%I
%S 12,24,25,36,37,38,48,49,50,51,60,61,62,63,64,72,73,74,75,76,77,84,85,
%T 86,87,88,89,90,96,97,98,99,100,101,102,103,108,109,110,111,112,113,
%U 114,115,116,120,121,122,123,124,125,126,127,128,129,132,133,134
%N Numbers whose base12 digits have greater downvariation than 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 A296749 first at 168 = 120_12, which is in not in A296749 because it has the same number of rises and falls, but in here because DV(168,12) =2 > UV(168,12) =1.  _R. J. Mathar_, Jan 23 2018
%H Clark Kimberling, <a href="/A297276/b297276.txt">Table of n, a(n) for n = 1..10000</a>
%e 134 in base12: 11,2, having DV = 9, UV = 0, so that 134 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 = 12; 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] (* A297276 *)
%t Take[Flatten[Position[w, 0]], 120] (* A297277 *)
%t Take[Flatten[Position[w, 1]], 120] (* A297278 *)
%Y Cf. A297330, A297277, A297278.
%K nonn,base,easy
%O 1,1
%A _Clark Kimberling_, Jan 16 2018
