A095144 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 11.

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, 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, 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

Offset: 0

Author

Robert G. Wilson v, May 29 2004

Keywords

• easy
• nonn
• tabl

Comments

{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

References

• Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Links

• Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)
• Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see pp. 130-132.
• Zubeyir Cinkir and Aysegul Ozturkalan, An extension of Lucas Theorem, arXiv:2309.00109 [math.NT], 2023. See Figures 3 and 4 p. 6.
• Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
• A. M. Reiter, Determining the dimension of fractals generated by Pascal's triangle, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
• Index entries for triangles and arrays related to Pascal's triangle

Crossrefs

Cf. A007318, A047999, A083093, A034931, A095140, A095141, A095142, A034930, A095143, A008975, A095145, A034932.

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).

Programs

Maple

R[0]:= 1:
for  n from 1 to 20 do
  R[n]:= op([R[n-1],0] + [0,R[n-1]] mod 11);
od:
for n from 0 to 20 do R[n] od; # Robert Israel, Jan 02 2019

Mathematica

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

Python

from math import isqrt, comb
def A095144(n):
    def f(m,k):
        if m<11 and k<11: return comb(m,k)%11
        c,a = divmod(m,11)
        d,b = divmod(k,11)
        return f(c,d)*f(a,b)%11
    return f(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Apr 30 2025

Formula

T(i, j) = binomial(i, j) mod 11.

From Robert Israel, Jan 02 2019: (Start)

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

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)