 111....1111 should officially be called the "unary expansion", since in base 1 only the digit 0 may appear.

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

%F From _Elmo R. Oliveira_, Jun 15 2024: (Start)

%F G.f.: x*(11111-11010*x-89*x^2-x^3-x^4-5*x^5)/(1-x).

%F a(n) = 5 for n > 5. (End)

%Y Cf. A001735, A063432, A035613, A094325, A094326, A094327, A094232, etc.

%K nonn,base,easy

%O 1,1

%A _N. J. A. Sloane_, Jun 04 2004