A387269 Numbers whose binary expansion consists of alternating runs of 1's and 0's where each run of 0's is exactly one longer than the preceding run of 1's, and the expansion ends with a 0-run.

4, 24, 36, 112, 152, 196, 292, 480, 624, 792, 900, 1176, 1220, 1572, 1984, 2340, 2528, 3184, 3608, 3844, 4720, 4888, 4996, 6296, 6340, 7204, 8064, 9368, 9412, 9764, 10176, 12580, 12768, 14448, 15384, 15876, 18724, 18912, 19568, 19992, 20228, 25200, 25368, 25476

Offset: 1

Author

Ahmet Caglar Saygili, Aug 24 2025

Keywords

• base
• nonn
• easy
• new

Comments

Every term is even (since the binary ends with a 0-run).

Writing the binary word as consecutive pairs 1^{a_1}0^{a_1+1}1^{a_2}0^{a_2+1} ..., the total number of 0's exceeds the total number of 1's by the number of pairs.

Single-pair subfamily 1^{k}0^{k+1}_2, with k >= 1, is A059153(k-1).

Let f(n) = n & (2n) (bitwise AND; A213370). If x has binary run pairs 1^{L_i+1}0^{L_i} (MSB->LSB) as in A387270, then f(x) has pairs 1^{L_i}0^{L_i+1}. Thus f maps A387270 bijectively onto this sequence. Reason: n & (2n) has 1-bits exactly where n has adjacent 1-bits, which shortens every 1-run by 1 and lengthens the following 0-run by 1. Conversely, g(n) = n | floor(n/2) increases each 1-run by 1 and shortens the following 0-run by 1, mapping this sequence back to A387270. Examples: f(6) = 4 (110_2 -> 100_2), f(28) = 24 (11100_2 -> 11000_2); g(24) = 28 (11000_2 -> 11100_2), g(36) = 54 (100100_2 -> 110110_2).

Examples

			36 = 100100_2 is a term since its pairs of (1 then 0) runs are (1,2), (1,2).
		

Links

• Ahmet Caglar Saygili, Table of n, a(n) for n = 1..10000
• Sean A. Irvine, Java program (github)

Crossrefs

Cf. A059153, A166751, A387270, A213370.

Programs

Julia

function ok(n::Integer)::Bool
    n > 0 && iseven(n) || return false
    x = unsigned(n)
    while x != 0
        z = trailing_zeros(x); x >>= z
        o = trailing_ones(x)
        z == o + 1 || return false
        x >>= o
    end
    true
end
[k for k in 0:10^5 if ok(k)]

PARI

isok(k) = if (!(k%2), my(b=binary(k), pos=1, d, dd); for (i=1, #b-1, if (b[i] != b[i+1], if (b[i], d = i-pos+1; pos = i+1, dd = i-pos+1; pos = i+1; if (dd != d+1, return(0))))); dd = #b - pos+1; if (dd != d+1, return(0)); return(1);); \\ Michel Marcus, Aug 26 2025

Python

from itertools import groupby
def ok(n):
    L = [len(list(g)) for k, g in groupby(bin(n)[2:])]
    return (m:=len(L))&1 == 0 and all(L[2*j]+1 == L[2*j+1] for j in range(m>>1))
print([k for k in range(10**5) if ok(k)]) # Michael S. Branicky, Aug 25 2025