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%I #7 Jan 27 2023 19:24:14
%S 1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99,101,102,103,104,105,106,
%T 107,108,109,111,120,121,130,131,132,140,141,142,143,150,151,152,153,
%U 154,160,161,162,163,164,165,170,171,172,173,174,175,176,180,181
%N Numbers whose base-10 digits d(m), d(m-1), ..., d(0) have #(rises) = #(falls); see Comments.
%C A rise is an index i such that d(i) < d(i+1); a fall is an index i such that d(i) > d(i+1). The sequences A296712-A296714 partition the natural numbers.
%C ****
%C Guide to related sequences:
%C Base #(rises) = #(falls) #(rises) > #(falls) #(rises) < #(falls)
%C 2 A005408 (none) A005843
%C 3 A296691 A296692 A296693
%C 4 A296694 A296695 A296696
%C 5 A296697 A296698 A296699
%C 6 A296700 A296701 A296702
%C 7 A296703 A296704 A296705
%C 8 A296706 A296707 A296708
%C 9 A296709 A296710 A296711
%C 10 A296712 A296713 A296714
%C 11 A296744 A296745 A296746
%C 12 A296747 A296748 A296749
%C 13 A296750 A296751 A296752
%C 14 A296753 A296754 A296755
%C 15 A296756 A296757 A296758
%C 16 A296759 A296760 A296761
%C 20 A296762 A296763 A296764
%C 60 A296765 A296766 A296767
%H Clark Kimberling, <a href="/A296712/b296712.txt">Table of n, a(n) for n = 1..10000</a>
%e The base-10 digits of 181 are 1,8,1; here #(rises) = 1 and #(falls) = 1, so 181 is in the sequence.
%t z = 200; b = 10; d[n_] := Sign[Differences[IntegerDigits[n, b]]];
%t Select[Range [z], Count[d[#], -1] == Count[d[#], 1] &] (* A296712 *)
%t Select[Range [z], Count[d[#], -1] < Count[d[#], 1] &] (* A296713 *)
%t Select[Range [z], Count[d[#], -1] > Count[d[#], 1] &] (* A296714 *)
%Y Cf. A296713, A296714, A296712.
%K nonn,base,easy
%O 1,2
%A _Clark Kimberling_, Jan 08 2018
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