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A349221
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Triangle read by rows: T(n, k) = 1 if k divides binomial(n-1, k-1), T(n, k) = 0 otherwise (n >= 1, 1 <= k <= n).
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3
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1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0
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OFFSET
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1,1
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COMMENTS
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Similar to A054521 as gcd(n, k) = 1 => k divides binomial(n-1, k-1) but not equivalent as the converse is not true, the earliest example being T(10,4) where 4 divides binomial(9,3) = 84 but gcd(10,4) is not 1. Question: What characterizes the cases where this triangle differs from A054521?
The period of the k-th column is given by A349593(k-1, k) = k * Product_{prime p|k} p^(floor(log(k-1)/log(p))). - Jianing Song, Nov 29 2021
{T(n, k)} is the sum of triangles [k|binomial(n-1, k-1) AND gcd(n, k) = j], n >= 1, 1 <= k <= n, j >= 1, where [] is the Iverson bracket. For j > 1, bitmaps of these triangles suggest simpler fractal gaskets that combine to produce the "shadowing" effect observed in the bitmap of {T(n, k)} provided in the LINKS section. For prime j, the bitmaps suggest a fractal (Hausdorff) dimension of log(A000217(j)/log(j) = log(j(j + 1)/2)/log(j), which is the same as that of the gasket formed by taking the Pascal triangle (A007318) mod j (see Bondarenko reference). - Richard L. Ollerton, Dec 10 2021
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REFERENCES
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Bondarenko, B. A. Generalized Pascal Triangles and Pyramids. Santa Clara, Calif: The Fibonacci Association, 1993, pp. 130-132.
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LINKS
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Michael De Vlieger, Bitmap of rows 1 <= n <= 2^10, showing 1 as black and 0 as white.
Michael De Vlieger, Table of b(n) for n = 1..3322, where b(n) is the compactification of row n of a(n) as a binary number.
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FORMULA
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T(n, k) = [k|binomial(n-1, k-1)] = Sum_{j>=1} [k|binomial(n-1, k-1) AND gcd(n, k) = j], n >= 1, 1 <= k <= n, where [] is the Iverson bracket. (The j = 1 case is A054521.)
T(n, k) = T(n, n-k), n > 1, 1 <= k < n.
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EXAMPLE
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The triangle T(n, k) begins:
n\k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
1: 1
2: 1 0
3: 1 1 0
4: 1 0 1 0
5: 1 1 1 1 0
6: 1 0 0 0 1 0
7: 1 1 1 1 1 1 0
8: 1 0 1 0 1 0 1 0
9: 1 1 0 1 1 0 1 1 0
10: 1 0 1 1 0 1 1 0 1 0
11: 1 1 1 1 1 1 1 1 1 1 0
12: 1 0 0 0 1 1 1 0 0 0 1 0
13: 1 1 1 1 1 1 1 1 1 1 1 1 0
14: 1 0 1 0 1 0 0 0 1 0 1 0 1 0
15: 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0
...
Differences between this example and that for A054521 occur at (n,k) = (10,4), (10,6), and (12,6).
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MATHEMATICA
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Table[Boole[Mod[Binomial[n - 1, k - 1], k] == 0], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Nov 11 2021 *)
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PROG
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(PARI) row(n) = vector(n, k, !(binomial(n-1, k-1) % k)); \\ Michel Marcus, Nov 11 2021
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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