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

0,5

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.

Formula

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

Example

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;
  ...

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 *)

Prog

(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)<int(w[0:1], 3): d += e
            c0, d0 = g1(s[1:], w[1:], e)
            c = c*c0%27
            d += d0
        return c, d
    g = (m:=isqrt(f:=n+1<<1))-(f<=m*(m+1))
    k = n-comb(g+1, 2)
    s, w = digits(g, 3), digits(k, 3)
    if sum(int(d) for d in w)+sum(int(d) for d in digits(g-k, 3))-sum(int(d) for d in s)>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

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).

Sequence in context: A034932 A374378 A180183 * A273914 A094495 A374452

Adjacent sequences: A386438 A386439 A386440 * A386442 A386443 A386444

Keyword

nonn,tabl,easy

Author

Chai Wah Wu, Jul 21 2025

Status

approved