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%I #20 Dec 17 2021 20:35:59
%S 1,1,1,1,2,1,1,3,3,1,1,4,1,4,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,2,1,0,0,1,
%T 2,1,1,3,3,1,0,1,3,3,1,1,4,1,4,1,1,4,1,4,1,1,0,0,0,0,2,0,0,0,0,1,1,1,
%U 0,0,0,2,2,0,0,0,1,1,1,2,1,0,0,2,4,2,0,0,1,2,1,1,3,3,1,0,2,1,1,2,0,1,3,3,1
%N Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 5.
%C {T(n,k)} is a fractal gasket with fractal (Hausdorff) dimension log(A000217(5))/log(5) = log(15)/log(5) = 1.68260... (see Reiter reference). Replacing values greater than 1 with 1 produces a binary gasket with the same dimension (see Bondarenko reference). - _Richard L. Ollerton_, Dec 14 2021
%D B. A. Bondarenko, Generalized Pascal Triangles and Pyramids, Santa Clara, Calif.: The Fibonacci Association, 1993, pp. 130-132.
%H Ilya Gutkovskiy, <a href="/A275198/a275198.pdf">Illustrations (triangle formed by reading Pascal's triangle mod m)</a>
%H A. M. Reiter, <a href="https://www.fq.math.ca/Issues/31-2.pdf">Determining the dimension of fractals generated by Pascal's triangle</a>, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F T(i, j) = binomial(i, j) mod 5.
%t Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 5]
%Y Cf. A007318, A047999, A083093, A034931, A095141, A095142, A034930, A095143, A008975, A095144, A095145, A034932.
%Y Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), (this sequence) (m = 5), A095141 (m = 6), A095142 (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).
%K easy,nonn,tabl
%O 0,5
%A _Robert G. Wilson v_, May 29 2004
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