A062296 a(n) = number of entries in n-th row of Pascal's triangle divisible by 3.

0, 0, 0, 2, 1, 0, 4, 2, 0, 8, 7, 6, 9, 6, 3, 10, 5, 0, 16, 14, 12, 16, 11, 6, 16, 8, 0, 26, 25, 24, 27, 24, 21, 28, 23, 18, 33, 30, 27, 32, 25, 18, 31, 20, 9, 40, 35, 30, 37, 26, 15, 34, 17, 0, 52, 50, 48, 52, 47, 42, 52, 44, 36, 58, 53, 48, 55, 44, 33, 52, 35, 18, 64, 56, 48, 58, 41

Offset: 0

Author

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 02 2001

Keywords

• nonn
• look

Comments

Number of zeros in row n of triangle A083093. - Reinhard Zumkeller, Jul 11 2013

Examples

			When n=3 the row is 1,3,3,1 so a(3) = 2.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• 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.

Crossrefs

Cf. A006047, A083093, A206424, A227428.

Cf. A062756, A081603.

Programs

Haskell

a062296 = sum . map ((1 -) . signum) . a083093_row
-- Reinhard Zumkeller, Jul 11 2013

Maple

p:=proc(n) local ct, k: ct:=0: for k from 0 to n do if binomial(n,k) mod 3 = 0 then else ct:=ct+1 fi od: end: seq(n+1-p(n),n=0..83); # Emeric Deutsch

Mathematica

a[n_] := Count[(Binomial[n, #] & )[Range[0, n]], _?(Divisible[#, 3] & )];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jan 26 2018 *)
Table[n + 1 - 2^(DigitCount[n, 3, 1])*3^(DigitCount[n, 3, 2]), {n, 0, 76}] (* Shenghui Yang, Jan 08 2025 *)

Python

from sympy.ntheory import digits
def A062296(n):
    s = digits(n,3)[1:]
    return n+1-(3**s.count(2)<Chai Wah Wu, Jul 24 2025

Formula

a(n) = n + 1 - A006047(n).

a(n) = n + 1 - A206424(n) - A227428(n). - Reinhard Zumkeller, Jul 11 2013

a(n) = n + 1 - 2^A062756(n)*3^A081603(n). - Shenghui Yang, Jan 08 2025

Extensions

More terms from Emeric Deutsch, Feb 03 2005