A166751 -id:A166751 - cp's OEIS frontend

A386705 a(n) = sum of the 2^(n-1) even positive integers having bit length 2*n and in which, when written in binary, each run of 0's is of exactly the same length as the run of 1's immediately before it.

2, 22, 192, 1576, 12704, 101856, 815360, 6524032, 52194816, 417564160, 3340525568, 26724231168, 213793906688, 1710351376384, 13682811273216, 109462490742784, 875699927121920, 7005599419465728, 56044795360968704, 448358362898759680, 3586866903213146112, 28694935225753403392

Offset: 1

Paolo Xausa, Aug 28 2025

Row sums of A166751, when viewed as an irregular triangle whose row terms have the same number of bits (see the Example section there).

			For n = 3, the 2^(n-1) terms with bit length 2*n = 6 satisfying the criteria are (in binary): 101010, 101100, 110010 and 111000, corresponding (in decimal) to 42, 44, 50 and 56, giving a sum of 192.

Cf. A166751.

A386705[n_] := With[{b = Array[IntegerDigits[4^# - 2^#, 2] &, n]}, Total[Map[FromDigits[Flatten[#], 2] &, Map[b[[#]] &, Map[Permutations, IntegerPartitions[n]], {2}], {2}], 2]];
Array[A386705, 20]

Empirical: a(n) = 12*a(n-1) - 36*a(n-2) + 32*a(n-3), with a(1) = 2, a(2) = 22, a(3) = 192.

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.

Original entry on oeis.org

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

Ahmet Caglar Saygili, Aug 24 2025

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).

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

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

Cf. A059153, A166751, A387270, A213370.

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)]

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

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

cp's OEIS Frontend

A386705 a(n) = sum of the 2^(n-1) even positive integers having bit length 2*n and in which, when written in binary, each run of 0's is of exactly the same length as the run of 1's immediately before it.

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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.

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