A296858 Numbers whose base-2 digits have #(pits) = #(peaks); see Comments.

1, 2, 3, 4, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 19, 20, 24, 25, 26, 28, 30, 31, 32, 33, 35, 37, 38, 39, 40, 41, 42, 48, 49, 51, 52, 56, 57, 58, 60, 62, 63, 64, 65, 67, 69, 70, 71, 75, 76, 78, 79, 80, 81, 83, 84, 96, 97, 99, 101, 102, 103, 104, 105, 106, 112

Offset: 1

Author

Clark Kimberling, Jan 08 2018

Keywords

• nonn
• base
• easy

Comments

A pit is an index i such that d(i-1) > d(i) < d(i+1); a peak is an index i such that d(i-1) < d(i) > d(i+1). The sequences A296858-A296860 partition the natural numbers. See the guides at A296882 and A296712.

Examples

			The base-2 digits of 112 are 1,1,1,0,0,0,0; here #(pits) = 0 and #(peaks) = 0, so that 112 is in the sequence.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A296882, A296712, A296859, A296860.

Programs

Mathematica

z = 200; b = 2;
d[n_] := Differences[Sign[Differences[IntegerDigits[n, b]]]];
Select[Range [z], Count[d[#], -2] == Count[d[#], 2] &]   (* A296858 *)
Select[Range [z], Count[d[#], -2] < Count[d[#], 2] &]    (* A296859 *)
Select[Range [z], Count[d[#], -2] > Count[d[#], 2] &]    (* A296860 *)

Python

def cwo(subs, s): # count with overlaps allowed
  c = i = 0
  while i != -1:
    i = s.find(subs, i)
    if i != -1: c += 1; i += 1
  return c
def ok(n): b = bin(n)[2:]; return cwo('101', b) == cwo('010', b)
print(list(filter(ok, range(1, 113)))) # Michael S. Branicky, May 11 2021