A095144 - OEIS

%I #40 Feb 06 2025 23:44:56

%S 1,1,1,1,2,1,1,3,3,1,1,4,6,4,1,1,5,10,10,5,1,1,6,4,9,4,6,1,1,7,10,2,2,

%T 10,7,1,1,8,6,1,4,1,6,8,1,1,9,3,7,5,5,7,3,9,1,1,10,1,10,1,10,1,10,1,

%U 10,1,1,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,2,1,0,0,0,0,0,0,0,0,1,2,1

%N Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 11.

%C {T(n,k)} is a fractal gasket with fractal (Hausdorff) dimension log(A000217(11))/log(11) = log(66)/log(11) = 1.74722... (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 Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

%H Robert Israel, <a href="/A095144/b095144.txt">Table of n, a(n) for n = 0..10010</a> (rows 0 to 140, flattened)

%H Boris A. Bondarenko, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/pascal.html">Generalized Pascal Triangles and Pyramids</a>, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see pp. 130-132.

%H Zubeyir Cinkir and Aysegul Ozturkalan, <a href="https://arxiv.org/abs/2309.00109">An extension of Lucas Theorem</a>, arXiv:2309.00109 [math.NT], 2023. See Figures 3 and 4 p. 6.

%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://web.archive.org/web/2024*/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 11.

%F From _Robert Israel_, Jan 02 2019: (Start)

%F T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod 11 with T(n,0) = 1.

%F T(n,k) = (Product_i binomial(n_i, k_i)) mod 11, where n_i and k_i are the base-11 digits of n and k. (End)

%p R[0]:= 1:

%p for  n from 1 to 20 do

%p   R[n]:= op([R[n-1],0] + [0,R[n-1]] mod 11);

%p od:

%p for n from 0 to 20 do R[n] od; # _Robert Israel_, Jan 02 2019

%t Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 11]

%Y Cf. A007318, A047999, A083093, A034931, A095140, A095141, A095142, A034930, A095143, A008975, 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), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930 (m = 8), A095143 (m = 9), A008975 (m = 10), (this sequence) (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).

%K easy,nonn,tabl,changed

%O 0,5

%A _Robert G. Wilson v_, May 29 2004