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 A339607 a(1) = 1, a(n) is the least m not already in the sequence that contains the binary expansion of the binary weight of a(n-1) anywhere within its own binary expansion. 4
 1, 2, 3, 4, 5, 6, 8, 7, 11, 12, 9, 10, 13, 14, 15, 16, 17, 18, 19, 22, 23, 20, 21, 24, 25, 26, 27, 28, 29, 32, 30, 33, 34, 35, 31, 37, 38, 39, 36, 40, 41, 43, 44, 45, 48, 42, 46, 49, 47, 52, 50, 51, 56, 53, 57, 60, 64, 54, 65, 55, 58, 66, 59, 61, 69, 62, 74, 63 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: permutation of the natural numbers. 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, Plot of (n, b(n)) with b(n) = a(n)-n for 1 <= n <= 2^14, highlighting and labeling maxima and local minima in b(n). Michael De Vlieger, Plot of (n, b(n)) with b(n) = a(n)-n for 1 <= n <= 2^14, color function indicating wt(a(n-1)). Michael De Vlieger, Plot of (n, b(n)) with b(n) = a(n)-n for 1 <= n <= 2^14, color function representing degree of consecutive repetition (persistence) of binary weight. 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 both first and last bit of yet unused 3 = "11"_2. a(4) = 4 since wt(3) = 2 = "10"_2, we find yet unused 4 = "100"_2 starts with "10"_2. a(5) = 5 since wt(4) = 1, and yet unused 5 = "101"_2 both starts and ends with 1. a(6) = 6 since wt(5) = 2 = "10"_2, we find yet unused 6 = "110"_2 ends with "10"_2. a(7) = 8 since wt(6) = 2 = "10"_2, we find that the least unused m = 7 only contains 1s in binary. The next term m = 8 furnishes "10"_2 at the start of its binary expansion "1000"_2. a(8) = 7 since wt(8) = 1, we find "1" in three places in the least unused number m = 7 = "111"_2. a(9) = 11 since wt(7) = 3 = "11"_2, The next unused numbers 9 and 10 are written "1001"_2 and "1010"_2, respectively. Only when we reach m = 11 = "1011"_2 do we find an unused binary number that contains the word "11"_2, etc. MATHEMATICA Nest[Append[#, Block[{k = 1, r = IntegerDigits[DigitCount[#[[-1]], 2, 1], 2]}, While[Nand[FreeQ[#, k], SequenceCount[IntegerDigits[k, 2], r] > 0], k++]; k]] & @@ {#, Length@ #} &, {1}, 2^7] PROG (Python) def aupto(n):   alst, used = , {1}   for i in range(2, n+1):     binprev = bin(alst[-1])[2:]     binwt = binprev.count("1")     targetstr = bin(binwt)[2:]     morebits, extra, ai = 0, 0, binwt     while ai in used:       morebits += 1       found = False       for k in range(2**morebits):         binstrk = bin(k)[2:]         binstrk = "0"*(morebits-len(binstrk)) + binstrk # pad to length         for msbs in range(morebits+1):           trystr = binstrk[:msbs] + targetstr + binstrk[msbs:]           if trystr == "0": continue           trynum = int(trystr, 2)           if trynum not in used:             if not found: ai = trynum; found = True             else: ai = min(ai, trynum)       if found: break     alst.append(ai); used.add(ai)   return alst    # use alst[n-1] for a(n) print(aupto(68)) # Michael S. Branicky, Dec 16 2020 CROSSREFS Cf. A000120, A338209, A339024. Sequence in context: A297440 A232895 A274607 * A262374 A299442 A299440 Adjacent sequences:  A339604 A339605 A339606 * A339608 A339609 A339610 KEYWORD nonn,base,easy AUTHOR Michael De Vlieger, Dec 16 2020 STATUS approved

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Last modified June 13 00:04 EDT 2021. Contains 344980 sequences. (Running on oeis4.)