OFFSET
1,1
COMMENTS
Can also be defined as "Even numbers whose binary expansion is a concatenation of one or more blocks of the form 1^k 0^{k-1} with k >= 2 (i.e., alternating 1- and 0-runs where each 0-run is exactly one shorter than the preceding 1-run, and the expansion ends with a 0-run)".
Let the binary expansion of n (without leading zeros) have alternating runs of 1's and 0's and end in a 0-run. Then n is in the sequence iff, reading from the most significant bit, it can be written as a concatenation of blocks (1^{k_1} 0^{k_1-1})(1^{k_2} 0^{k_2-1})...(1^{k_t} 0^{k_t-1}) with t >= 1 and all k_i >= 2.
Equivalently, reading from the least significant bit, the trailing 0-run has length r_1 >= 1, the following 1-run has length r_1+1, then a 0-run of length r_2 >= 1, a 1-run of length r_2+1, etc., until the bits exhaust.
Every term is even because the expansion ends with a 0-run, hence the least significant bit is 0.
The single-block numbers 1^k 0^{k-1} are in the sequence for every k >= 2, so there are infinitely many terms.
The numbers with exactly one block 1^k 0^{k-1} are 2^{k-1} (2^k - 1) (for k >= 2), which is A006516 (shifted to start at 1 rather than 0). In particular, when 2^k - 1 is prime, these are the even perfect numbers (A000396).
If L is the bit-length of n, then L is the sum of the odd integers (2*k_i - 1) contributed by each block. Thus, the number of terms of bit-length exactly L equals the number of compositions of L into odd parts >= 3, which is A000931 evaluated at L.
Let v_2(n) be the 2-adic valuation (trailing zero count). Set n_0 = n. For i >= 0, let r_i = v_2(n_i) (so r_i >= 1), and define m_i = n_i / 2^{r_i}. The current block is valid iff m_i == 2^{r_i+1} - 1 (mod 2^{r_i+1}), i.e., after removing the trailing r_i zeros, the next r_i+1 low bits are all 1's. Then remove the entire block 1^{r_i+1} 0^{r_i} by n_{i+1} = floor(n_i / 2^{r_i+1 (equivalently, n_{i+1} = floor(m_i / 2^{r_i+1})). Continue until n_t = 0.
The set of binary strings here is { 1^k 0^{k-1} : k >= 2 }^+. (Compare with A166751, where each 0-run equals the preceding 1-run, not one shorter.)
LINKS
Ahmet Caglar Saygili, Table of n, a(n) for n = 1..10000
FORMULA
If the blocks (from most significant to least significant) have 1-run lengths k_1,...,k_t with k_i >= 2, then letting S_i = Sum_{j=i+1..t} (2*k_j - 1) (total length of blocks to the right of block i), we have a = Sum_{i=1..t} 2^{S_i + (k_i - 1)} * (2^{k_i} - 1). (Each block contributes 1^{k_i} 0^{k_i-1} = (2^{k_i} - 1) * 2^{k_i-1}, shifted left by S_i bits.)
EXAMPLE
220 = 11011100_2 is a term. Its runs are 11,0,111,00 and the adjacent pairs have lengths (2,1) and (3,2), where the 0-run length equals the preceding 1-run length minus 1 for each pair.
MATHEMATICA
ok[n_]:=Module[{L, m}, L=Length/@ Split[IntegerDigits[n, 2]]; m = Length[L]; EvenQ[m] && AllTrue[Range[1, m/2], L[[2*#-1]]-1==L[[2*#]] &]]; Select[Range[0, 29999], ok] (* James C. McMahon, Sep 10 2025 *)
PROG
(Julia)
function ok(n::Integer)::Bool
n > 0 || return false
x = n
while x > 0
z = 0
while (x & 1) == 0
z += 1
x >>= 1
end
z == 0 && return false
o = 0
while (x & 1) == 1
o += 1
x >>= 1
end
o == z + 1 || return false
end
true
end
[k for k in 0:10^5 if ok(k)]
(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(30000) if ok(k)]) # Michael S. Branicky, Aug 26 2025
CROSSREFS
KEYWORD
base,nonn,easy
AUTHOR
Ahmet Caglar Saygili, Aug 24 2025
STATUS
approved