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%I #8 Jan 23 2018 19:44:27
%S 1,2,3,4,5,6,7,8,9,10,12,24,36,48,60,72,84,96,108,120,122,133,144,155,
%T 166,177,188,199,210,221,232,244,255,266,277,288,299,310,321,332,343,
%U 354,366,377,388,399,410,421,432,443,454,465,476,488,499,510,521
%N Numbers whose base-11 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 after the zero from A029956 first at 1343 = 1011_11, which is not a palindrome in base 11 but has DV(1343,11) = UV(1343,11) =1. - _R. J. Mathar_, Jan 23 2018
%H Clark Kimberling, <a href="/A297274/b297274.txt">Table of n, a(n) for n = 1..10000</a>
%e 521 in base-11: 4,3,4, having DV = 1, UV = 1, so that 521 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 = 11; 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] (* A297273 *)
%t Take[Flatten[Position[w, 0]], 120] (* A297274 *)
%t Take[Flatten[Position[w, 1]], 120] (* A297275 *)
%Y Cf. A297330, A297273, A297275.
%K nonn,base,easy
%O 1,2
%A _Clark Kimberling_, Jan 16 2018
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