A206424 The number of 1's in row n of Pascal's Triangle (mod 3).

1, 2, 2, 2, 4, 4, 2, 4, 5, 2, 4, 4, 4, 8, 8, 4, 8, 10, 2, 4, 5, 4, 8, 10, 5, 10, 14, 2, 4, 4, 4, 8, 8, 4, 8, 10, 4, 8, 8, 8, 16, 16, 8, 16, 20, 4, 8, 10, 8, 16, 20, 10, 20, 28, 2, 4, 5, 4, 8, 10, 5, 10, 14, 4, 8, 10, 8, 16, 20, 10, 20, 28, 5, 10, 14, 10, 20, 28

Offset: 0

Author

Marcus Jaiclin, Feb 07 2012

Keywords

• nonn
• easy

Comments

A006047(n) = a(n) + A227428(n).

a(n) = n + 1 - A062296(n) - A227428(n); number of ones in row n of triangle A083093. - Reinhard Zumkeller, Jul 11 2013

Examples

			Rows 0-8 of Pascal's Triangle (mod 3) are:
  1                   So a(0) = 1
  1 1                 So a(1) = 2
  1 2 1               So a(2) = 2
  1 0 0 1                 .
  1 1 0 1 1               .
  1 2 1 1 2 1             .
  1 0 0 2 0 0 1
  1 1 0 2 2 0 1 1
  1 2 1 2 1 2 1 2 1
		

Links

• Reinhard Zumkeller (terms 0..1000) & Antti Karttunen, Table of n, a(n) for n = 0..19683
• R. Garfield and H. S. Wilf, The distribution of the binomial coefficients modulo p, J. Numb. Theory 41 (1) (1992) 1-5
• Marcus Jaiclin, et al. Pascal's Triangle, Mod 2,3,5
• D. L. Wells, Residue counts modulo three for the fibonacci triangle, Appl. Fib. Numbers, Proc. 6th Int Conf Fib. Numbers, Pullman, 1994 (1996) 521-536.
• Avery Wilson, Pascal's Triangle Modulo 3, Mathematics Spectrum, 47-2 - January 2015, pp. 72-75.

Crossrefs

Cf. A083093, A062296, A006047, A062756, A081603, A206427, A227428.

Programs

Haskell

a206424 = length . filter (== 1) . a083093_row
-- Reinhard Zumkeller, Jul 11 2013

Mathematica

Table[Count[Mod[Binomial[n, Range[0, n]], 3], 1], {n, 0, 99}] (* Alonso del Arte, Feb 07 2012 *)

PARI

A206424(n) = sum(k=0,n,1==(binomial(n,k)%3)); \\ (naive way) Antti Karttunen, Jul 26 2017

Python

from sympy.ntheory import digits
def A206424(n):
    s = digits(n,3)[1:]
    return (3**s.count(2)+1)<>1 # Chai Wah Wu, Jul 24 2025

Scheme

(define (A206424 n) (* (+ (A000244 (A081603 n)) 1) (A000079 (- (A062756 n) 1)))) ;; (fast way) Antti Karttunen, Jul 27 2017

Formula

From Antti Karttunen, Jul 27 2017: (Start)

a(n) = (3^k + 1)*2^(y-1), where y = A062756(n) and k = A081603(n). [See e.g. Wells or Wilson references.]

a(n) = A006047(n) - A227428(n).

(End)

From David A. Corneth and Antti Karttunen, Jul 27 2017: (Start)

Based on the first formula above, we have following identities:

a(3n) = a(n).

a(3n+1) = 2*a(n).

a(9n+4) = 4*a(n).

(End)

a(n) = (1/2)*Sum_{k = 0..n} mod(C(n,k) + C(n,k)^2, 3). - Peter Bala, Dec 17 2020