A385575 Numbers whose binary indices have the same number of adjacent parts differing by 1 as adjacent parts differing by more than 1.

1, 2, 4, 8, 11, 13, 16, 19, 22, 25, 26, 32, 35, 38, 44, 49, 50, 52, 64, 67, 70, 76, 87, 88, 91, 93, 97, 98, 100, 104, 107, 109, 117, 128, 131, 134, 140, 151, 152, 155, 157, 167, 174, 176, 179, 182, 185, 186, 193, 194, 196, 200, 203, 205, 208, 211, 214, 217

Offset: 1

Author

Gus Wiseman, Jul 04 2025

Keywords

• nonn

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

Examples

			The terms together with their binary expansions and binary indices begin:
    1:       1 ~ {1}
    2:      10 ~ {2}
    4:     100 ~ {3}
    8:    1000 ~ {4}
   11:    1011 ~ {1,2,4}
   13:    1101 ~ {1,3,4}
   16:   10000 ~ {5}
   19:   10011 ~ {1,2,5}
   22:   10110 ~ {2,3,5}
   25:   11001 ~ {1,4,5}
   26:   11010 ~ {2,4,5}
   32:  100000 ~ {6}
   35:  100011 ~ {1,2,6}
   38:  100110 ~ {2,3,6}
   44:  101100 ~ {3,4,6}
   49:  110001 ~ {1,5,6}
   50:  110010 ~ {2,5,6}
   52:  110100 ~ {3,5,6}
   64: 1000000 ~ {7}
   67: 1000011 ~ {1,2,7}
   70: 1000110 ~ {2,3,7}
   76: 1001100 ~ {3,4,7}
   87: 1010111 ~ {1,2,3,5,7}
   88: 1011000 ~ {4,5,7}
   91: 1011011 ~ {1,2,4,5,7}
   93: 1011101 ~ {1,3,4,5,7}
   97: 1100001 ~ {1,6,7}
   98: 1100010 ~ {2,6,7}
  100: 1100100 ~ {3,6,7}
		

Crossrefs

The LHS rank statistic is A069010, counted by A034839 (for partitions A384881, A116674).

The RHS rank statistic is A384890, counted by A384893 (for partitions A268193, A384905).

Subsets of this type are counted by A385572, with n A217615.

A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.

A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.

Cf. A000079, A010027, A053538, A210034, A384177, A384879, A384889.

Programs

Mathematica

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Select[Range[100],Length[Split[bpe[#],#2==#1+1&]]==Length[Split[bpe[#],#2!=#1+1&]]&]

PARI

is_ok(n)=hammingweight(n)==2*hammingweight(bitand(n,n>>1))+1 \\ Christian Sievers, Jul 18 2025