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A343071
Irregular triangle read by rows: T(n,k) = 2^(2n+1) * L(n,k), where L(n,k) is the k-th left endpoint after the n-th step of removal in the construction of the Smith-Volterra-Cantor set (SVC), 0 <= k <= 2^n - 1.
2
0, 0, 5, 0, 7, 20, 27, 0, 11, 28, 39, 80, 91, 108, 119, 0, 19, 44, 63, 112, 131, 156, 175, 320, 339, 364, 383, 432, 451, 476, 495, 0, 35, 76, 111, 176, 211, 252, 287, 448, 483, 524, 559, 624, 659, 700, 735, 1280, 1315, 1356, 1391, 1456, 1491, 1532, 1567, 1728, 1763, 1804, 1839, 1904, 1939, 1980, 2015
OFFSET
0,3
COMMENTS
The SVC (or fat Cantor set) is a subset of [0, 1] similar to the standard Cantor set, constructed as follows:
We start from C_0 = [0, 1], which is written as [0/2, 2/2] for convenience.
- Step 1: Remove the middle open interval of length 1/4 in C_0 (namely (3/8, 5/8)), leaving the union of 2 closed intervals C_1 = [0/8, 3/8] U [5/8, 8/8];
- Step 2: Remove the middle open interval of length 1/4^2 in each of the closed interval of C_1 (namely (5/32, 7/32) and (27/32, 29/32)), leaving the union of 4 closed intervals C_2 = [0/32, 5/32] U [7/32, 12/32] U [20/32, 27/32] U [29/32, 32/32].
...
- Step n: Remove the middle open interval of length 1/4^n in each of the closed interval of C_{n-1}, leaving the union of 4 closed intervals C_n.
The SVC is thereby defined as the intersection of all {C_n} for n >= 0.
After the n-th step, the k-th left closed interval (from left to right) is [L(n,k), R(n,k)] = [T(n,k)/2^(2n+1), A343072(n,k)/2^(2n+1)].
Like the standard Cantor set, the SVC is perfect (i.e., closed and every point inside is an accumulation point of itself) and has empty interior. The difference between the SVC and the standard Cantor set is that the SVC has positive Lebesgue measure, namely 1 - (1/4) - (1/4^2)*2 - (1/4^3)*2^2 - ... = 1/2.
The construction of the famous Volterra's function (a function differentiable everywhere on [0, 1] whose derivative is bounded yet not Riemann integratable) is based on the SVC.
LINKS
FORMULA
A343072(n,k) - T(n,k) = 2^n + 1, which corresponds to the fact that each closed interval of C_n is of length (2^n + 1)/2^(2n+1).
For n >= 0, 0 <= k <= 2^(n-1) - 1, T(n,2k) = 4*T(n-1,k), T(n,2k+1) = 4*A343072(n-1,k) - (2^n + 1) = 4*T(n-1,k) + (2^n + 3).
For k = Sum_{i=0..n-1} (b_i) * 2^i, b_i = 0 or 1, T(n,k) = Sum_{i=0..n-1} (b_i) * 4^i * (2^(n-i) + 3).
T(n,k) = 2^(2n+1) - A343072(n,(2^n-1)-k).
EXAMPLE
Table begins
0,
0, 5,
0, 7, 20, 27,
0, 11, 28, 39, 80, 91, 108, 119,
0, 19, 44, 63, 112, 131, 156, 175, 320, 339, 364, 383, 432, 451, 476, 495,
0, 35, 76, 111, 176, 211, 252, 287, 448, 483, 524, 559, 624, 659, 700, 735, 1280, 1315, 1356, 1391, 1456, 1491, 1532, 1567, 1728, 1763, 1804, 1839, 1904, 1939, 1980, 2015,
...
After the n-th step of construction, we have
C_0 = [0/2, 2/2],
C_1 = [0/8, 3/8] U [5/8, 8/8],
C_2 = [0/32, 5/32] U [7/32, 12/32] U [20/32, 27/32] U [29/32, 32/32],
C_3 = [0/128, 9/128] U [11/128, 20/128] U [28/128, 37/128] U [39/128, 48/128] U [80/128, 89/128] U [91/128, 100/128] U [108/128, 117/128] U [119/128, 128/128],
...
PROG
(PARI) T(n, k) = my(t=0); for(i=0, n-1, t+=(k%2)*4^i*(2^(n-i)+3); k=k\2); t
CROSSREFS
Cf. A343072 (the right endpoints).
Sequence in context: A084248 A201417 A147666 * A215892 A200643 A200231
KEYWORD
nonn,tabf,easy
AUTHOR
Jianing Song, Apr 04 2021
STATUS
approved