A386441 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 27.

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 8, 8, 21, 7, 1, 1, 8, 1, 2, 16, 2, 1, 8, 1, 1, 9, 9, 3, 18, 18, 3, 9, 9, 1, 1, 10, 18, 12, 21, 9, 21, 12, 18, 10, 1, 1, 11, 1, 3, 6, 3, 3, 6, 3, 1, 11, 1, 1, 12, 12, 4, 9, 9, 6, 9, 9, 4, 12, 12, 1, 1, 13, 24, 16, 13, 18, 15, 15, 18, 13, 16, 24, 13, 1

Offset: 0

Author

Chai Wah Wu, Jul 21 2025

Keywords

• nonn
• tabl
• easy

Examples

			Triangle begins:
               1;
             1,  1;
           1,  2,  1;
         1,  3,  3,  1;
       1,  4,  6,  4,  1;
     1,  5,  10,  10,  5,  1;
   1,  6,  15,  20,  15,  6,  1;
 1,  7,  21,  8,   8,  21,  7,  1;
  ...
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)
• Kenneth S. Davis and William A. Webb, Lucas' Theorem for Prime Powers, Europ. J. Combinatorics, Vol. 11, No. 3 (1990), 229-233.

Crossrefs

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

Programs

Mathematica

T[i_,j_]:=Mod[Binomial[i,j],27]; Table[T[n,k],{n,0,13},{k,0,n}]//Flatten (* Stefano Spezia, Jul 22 2025 *)

Python

from math import isqrt, comb
from sympy import multiplicity
from gmpy2 import digits
def A386441(n):
    def g1(s,w,e):
        c, d = 1, 0
        if len(s) == 0: return c, d
        a, b = int(s,3), int(w,3)
        if a>=b:
            k = comb(a,b)%27
            j = multiplicity(3,k)
            d += j*e
            k = k//3**j
            c = c*pow(k,e,27)%27
        else:
            if int(s[0:1],3)4: return 0
    s = s.zfill(3)
    w = w.zfill(l:=len(s))
    c, d = g1(s[:3],w[:3],1)
    for i in range(1,l-2):
        c0, d0 = g1(s[i:i+3],w[i:i+3],1)
        c1, d1 = g1(s[i:i+2],w[i:i+2],-1)
        c = c*c0*c1%27
        d += d0+d1
    return c*3**d%27

Formula

T(i, j) = binomial(i, j) mod 27.