OEIS A119435
A sequence by Leroy Quet (2006 0519), this page written by Michael Thomas De Vlieger, St. Louis, Missouri, 2022 0420.
Abstract.
We are inspired by the curious scatterplot of A119435 and the tantalizing bitmaps of A353035 and A353036 to write a few simple theorems about each.
Introduction.
Let A = A119435. Let U(n, j) > 0 be the j-th smallest number missing from A(1..n−1), hence, the set Un contains m ∉ A(1…n−1). Let u = U(n, 1) be the smallest missing number in A(1..n−1), that is, a local minimum of A.
We may define A as follows:
A(n) = m = U(n, A030101(n)),
[Formula 1]
where A030101(n) is the bit-reversal of n. Example: A030101(12) = 3 since 12 expanded in binary is “1100”; reversing this gives us binary “0011”, i.e., 2¹ + 2⁰ = 2+1 = 3.
The sequence begins as follows (See Codes 1, 2, and 3, Figures 1 and 2):
1, 2, 5, 3, 9, 7, 13, 4, 17, 12, 23, 10, 22, 18, 29, 6, 33, 24, 43, 16, 40, 31, 51, 14, 41, 30, 53, 25, 49, 38, 61, 8, 65, 45, 83, 32, 76, 58, 95, 21, 74, 55, 94, 42, 87, 68, 107, 19, 78, 56, 100, 39, 91, 70, 113, 34, 89, 66, 112, 52, 104, 81, 125, 11, 129, 86, 163, 60, 148, 114, 183, 44, 142, 103, 178, 77, 162, 127, 199, 28, 143, 101, 181, 72, 165, 126, 203, 57, 157, 118, 197, 92, 180, 140, 219, 26, 151, 106, 193, 73, 174, 133, 216, 54, 168, 122, 211, 96, 192, 149, 233, 46, 169, 120, 214, 88, 195, 147, 237, 69, ...
A comment at A119435 claims the sequence is a permutation of the natural numbers. Sloane added a proof 20 April 2022 based on the logic of Theorem 2 that shows the sequence is infinite. Given this proof and the fact A is a lexically-eariest sequence (LES), we show that the sequence indeed is a permutation.
Axioms.
The first 2 axioms come about since A is a LES:
Lexical Axiom: A(n) = m : m ∈ ℕ ∧ m ∉ A(1…n−1).
Earliest Axiom:
If M > m and M also satisfies the other axioms, A(n) = m.
Selection Axiom: A(n) = U(n, A030301(n)).
Unused Axiom: A(n) = m ⇒Un \ {m}.
Consider a characteristic of all lexically earliest sequences. That is, the range of A(n) = m is divided into 3 intervals:
1. The saturated zone m < u. Here, m ∈ A(1…n−1), hence there is no need to consider such m.
2. The open zone m > r. Here, m ∉ A(1…n−1) by definition of record r. Therefore, iff we arrived at m through a greedy algorithm, i.e., satisfying all the other axioms and approaching from below, then A(n) = m.
3. The mixed zone u ≤ m < r. If we arrived at m through a greedy algorithm, we have to test whether m ∈ A(1…n−1). If false, A(n) = m. We may include r in this zone, though we know if m = r, then m ∈ A(1…n−1) and is banned from a second entry into A.
Theorem 1: In this sequence, both axioms entailed by lexically earliest approach are embodied in the Selection Axiom.
Proof 1: Suppose that U(n, A030301(n))) → A(n) = A(i) = m, i < n. Then A(i) =m ⇒ Un \ {m}, a contradiction. Also, suppose that U(n, A030301(n))) → M > m. This would entail that it were possible to find both M and m via U(n, A030301(n))), or that A030301(n) might have 2 values, both of which is clearly impossible. ∎
Consequently, Formula 1 is an adequate definition of the sequence.
Maxima and minima in A119435.
Theorem 2: A(2k) is a local minimum.
Proof 2: Observe that A030101(2k) = 1. 2k expressed in binary is 1 followed by zeros. When we reverse this number, the leading zeros are trivial and we read the number 1 in the 2⁰ place. Therefore we set A(n) = U(2k, 1), which by definition is the smallest missing number m ∉ A(1…n−1). ∎
Corollary 2.1: Since A(2k) = u, we may designate u instead as uk. Hence,un = u⌊log₂ n⌋ = uk for 2k ≤ n < 2(k+1).
We define A353035(n) = A(2n), the set of local minima in A. (See Code 5 and Figure 3.) Rémy Sigrist has expanded this to 45 terms:
1, 2, 3, 4, 6, 8, 11, 15, 20, 27, 36, 48, 64, 85, 116, 153, 208, 273, 366, 493, 649, 888, 1161, 1579, 2092, 2784, 3783, 4946, 6772, 8875, 11977, 16065, 21193, 28979, 37823, 51633, 68117, 91045, 123377, 161622, 221441, 289493, 392259, 523328, 692771, 945393, ...
We define A353036 as the set of records r in A, and A353037 the indices of records in A.
Therefore,
A(A353037(n)) = A353036(n) is the n-th record in A.
The set of records A353036 begins as follows (See Code 4 and Figure 4):
1, 2, 5, 9, 13, 17, 23, 29, 33, 43, 51, 53, 61, 65, 83, 95, 107, 113, 125, 129, 163, 183, 199, 203, 219, 233, 237, 253, 257, 323, 359, 383, 407, 419, 443, 449, 473, 485, 509, 513, 643, 711, 751, 783, 791, 823, 851, 859, 891, 913, 921, 953, 981, 989, 1021, 1025, ...
The sequence of their indices A353037 begins as follows (See Code 4 and Figure 5):
1, 2, 3, 5, 7, 9, 11, 15, 17, 19, 23, 27, 31, 33, 35, 39, 47, 55, 63, 65, 67, 71, 79, 87, 95, 111, 119, 127, 129, 131, 135, 143, 159, 175, 191, 207, 223, 239, 255, 257, 259, 263, 271, 287, 303, 319, 351, 367, 383, 415, 431, 447, 479, 495, 511, 513, 515, 519, 527, ...
The bitmaps seen in Figures 4 and 5, and their interpretation in Table 1 inspires us to attempt to prove a few things about the records transform of A119435.
Theorem 3: 2k±1 ∈ A353037 for k > 0.
Proof 3: m = 2k – 1 is a repunit, the largest k-bit number, thus A030101(m) = m. Let r = max(A(1…m−1). Since the function U(n, A030101(n)) appends a missing number to A at each iteration, any previous record r < U(m, m). Therefore, U(m, m) is the largest missing number and sets a new record in A. ∎
M = 2k + 1 is the second-smallest k-bit number with 1 in the least and most significant places and all the rest of the bits are 0, hence A030101(M) = M. The previous record r = U(m, m), since A(2k) = U(2k,1) is a local minimum. Since M > m, U(M, M) > U(m, m). ∎
Corollary 3.1: A(2k + 1) = A(2k − 1) + 4, i.e., U(M, M) = U(m, m) + 4, since M = 2k+1 = (2k − 1)+2 = m + 2, and 2 additional missing numbers entered A since A(2k − 2).
Theorem 4: A(2k − 1) + 3 = 2(k+1) ∧ A(2k + 1) − 1 = 2(k+1) : k > 1. This is tantamount to proving A(2k − 1) = 2(k+1) – 3 and A(2k + 1) = 2(k+1) + 1.
Through mechanics of A119435, the sequence begins as follows, highlighting A(2k):
1, 2, 5, 3, 9, 7, 13, 4, 17, 12, 23, 10, …
Proof 4: We note that A(2k + 1) = A(2k − 1) + 4 per Corollary 3.1. A(2²−1) = A(3) = 5, since U(3) = {3, …}, and since A030101(3) = 3, knowing U(3, 3) is a record, we have A(3) = U(3, 3) = 5. We have proved A(2k±1) set records in A by Theorem 3. Therefore it merely remains to show that A(2(k+1)±1) = A(2k±1) + 2(k+1). Indeed, via A(n) = m ⇒ Un\ {m}, we lose 1 element as n increments. But then n has increased as well each time. Hence we see that indeed, A(2k − 1) + 3 = 2(k+1) ∧ A(2k + 1) − 1 = 2(k+1) : k > 1. ∎
Corollary 4.1: (2(k+1) – 1) ± 2 ∈ A353036.
The nature of Un.
Let U(n, R) = r + 1, where r is the latest record in A. In other words, Rn is that index in the list Un of m ∉ A(1…n–1), that is, U(n, Rn), where U(n, Rn+ 1) − U(n, Rn) = 1, generally, U(n, Rn+ j + 1) − U(n, Rn+ j) = 1 for j ≥ 0.
The sequence R = {Rn} begins as follows (See Code 6.):
1, 1, 3, 2, 5, 4, 7, 6, 9, 8, 13, 12, 11, 10, 15, 14, 17, 16, 25, 24, 23, 22, 29, 28, 27, 26, 27, 26, 25, 24, 31, 30, 33, 32, 49, 48, 47, 46, 57, 56, 55, 54, 53, 52, 51, 50, 61, 60, 59, 58, 57, 56, 55, 54, 59, 58, 57, 56, 55, 54, 53, 52, 63, 62, 65, 64, 97, 96, 95, 94, 113, 112, 111, 110, 109, 108, 107, 106, 121, 120, 119, 118, 117, 116, 115, 114, 117, 116, 115, 114, 113, 112, 111, 110, 125, 124, 123, 122, 121, 120, 119, 118, 117, 116, 115, 114, 113, 112, 111, 110, 123, 122, 121, 120, 119, 118, 117, 116, 119, 118, ...
Figure 7 plots R in green against A in black and shows the most recent record r in red.
Theorem 5: U(n, Rn+ j) = r + j + 1. (See Figure 7).
Proof 5: By definition of record r = max(A(1…n−1)), it is clear that r + 1 = m ∉ A(1…n–1), and since r + j +1 > r + 1 for j ≥ 0, we see that we can describe the open zone of m ∉ A(1…n−1) through induction on j. ∎
Corollary 5.1: |Un | = ℵ₀. This, since Un = ℕ \ A(1…n−1), where | ℕ | = ℵ₀ and |A(1…n−1)| = n−1.
Corollary 5.2: We can partition Un into 2 zones, a finite Wn and an infinite Vn, using Rn.
- The open zone V(n, j) = U(n, Rn+ j − 1) : j ≥ 1 (resetting j from Theorem 5).
Vn ∋ {m > r : m ∉ A(1…n−1) ∧ r = max(A(1…n−1))}.
V(n, j+1) − V(n, j) = 1 : j > 1, hence the open zone Vn has first difference D = 1. Therefore, whereupon A(n) = m > r, Vn \ {r+1…m}. Since Vn = { (r+1)…∞ }, | Vn | = ℵ₀. - The mixed zone W(n, j) = U(n, j) : 1 ≤ j < Rn.
Wn ∋ {m ∉ A(1…n−1) : u ≤ m < r}, thus has finite cardinality (Rn−1).
The mixed zone has first differences D ≥ 1.
Corollary 5.3: R(n) decreases monotonically as n increases, unless A(n) = r' > r. If A(n) = r' > r, then R(n+d) = Rn+ (r' − r). Figure 7 shows how Rn (in green) decrements as n increments, unless we have a new record, moving the red line of r up; green also moves up. And it is clear that the blue line r − Rn = (n − 1), since 1 term m enters A as n increases, while Un loses the term m.
Corollary 5.4: Rn = r − n + 1, thus, sequence R + ℕ − 1 = A.
Corollary 5.5: A030301(n) ≥ Rn ⇒A(n) = r' > r. This follows from the definition of record and the definition of Rn.
Conjecture 6: We expand on Theorem 3, tantalized by Figure 6, to suggest the following regarding S = A353037:
2(k+1) − (A043569(A152948(n) − 1 … A152948(n+1) − 2) ∈ S : k ≥ 1.
This latter conjecture generates the following numbers that are known to be in A353037:
3
5 7
9 11 15
17 19 23 31
33 35 39 47 63
65 67 71 79 95 127
129 131 135 143 159 191 255
257 259 263 271 287 319 383 511
513 515 519 527 543 575 639 767 1023
...
A proof remains to be written, but likely has to do with the nature of Un, the values of local minima A(2k) and the sandwiching local maxima A(2k±1).
Conjecture 7: Let irregular triangle T(k, i) = A353037 partitioned by k = ⌊log₂ m⌋ : m ∈ A353037. Recall that A353037 is a list of indices of records in A. Theorem 3 proved 2k±1 ∈ A353037 for k > 0, therefore we begin row k with 2k+1 and end it with 2(k+1)−1.
The triangle begins:
0: 1
1: 2 3
2: 5 7
3: 9 11 15
4: 17 19 23 27 31
5: 33 35 39 47 55 63
6: 65 67 71 79 87 95 111 119 127
7: 129 131 135 143 159 175 191 207 223 239 255
8: 257 259 263 271 287 303 319 351 367 383 415 431 447 479 495 511
Row lengths ℓ of this triangle are as follows:
1, 2, 2, 3, 5, 6, 9, 11, 16, 20, 29, 37, 54, 70, 103, 135, 200, 264, 393, ...
We take first differences δ so as to note a pattern, prepending the row with the first term:
0: 1
1: 2 1
2: 5 2
3: 9 2 4
4: 17 2 4 4 4
5: 33 2 4 8 8 8
6: 65 2 4 8 8 8 16 8 8
7: 129 2 4 8 16 16 16 16 16 16 16
8: 257 2 4 8 16 16 16 32 16 16 32 16 16 32 16 16
So as to condense this table, we write only log₂ δ and drop the first column:
1: 0
2: 1
3: 12
4: 1222
5: 12333
6: 12333433
7: 1234444444
8: 123444544544544
9: 1234555555555555555
10: 1234555655655655655655655655
11: 12345666666666666666666666666666666
12: 12345666766766766766766766766766766766766766766766766
Hence it seems that we can write some method of generating A353037.
We can match 1360 terms of A353037 resulting from A(1…2¹⁹) using the following algorithm. The algorithm derives from observations of the tables immediately above and Figure 5.
1.) For k = 0, row k = {1}, and for k = 1, we prepend 2 to the algorithmically-generated row.
2.) Do loop using iterator k = 1.
a.) Set m = 2k, set counter c and exponent j to −1. Set maximum exponent e = ⌈k/2⌉.
b.) While log₂ m < k + 1, sow m, increment c, and if j < e, increment j. Set m = 2(j + x).
Note: x = [IsInteger(k/2+3) ∧ c ≥ k/2+3 ∧ c mod 3 ≡ (k/2) mod 3], Iverson brackets.
3. Reap the sown values of m to comprise row k.
Setting k = 18, we obtain a dataset T(1..1360) = A353037(1..1360), verified using T(n) XOR A353037(n). See Code 9.
A proof remains to be written that verifies this conjectural algorithm approach.
If the approach confirms indices of records in A119435, then the extended list of row lengths ℓ is as follows:
1, 2, 2, 3, 5, 6, 9, 11, 16, 20, 29, 37, 54, 70, 103, 135, 200, 264, 393, 521, 778, 1034, 1547, 2059, 3084, 4108, 6157, 8205, 12302, 16398, 24591, 32783, 49168, 65552, 98321, 131089, 196626, 262162, 393235, 524307, 786452, ...
This sequence does not appear in OEIS (and probably shouldn’t).
Conjecture 8: We can compute records r (in A353036) from their indices n (in A353037). It seems to follow that we might be able to generate records from their indices because of the axioms. Indeed, the sequence A119435 might either be seen as the recursive mapping of U(n, A030101(n)), or it may be seen as A(n) → U \ {m} ↦ A030101 : m ∈ A030101.
We base this conjecture on the bitmaps of A353036 and A353037, specifically, when they are overlaid as in Figure 6. The last-mentioned figure gives us the impression that the bits in r are a function of the bits in n, or, rather, that the algorithm that generates n (Code 9) could also produce r.
(This segment is in progress; we have found means of manipulating the bits in certain n to furnish r.)
Conclusion.
Therefore we may state the following facts:
- A119435(2k) = U(n, 1), ∴ A119435(2k) is a local minimum of A,
- 2k ± 1 ∈ A353037 : k > 0,
- A119435(2k + 1) = A119435(2k − 1) + 4,
- A119435(2k − 1) = 2(k+1) − 3 ∧ A119435(2k + 1) = 2(k+1) + 1 ∴ (2(k+1) − 1) ± 2 ∈ A353036.
We restate the above in plain English so as to be clear.
- A119435(2k) is the smallest missing number u, thus, A119435(2k) is a local minimum of A and appears in A353035.
- A119435(2k ± 1) with k > 0 are records and thus appear in A353037,
- A119435(2k + 1) = A119435(2k − 1) + 4,
- A119435(2k − 1) = 2(k+1) − 3 and A119435(2k + 1) = 2(k+1) + 1, therefore (2(k+1) − 1) ± 2 are records in A119435 and appear in A353036.
Furthermore it appears we have a means via Code 9 of generating A353037 independent of taking a records transform of A.
Some open questions:
- Is there a way to ascertain A119435(2k) = u, that is, the terms in A353035 without generating A119435? We are tantalized by being able to compute A119435(2k ± 1) and Rn given A353036 and A353037. Figure 3 doesn’t seem to help.
- Can we use Conjecture 6 and the patterns seen in Figures 5 through 8 to compute A(n) in a manner other than recursively mapping U(n, A030101(n))?
- If we have a means of computing A353036 in addition to A353037 using Conjecture 6 or the patterns mentioned above, is such sufficient to compute A?
Figure 6 seems to suggest that, given A353037(n), we ought to be able to generate A353036(n). If this is possible, then we would be able to generate the records transform of A without generating A itself. Furthermore, if we have both A353036 and A353037, we can generate R.
This concludes our examination.
Acknowledgement.
I would like to acknowledge Dr. Neil J. A. Sloane for mentioning A119435 on 17 April 2022 in a request to extend the b-file.
Appendix.
Table 1: Binary expansion of A353036(n) and A353037(n) for n = 1..56. Let ℓ = 1+⌊log₂ m⌋ where m is either A353036(n) or A353037(n). Then 2ℓ−1 is the largest number with the same number of bits as m. We define N = 2ℓ− m − 1. (See Code 7.)
n A353036(i) 2^L-1 N A353037(i) 2^L-1 N
-------------------------------------------------------------
1 1 1 1 0 1 1 1 0
2 2 1. 3 1 2 1. 3 1
3 5 1.1 7 2 3 11 3 0
4 9 1..1 15 6 5 1.1 7 2
5 13 11.1 15 2 7 111 7 0
6 17 1...1 31 14 9 1..1 15 6
7 23 1.111 31 8 11 1.11 15 4
8 29 111.1 31 2 15 1111 15 0
9 33 1....1 63 30 17 1...1 31 14
10 43 1.1.11 63 20 19 1..11 31 12
11 51 11..11 63 12 23 1.111 31 8
12 53 11.1.1 63 10 27 11.11 31 4
13 61 1111.1 63 2 31 11111 31 0
14 65 1.....1 127 62 33 1....1 63 30
15 83 1.1..11 127 44 35 1...11 63 28
16 95 1.11111 127 32 39 1..111 63 24
17 107 11.1.11 127 20 47 1.1111 63 16
18 113 111...1 127 14 55 11.111 63 8
19 125 11111.1 127 2 63 111111 63 0
20 129 1......1 255 126 65 1.....1 127 62
21 163 1.1...11 255 92 67 1....11 127 60
22 183 1.11.111 255 72 71 1...111 127 56
23 199 11...111 255 56 79 1..1111 127 48
24 203 11..1.11 255 52 87 1.1.111 127 40
25 219 11.11.11 255 36 95 1.11111 127 32
26 233 111.1..1 255 22 111 11.1111 127 16
27 237 111.11.1 255 18 119 111.111 127 8
28 253 111111.1 255 2 127 1111111 127 0
29 257 1.......1 511 254 129 1......1 255 126
30 323 1.1....11 511 188 131 1.....11 255 124
31 359 1.11..111 511 152 135 1....111 255 120
32 383 1.1111111 511 128 143 1...1111 255 112
33 407 11..1.111 511 104 159 1..11111 255 96
34 419 11.1...11 511 92 175 1.1.1111 255 80
35 443 11.111.11 511 68 191 1.111111 255 64
36 449 111.....1 511 62 207 11..1111 255 48
37 473 111.11..1 511 38 223 11.11111 255 32
38 485 1111..1.1 511 26 239 111.1111 255 16
39 509 1111111.1 511 2 255 11111111 255 0
40 513 1........1 1023 510 257 1.......1 511 254
41 643 1.1.....11 1023 380 259 1......11 511 252
42 711 1.11...111 1023 312 263 1.....111 511 248
43 751 1.111.1111 1023 272 271 1....1111 511 240
44 783 11....1111 1023 240 287 1...11111 511 224
45 791 11...1.111 1023 232 303 1..1.1111 511 208
46 823 11..11.111 1023 200 319 1..111111 511 192
47 851 11.1.1..11 1023 172 351 1.1.11111 511 160
48 859 11.1.11.11 1023 164 367 1.11.1111 511 144
49 891 11.1111.11 1023 132 383 1.1111111 511 128
50 913 111..1...1 1023 110 415 11..11111 511 96
51 921 111..11..1 1023 102 431 11.1.1111 511 80
52 953 111.111..1 1023 70 447 11.111111 511 64
53 981 1111.1.1.1 1023 42 479 111.11111 511 32
54 989 1111.111.1 1023 34 495 1111.1111 511 16
55 1021 11111111.1 1023 2 511 111111111 511 0
56 1025 1.........1 2047 1022 513 1........1 1023 510
Figure 1: Log-log scatterplot of A(n), n = 1..2¹², showing odd n in blue and even n in red.
Figure 2: Bitmap of A(n), n = 1..2¹⁰, 24× vertical exaggeration. Here we show 1 in black and 0 in white, with the least-significant bit at bottom.
Figure 3: Bitmap of A353035(n), n = 1..45, 4× exaggeration. Here we show 1 in black and 0 in white, with the least-significant bit at bottom.
Figure 4: Bitmap of A353036(n), n = 1..967, 12× vertical exaggeration. Here we show 1 in black and 0 in white, with the least-significant bit at bottom. (See Code 8.)
Figure 5: Bitmap of A353037(n), n = 1..967, 12× vertical exaggeration. Here we show 1 in black and 0 in white, with the least-significant bit at bottom. (Similar to Code 8.)
Figure 6: Bitmap of A353036(n) with 1s in red overlaid on A353037(n) with 1s in blue, n = 1..142, 4× exaggeration. We add the RGB values of the pixels such that where the two sequences agree, we have magenta.
Figure 7: Log-log scatterplot of A(n), n = 1..2¹⁰, overlaying the latest record r in red, R in green, and r − R in blue.
References:
Code 1: Syntactically simple list method:
a119435 = Block[{a = {1}, nn = 2^14},
Monitor[Do[
AppendTo[a, Complement[Range[i + 2 nn], a][[IntegerReverse[i, 2]]]],
{i, 2, nn}], i]; a];
Code 2: Gap incrementation method (slow):
Block[{nn = 2^12, a, c, j, k, r, m, u}, c[_] = 0; u = 1;
Monitor[Do[r = IntegerReverse[i, 2]; j = 1; k = u;
While[Nand[c[k] == 0, j == r], j += Boole[c[k] == 0]; k++];
Set[{a[i], c[k]}, {k, i}];
If[k == u, While[c[u] > 0, u++]], {i, nn}], i]; Array[a, nn]]
Code 3: Literal U(n, j) approach:
Block[{nn = 2^16, a, c, j, k, r, m, u, w}, c[_] = 0; u = 1;
w = Range[16];
Monitor[Do[
If[IntegerQ[#],
Set[w, DeleteCases[
Union[w, Range[Max[w], 2^(# + 2)]], _?(# < u &)]]] &@ Log2[i];
m = IntegerReverse[i, 2]; j = 1; k = w[[m]];
Set[{a[i], c[k]}, {k, i}];
Set[w, DeleteCases[w, k]];
If[k == u, While[c[u] > 0, u++]], {i, nn}], i]; Array[a, nn]]
Set[{a353036, a353037},
Block[{r = 0, s = a119435},
Transpose@ Reap[Do[If[# > r, r = #; Sow[{r, i}]] &@ s[[i]],
{i, Length@ s}]][[-1, -1]] ] ];
Code 5: Local minima transform:
Set[{a353035, a353035i},
Block[{u = 1, c, s = a119435}, c[_] = 0;
Transpose@ Reap[Do[Set[c[s[[i]]], i];
If[s[[i]] == u, Sow[{u, i}]; While[c[u] > 0, u++]],
{i, Length@ s}]][[-1, -1]]]];
Code 6: Calculate R from a dataset a119435.
a353036r = Block[{c, u, r, s = a119435[[1 ;; 2^10]]},
c[_] = 0; u = r = 0;
Reap[Do[c[s[[i]]] = i;
Which[# > r, r = #, # == u, While[c[u] > 0, u++]] &@ s[[i]];
Sow[Count[Range[u, r - 1], _?(c[#] == 0 &)]], {i, Length@ s}]][[-1, -1]]];
Table[{#1, #2, StringReplace[IntegerString[#2, 2], "0" -> "." ], #4 - 1, Total[2^(#5)], #6,
StringReplace[IntegerString[#6, 2], "0" -> "." ], #8 - 1, Total[2^(#9)]} & @@
{#1, #2, #3, 2^Length[#3], Position[#3, 0][[All, 1]] - 1, #4, #5,
2^Length[#5], Position[#5, 0][[All, 1]] - 1} & @@
{n, a353036[[n]], Reverse@ IntegerDigits[a353036[[n]], 2], a353037[[n]],
Reverse@ IntegerDigits[a353037[[n]], 2]}, {n, 56}] // TableForm
Code 8: Produce Figure 4 (for Figure 5, substitute “a353037” for “a353036”:
Block[{s = a353036, w},
w = Function[{t, m}, Map[PadLeft[#, m] &, t]] @@ {#, Max[Length /@ #]} &@
Array[IntegerDigits[s[[#]], 2] &, Length[s]];
ImageReflect@ ImageRotate@
ImageResize[ColorNegate@ Image@ w, {Scaled[12], All}, Resampling -> "Nearest"] ]
Code 9: Generate A353037 via Conjecture 7:
Block[{r, nn = 18, c, j, k, m},
Monitor[Do[c = j = -1; k = 2^i; m = Ceiling[i/2];
r[i] = Rest@ Reap[
While[Log2[k] < i + 1, Sow[k]; c++; If[j < m, j++]; k += 2^(j +
Boole[And[IntegerQ[#], c >= #, Mod[c, 3] == Mod[#, 3]]] &[i/2 + 3])]][[-1, -1]],
{i, nn}], i];
{1, 2}~Join~Flatten@ Array[r, nn]]
Concerns sequences:
A030101: Bit-reversal of n.
A043569: Numbers m whose binary expansion has exactly 2 runs.
A119435: A(n) = (binary reversal of n)-th integer among those positive integers not occurring earlier in the sequence.
A152948: A152948(n) = (n² − 3n + 6)/2.
A353035: Local minima in A; A353035(n) = A(2n).
A353036: Maxima in A.
A353037: Positions of maxima in A.
A353069: R(n) = index of (r+1) in Un.
Document Revision Record.
2022 0420 1545 Draft.
2022 0420 2115 Publication.
2022 0421 2200 Conjecture 7.
2022 0422 2215 Conjecture 8.