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Start with 1, then at each step prepend 10 and append 01.
1

%I #18 Mar 11 2024 11:32:20

%S 1,10101,101010101,1010101010101,10101010101010101,

%T 101010101010101010101,1010101010101010101010101,

%U 10101010101010101010101010101,101010101010101010101010101010101

%N Start with 1, then at each step prepend 10 and append 01.

%C Bisection of A094028. - _Omar E. Pol_, Nov 12 2008

%C a(n) is also A144864(n) written in base 2. - _Omar E. Pol_, Nov 13 2008

%C Quadrisection of A147759. - _Omar E. Pol_, Nov 16 2008

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10001,-10000).

%F a(n) = (-1+100^(-1+2*n))/99.

%F If a(n) is interpreted as binary number, (-4+16^n)/12 gives the decimal representation of a(n).

%F a(n) = 10000*a(n-1)+101, n>1.

%F G.f.: x*(1+100*x) / ( (10000*x-1)*(x-1) ).

%t a = {}; k = {1}; Do[x = FromDigits[k, 2]; AppendTo[a, FromDigits[RealDigits[x, 2]]]; AppendTo[k, 0]; AppendTo[k, 1]; PrependTo[k, 0]; PrependTo[k, 1], {n, 1, 100}];

%t Table[FromDigits[RealDigits[1/12 (-4 + 16^n), 2]], {n, 1, 10}]

%t a = {}; k = 1; Do[AppendTo[a, k]; k = 10000 k + 101, {n, 1, 10}]; a

%t Table[1/99 (-1 + 100^(-1 + 2 n)), {n, 1, 20}]

%t LinearRecurrence[{10001,-10000},{1,10101},20] (* _Harvey P. Dale_, Aug 22 2014 *)

%Y Cf. A056830, A094028, A135576, A144864, A147759.

%K base,nonn,easy

%O 1,2

%A _Artur Jasinski_, Sep 23 2008, Sep 25 2008