A385741 a(n) = Sum_{k=0..n} (binomial(n, k) mod 9).

1, 2, 4, 8, 16, 14, 28, 38, 31, 8, 16, 32, 28, 56, 49, 62, 52, 68, 28, 56, 76, 62, 79, 122, 91, 92, 112, 8, 16, 32, 28, 56, 76, 80, 124, 140, 28, 56, 103, 80, 142, 158, 145, 146, 184, 62, 124, 158, 100, 146, 184, 188, 232, 230, 28, 56, 76, 80, 151, 158, 136, 236

Offset: 0

Author

Chai Wah Wu, Jul 09 2025

Keywords

• nonn

Comments

Sum of n-th row of Pascal's triangle mod 9, A095143.

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..10000
• James G. Huard, Blair K. Spearman, and Kenneth S. Williams, Pascal's triangle (mod 9), Acta Arithmetica, 78 (1997), 331-349.

Crossrefs

Cf. A001316, A051638, A384715, A385285.

Programs

Mathematica

a[n_]:=Sum[Mod[Binomial[n,k],9],{k,0,n}];Table[a[n],{n,0,61}] (* James C. McMahon, Jul 10 2025 *)

PARI

a(n) = sum(k=0, n, binomial(n, k) % 9); \\ Michel Marcus, Jul 10 2025

Python

from gmpy2 import digits
import re, sympy
from sympy import S, I, sqrt, simplify, Rational
def A385741(n):
    s = digits(n,3)
    n1 = s.count('1')
    n2 = s.count('2')
    n01 = s.count('10')
    n02 = s.count('20')
    n11 = len(re.findall('(?=11)',s))
    n12 = s.count('21')
    n121 = len(re.findall('(?=121)',s))
    n122 = s.count('221')
    n21 = s.count('12')
    n22 = len(re.findall('(?=22)',s))
    x1 = (3*(3**n2*(12*n01+(n02<<4)+3*n11+(n12<<2))-(n01+n12<<2)+(n02<<4)+n11)<>3
    beta = S.Half*(I*sqrt(3)-1)
    def ind2(t): return (0,0,1,0,2,5,0,4,3)[t]
    def X(t): return beta**(ind2(t)-n11-n12+n121-n122)*(2-beta)**(n21-n121)*(3+beta)**(n2-n12-n21-n22+n121+n122)
    def Y(t): return beta**(n11-ind2(t))*(1-beta)**(n21-n121)*(2+beta)**(n2-n21-n22)*(1+2*beta)**n121
    def f(t): return ((3**n2<