OFFSET
0,5
COMMENTS
Start with [1], repeatedly apply the map 0 -> [000/000/000], 1 -> [111/120/100], 2 -> [222/210/200]. - Philippe Deléham, Apr 16 2009
{T(n,k)} is a fractal gasket with fractal (Hausdorff) dimension log(A000217(3))/log(3) = log(6)/log(3) = 1.63092... (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
B. A. Bondarenko, Generalized Pascal Triangles and Pyramids, Santa Clara, Calif.: The Fibonacci Association, 1993, pp. 130-132.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
LINKS
Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
J.-P. Allouche, F. von Haeseler, H.-O. Peitgen, and G. Skordev, Linear cellular automata, finite automata and Pascal's triangle, Disc. Appl. Math. 66 (1996) 1-22.
Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
Lin Jiu and Christophe Vignat, On Binomial Identities in Arbitrary Bases, arXiv:1602.04149 [math.CO], 2016.
Y. Moshe, The density of 0's in recurrence double sequences, J. Number Theory, 103 (2003), 109-121.
Y. Moshe, The distribution of elements in automatic double sequences, Discr. Math., 297 (2005), 91-103.
A. M. Reiter, Determining the dimension of fractals generated by Pascal's triangle, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
FORMULA
T(i, j) = binomial(i, j) mod 3.
T(n+1,k) = (T(n,k) + T(n,k-1)) mod 3. - Reinhard Zumkeller, Jul 11 2013
T(n,k) = Product_{i>=0} binomial(n_i,k_i) mod 3, where n = Sum_{i>=0} n_i*3^i and k = Sum_{i>=0} k_i*3^i, 0<=n_i, k_i <=2 [Allouche et al.]. - R. J. Mathar, Jul 26 2017
EXAMPLE
. Rows 0 .. 3^3:
. 0: 1
. 1: 1 1
. 2: 1 2 1
. 3: 1 0 0 1
. 4: 1 1 0 1 1
. 5: 1 2 1 1 2 1
. 6: 1 0 0 2 0 0 1
. 7: 1 1 0 2 2 0 1 1
. 8: 1 2 1 2 1 2 1 2 1
. 9: 1 0 0 0 0 0 0 0 0 1
. 10: 1 1 0 0 0 0 0 0 0 1 1
. 11: 1 2 1 0 0 0 0 0 0 1 2 1
. 12: 1 0 0 1 0 0 0 0 0 1 0 0 1
. 13: 1 1 0 1 1 0 0 0 0 1 1 0 1 1
. 14: 1 2 1 1 2 1 0 0 0 1 2 1 1 2 1
. 15: 1 0 0 2 0 0 1 0 0 1 0 0 2 0 0 1
. 16: 1 1 0 2 2 0 1 1 0 1 1 0 2 2 0 1 1
. 17: 1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1
. 18: 1 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 1
. 19: 1 1 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 1 1
. 20: 1 2 1 0 0 0 0 0 0 2 1 2 0 0 0 0 0 0 1 2 1
. 21: 1 0 0 1 0 0 0 0 0 2 0 0 2 0 0 0 0 0 1 0 0 1
. 22: 1 1 0 1 1 0 0 0 0 2 2 0 2 2 0 0 0 0 1 1 0 1 1
. 23: 1 2 1 1 2 1 0 0 0 2 1 2 2 1 2 0 0 0 1 2 1 1 2 1
. 24: 1 0 0 2 0 0 1 0 0 2 0 0 1 0 0 2 0 0 1 0 0 2 0 0 1
. 25: 1 1 0 2 2 0 1 1 0 2 2 0 1 1 0 2 2 0 1 1 0 2 2 0 1 1
. 26: 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1
. 27: 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 .
- Reinhard Zumkeller, Jul 11 2013
MAPLE
A083093 := proc(n, k)
modp(binomial(n, k), 3) ;
end proc:
seq(seq(A083093(n, k), k=0..n), n=0..10) ; # R. J. Mathar, Jul 26 2017
MATHEMATICA
Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 3] (* Robert G. Wilson v, Jan 19 2004 *)
PROG
(Haskell)
a083093 n k = a083093_tabl !! n !! k
a083093_row n = a083093_tabl !! n
a083093_tabl = iterate
(\ws -> zipWith (\u v -> mod (u + v) 3) ([0] ++ ws) (ws ++ [0])) [1]
-- Reinhard Zumkeller, Jul 11 2013
(Magma) /* As triangle: */ [[Binomial(n, k) mod 3: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 15 2016
(Python)
from sympy import binomial
def T(n, k):
return binomial(n, k) % 3
for n in range(21): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Jul 26 2017
CROSSREFS
Cf. A007318, A051638 (row sums), A090044, A047999, A034931, A034930, A008975, A034932, A062296, A006047.
Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), (this sequence) (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).
KEYWORD
AUTHOR
Benoit Cloitre, Apr 22 2003
STATUS
approved