OFFSET
1,2
COMMENTS
Define binary weight wt(n) as A000120(n), the number of 1s in the binary expansion of n. Let w = A000120(a(n-1)) the binary weight of the previous term. In other words, a(n) is the least m not already in the sequence such that m mod 2^k = w, where k = floor(1 + log_2 w).
Likely a permutation of the natural numbers.
The numbers m = 2^k with 0 <= k <= 3 appear at indices {1, 3, 11, 222}. The term 16 has not appeared for n <= 2^14 and may not until n approaches 2^16.
The numbers m = (2^k + 1) appear at indices {2, 4, 12, 223, ...}. The numbers m = 2^k or (2^k + 1) require n approximately equal to 2^m in order to appear in the sequence.
The numbers m = (2^k - 1) with 1 <= k <= 14 appear at indices {1, 2, 8, 10, 23, 43, 130, 278, 447, 758, 1390, 2525, 4719, 9333}, respectively.
The plot exhibits dendritic streams of residues r (mod 2^k). We can identify coordinates (x, y) = (n, a(n)) on the plot where the streams branch.
The branches of the tree in the plot contain m congruent to r (mod 2^k), where r is a term (except the last term) in row (k-1) of A049773.
Given 2^14 terms of this sequence, we see 2 or 3 successive invocations of w, otherwise, w appears just once before a different value succeeds it in the next term.
2^4 appears at index 47201. - Michael S. Branicky, Dec 16 2020
A permutation of the integers since n appears at or before index 2^n - 1, the first number with binary weight n. - Michael S. Branicky, Dec 16 2020
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..16385
Michael De Vlieger, Annotated plot (n, a(n)) for 1 <= n <= 2^8, color coded to show k (mod 8).
Michael De Vlieger, Enlarged plot (n, a(n)) for 1 <= n <= 2^14, color coded to show k (mod 8).
Michael De Vlieger, Log-log plot (n, a(n)) for 1 <= n <= 2^14, color coded to show k (mod 8), showing branching points.
Michael De Vlieger, Persistence plot (n, a(n)) for 1 <= n <= 2^8, color coded to show consecutive repetitions of the same binary weight.
Wikipedia, Hamming weight
Wolfram Research, Numbers in Pascal's triangle
EXAMPLE
a(2) = 3 since the binary weight of 1 is 1, and 3 = 1 (mod 2^1).
a(3) = 2 since wt(3) = 2, and 2 = 2 (mod 2^2).
a(4) = 5 since wt(2) = 1, 5 = 1 (mod 2^1), etc.
MATHEMATICA
Nest[Append[#, Block[{k = 1, r = DigitCount[#[[-1]], 2, 1], s}, s = IntegerLength[r, 2]; While[Nand[FreeQ[#, k], Mod[k, 2^s] == r], k++]; k]] & @@ {#, Length@ #} &, {1}, 2^7]]
PROG
(Python)
def aupto(n):
alst, used = [1], {1}
for i in range(2, n+1):
binprev = bin(alst[-1])[2:]
binwt = binprev.count("1")
pow2 = 2**(len(bin(binwt))-2)
while binwt in used: binwt += pow2
alst.append(binwt); used.add(binwt)
return alst # use alst[n-1] for a(n)
print(aupto(66)) # Michael S. Branicky, Dec 16 2020
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Michael De Vlieger, Dec 16 2020
STATUS
approved