%I
%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
%N Triangle formed by reading Pascal's triangle (A007318) mod 11.
%H Robert Israel, <a href="/A095144/b095144.txt">Table of n, a(n) for n = 0..10010</a> (rows 0 to 140, flattened)
%F T(i, j) = binomial(i, j) (mod 11).
%F From _Robert Israel_, Jan 02 2019: (Start)
%F T(n,k) = T(n1,k1) + T(n1,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 base11 digits of n and k. (End)
%p R[0]:= 1:
%p for n from 1 to 20 do
%p R[n]:= op([R[n1],0] + [0,R[n1]] 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), 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
