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%I #9 Jan 23 2018 19:56:32
%S 1,2,3,4,5,6,7,8,9,10,11,13,26,39,52,65,78,91,104,117,130,143,145,157,
%T 169,181,193,205,217,229,241,253,265,277,290,302,314,326,338,350,362,
%U 374,386,398,410,422,435,447,459,471,483,495,507,519,531,543,555
%N Numbers whose base-12 digits have equal down-variation and up-variation; see Comments.
%C Suppose that n has base-b digits b(m), b(m-1), ..., b(0). The base-b down-variation of n is the sum DV(n,b) of all d(i)-d(i-1) for which d(i) > d(i-1); the base-b up-variation of n is the sum UV(n,b) of all d(k-1)-d(k) for which d(k) < d(k-1). The total base-b variation of n is the sum TV(n,b) = DV(n,b) + UV(n,b). See the guide at A297330.
%C Differs first from A029957 after the zero for 1741 = 1011_12, which is not a palindrome in base 12 but has DV(1741,12) = UV(1741,12) =1. - _R. J. Mathar_, Jan 23 2018
%H Clark Kimberling, <a href="/A297277/b297277.txt">Table of n, a(n) for n = 1..10000</a>
%e 555 in base-12: 3,10,3, having DV = 7, UV = 7, so that 555 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, A297276, A297278.
%K nonn,base,easy
%O 1,2
%A _Clark Kimberling_, Jan 16 2018
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