%I
%S 0,1,2,3,6,7,10,11,19,20,23,30,33,34,57,60,61,69,70,91,92,100,101,104,
%T 172,173,181,182,185,208,209,212,273,276,277,300,303,304,312,313,516,
%U 519,520,543,546,547,555,556,624,627,628,636,637
%N Numbers whose base3 representation Sum_{i=0..m} d(i)*3^i has d(m) > d(m1) < d(m2) > ...
%C Every other base3 digit must be strictly less than its neighbors.  _M. F. Hasler_, Oct 05 2018
%C The terms can be generated in the following way: if A(n) are the terms with n digits in base 3, the terms with n+2 digits are obtained by prefixing them with '10' and with '20', and prefixing '21' to those starting with a digit '2'. It is easy to prove that #A(n) = A000045(n+2), since from the above we have #A(n+2) = 2*#A(n) + #A(n1) = #A(n) + #A(n+1). (The #A(n1) numbers starting with '2' are #A(n2) numbers prefixed with '20' and #A(n3) prefixed with '21'.)  _M. F. Hasler_, Oct 05 2018
%H M. F. Hasler, <a href="/A032858/b032858.txt">Table of n, a(n) for n = 1..5000</a>
%F a(A000071(n+3)) = floor(3^(n+1)/8) = A033113(n).  _M. F. Hasler_, Oct 05 2018
%e The base3 representation of the initial terms is 0, 1, 2, 10, 20, 21, 101, 102, 201, 202, 212, 1010, 1020, 1021, 2010, 2020, 2021, 2120, 2121, 10101, 10102, ...
%t sdQ[n_]:=Module[{s=Sign[Differences[IntegerDigits[n, 3]]]}, s==PadRight[{}, Length[s], {1, 1}]]; Select[Range[0, 700], sdQ] (* _Vincenzo Librandi_, Oct 06 2018 *)
%o (PARI) is(n,b=3)=!for(i=2,#n=digits(n,b),(n[i1]n[i])*(1)^i>0return) \\ _M. F. Hasler_, Oct 05 2018
%Y Cf. A032859 .. A032865 for base4 .. 10 variants.
%Y Cf. A000975 (or A056830 in binary) for the base2 analog.
%Y Cf. A306105 for these terms written in base 3.
%K nonn,base
%O 1,3
%A _Clark Kimberling_
%E Definition edited, crossreferences and a(1) = 0 inserted by _M. F. Hasler_, Oct 05 2018
