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

0,2

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.

Formula

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

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

Example

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

Mathematica

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

Prog

(Haskell)
a206424 = length . filter (== 1) . a083093_row
-- Reinhard Zumkeller, Jul 11 2013
(PARI) A206424(n) = sum(k=0, n, 1==(binomial(n, k)%3)); \\ (naive way) Antti Karttunen, Jul 26 2017
(Scheme) (define (A206424 n) (* (+ (A000244 (A081603 n)) 1) (A000079 (- (A062756 n) 1)))) ;; (fast way) Antti Karttunen, Jul 27 2017
(Python)
from sympy.ntheory import digits
def A206424(n):
    s = digits(n, 3)[1:]
    return (3**s.count(2)+1)<<s.count(1)>>1 # Chai Wah Wu, Jul 24 2025

Crossrefs

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

Sequence in context: A380272 A105080 A336299 * A117728 A172309 A130453

Adjacent sequences: A206421 A206422 A206423 * A206425 A206426 A206427

Keyword

nonn,easy

Author

Marcus Jaiclin, Feb 07 2012

Status

approved