A095140 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 5.
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
References
- Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
Links
- Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
- 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.
- 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, A095141, A095142, A034930, A095143, A008975, A095144, A095145, A034932.
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).
Programs
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Mathematica
Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 5]
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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
Formula
T(i, j) = binomial(i, j) mod 5.
Comments