OFFSET
1,2
COMMENTS
We define binary weight wt(n) = A000120(n) as the number of 1s in n_2, the number n expressed in binary. Let w = wt(a(n-1)) the binary weight of the previous term, where w_2 is w expressed in binary, and let interval I(j) = 2^j <= n <= (2^(j+1)-1).'
Likely a permutation of the natural numbers.
The plot (n, a(n)) is organized into streaky clouds that pertains to a "family" M(i) <= m < M(i+1) whose binary expansion begins with an odd "prefix" m/2^v, where v is the 2-adic valuation of m. There are thus 2^v numbers in this range.
The numbers in this range accommodate the binary weights wt(a(n-1)) = w with 1 <= w <= ceiling(log_2 a(n-1)) such that w_2 appears in part or all of the binary expansion of the prefix m/2^v, and perhaps an additional bit in m after the prefix.
Small values of w, for instance w = 1, may appear in any family, but large w require the entire prefix and potentially more (if even).
The w that cannot be found in a particular family are found in a different family that has M(i+1) as its least member.
The families M(i) belong in turn to classes according to odd prefixes. Thus, for example, we may find w = 1, 2, 4, and 9 in class 9, since "1", "10", "100", and "1001" can be found in numbers m that begin, "1001...".
For w in interval I(j), we have values 1 <= k <= j - 1 distributed binomially.
Permutation of the natural numbers. We can always find w in a number m in family M(i) that pertains to a class C of numbers that in binary start with the binary expansion of an odd number c.
Numbers m that begin with numbers that are formed of left-trimmed bits of c exhaust the numbers in M(i) before moving to M(i+1) in the same class C.
When we have recordsetting odd w, a new class C opens up based on the binary expansion of a larger odd number c.
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..16384
Michael De Vlieger, Plot (n, a(n)) for 1 <= n <= 2^10 color-coded to show wt(a(n-1)), with the first term in the family indicated.
Michael De Vlieger, Plot (n, a(n)/A007814(a(n)) for 1 <= n <= 2^11, color-coded to show wt(a(n-1)).
Wikipedia, Hamming weight
Wolfram Research, Numbers in Pascal's triangle
EXAMPLE
Let wt(n) = A000120(n).
a(2) = 2 since wt(a(1)) = wt(1) = 1, and we find "1" at the beginning of the binary expansion of the yet unused 2 = "10"_2.
a(3) = 3 since wt(2) = 1, we find "1" as first bit of yet unused 3 = "11"_2.
a(4) = 4 since wt(3) = 2 = "10"_2, we find "10" as first bits of yet unused 4 = "100"_2.
a(5) = 5 since wt(4) = 1, and yet unused 5 = "101"_2 starts with 1.
a(6) = 8 since wt(5) = 2 = "10"_2; we see that the yet unused 6 and 7 start with "11"_2, and it isn't until 8 that we have a number that when expressed in binary starts with "10"_2.
a(7) = 6 since wt(8) = 1, we can now apply the yet unused 6 = "110"_2 because it starts with 1, etc.
MATHEMATICA
Nest[Append[#, Block[{k = 1, r = IntegerDigits[DigitCount[#[[-1]], 2, 1], 2]}, While[Nand[FreeQ[#, k], Take[IntegerDigits[k, 2], Length@ r] == 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")
lsbs, extra = 0, 0
while binwt + extra in used:
lsbs += 1
binwt *= 2
for extra in range(2**lsbs):
if binwt + extra not in used: break
alst.append(binwt+extra); used.add(binwt+extra)
return alst # use alst[n-1] for a(n)
print(aupto(68)) # Michael S. Branicky, Dec 16 2020
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Michael De Vlieger, Dec 16 2020
STATUS
approved