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%I #63 Jan 27 2022 03:07:08
%S 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,
%T 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,
%U 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
%N 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).
%C 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?
%C 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
%C {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
%D Bondarenko, B. A. Generalized Pascal Triangles and Pyramids. Santa Clara, Calif: The Fibonacci Association, 1993, pp. 130-132.
%H Michael De Vlieger, <a href="/A349221/b349221.txt">Table of n, a(n) for n = 1..11325</a> (rows 1 <= n <= 150, flattened)
%H Michael De Vlieger, <a href="/A349221/a349221.png">Bitmap</a> of rows 1 <= n <= 2^10, showing 1 as black and 0 as white.
%H Michael De Vlieger, <a href="/A349221/a349221.txt">Table of b(n)</a> for n = 1..3322, where b(n) is the compactification of row n of a(n) as a binary number.
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F 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.)
%F T(n, k) = T(n, n-k), n > 1, 1 <= k < n.
%e The triangle T(n, k) begins:
%e n\k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
%e 1: 1
%e 2: 1 0
%e 3: 1 1 0
%e 4: 1 0 1 0
%e 5: 1 1 1 1 0
%e 6: 1 0 0 0 1 0
%e 7: 1 1 1 1 1 1 0
%e 8: 1 0 1 0 1 0 1 0
%e 9: 1 1 0 1 1 0 1 1 0
%e 10: 1 0 1 1 0 1 1 0 1 0
%e 11: 1 1 1 1 1 1 1 1 1 1 0
%e 12: 1 0 0 0 1 1 1 0 0 0 1 0
%e 13: 1 1 1 1 1 1 1 1 1 1 1 1 0
%e 14: 1 0 1 0 1 0 0 0 1 0 1 0 1 0
%e 15: 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0
%e ...
%e Differences between this example and that for A054521 occur at (n,k) = (10,4), (10,6), and (12,6).
%t Table[Boole[Mod[Binomial[n - 1, k - 1], k] == 0], {n, 12}, {k, n}] // Flatten (* _Michael De Vlieger_, Nov 11 2021 *)
%o (PARI) row(n) = vector(n, k, !(binomial(n-1,k-1) % k)); \\ _Michel Marcus_, Nov 11 2021
%Y Cf. A007318, A051731, A054521.
%K nonn,tabl,look
%O 1,1
%A _Richard L. Ollerton_, Nov 11 2021
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