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A095140
Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 5.
15
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, 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, 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
OFFSET
0,5
COMMENTS
{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
REFERENCES
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
LINKS
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.
Ashley Melia 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 5.
MATHEMATICA
Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 5]
PROG
(Python)
from math import isqrt, comb
def A095140(n):
def f(m, k):
if m<5 and k<5: return comb(m, k)%5
c, a = divmod(m, 5)
d, b = divmod(k, 5)
return f(c, d)*f(a, b)%5
return f(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)), n-comb(r+1, 2)) # Chai Wah Wu, Apr 30 2025
CROSSREFS
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).
Sequence in context: A370399 A095141 A177974 * A225043 A125605 A110570
KEYWORD
easy,nonn,tabl
AUTHOR
Robert G. Wilson v, May 29 2004
STATUS
approved