%I
%S 1,2,3,4,5,6,7,8,10,20,30,40,50,60,70,80,82,91,100,109,118,127,136,
%T 145,154,164,173,182,191,200,209,218,227,236,246,255,264,273,282,291,
%U 300,309,318,328,337,346,355,364,373,382,391,400,410,419,428,437,446
%N Numbers whose base9 digits have equal downvariation and upvariation; see Comments.
%C Suppose that n has baseb digits b(m), b(m1), ..., b(0). The baseb downvariation of n is the sum DV(n,b) of all d(i)d(i1) for which d(i) > d(i1); the baseb upvariation of n is the sum UV(n,b) of all d(k1)d(k) for which d(k) < d(k1). The total baseb variation of n is the sum TV(n,b) = DV(n,b) + UV(n,b). See the guide at A297330.
%C Differs from A029955 first at 739=1011_9 which is not a palindrome in base 9 but has DV(739,9)=UV(793,9) =1.  _R. J. Mathar_, Jan 23 2018
%H Clark Kimberling, <a href="/A297268/b297268.txt">Table of n, a(n) for n = 1..10000</a>
%e 446 in base9: 5,4,5, having DV = 1, UV = 1, so that 446 is in the sequence.
%t g[n_, b_] := Map[Total, GatherBy[Differences[IntegerDigits[n, b]], Sign]];
%t x[n_, b_] := Select[g[n, b], # < 0 &]; y[n_, b_] := Select[g[n, b], # > 0 &];
%t b = 9; z = 2000; p = Table[x[n, b], {n, 1, z}]; q = Table[y[n, b], {n, 1, z}];
%t w = Sign[Flatten[p /. {} > {0}] + Flatten[q /. {} > {0}]];
%t Take[Flatten[Position[w, 1]], 120] (* A297267 *)
%t Take[Flatten[Position[w, 0]], 120] (* A297268 *)
%t Take[Flatten[Position[w, 1]], 120] (* A297269 *)
%Y Cf. A297330, A297267, A297269.
%K nonn,base,easy
%O 1,2
%A _Clark Kimberling_, Jan 15 2018
