A083093 -id:A083093 - cp's OEIS frontend

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

Marcus Jaiclin, Feb 07 2012

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

			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

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.

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

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

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

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

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

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

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

A090171 Triangle read by rows, related to Pascal's triangle read mod 2, starting with 0, 1, 0.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0

Offset: 0

N. J. A. Sloane, Jan 19 2004

			Triangle begins:
0
1 0
1 1 0
0 1 0 0
1 1 1 1 0
...

Y. Moshe, The density of 0's in recurrence double sequences, J. Number Theory, 103 (2003), 109-121; see Fig. 2.

Cf. A007318, A083093.
Cf. A090172, A090173, A090174, A091533, A091562, A205575 (same recurrence).
a(n, k) = A090174(n-1, k), k

T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k) + T(n-2, k-1) + T(n-2, k-2) for n >= 2, k >= 0, with initial conditions specified by first two rows.

Edited and extended by Christian G. Bower, Jan 20 2004

A090172 Triangle read by rows, related to Pascal's triangle read mod 2.

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0

Offset: 0

N. J. A. Sloane, Jan 19 2004

			1; 0,0; 1,1,1; 1,0,0,1; 0,1,1,1,0; ...

Y. Moshe, The density of 0's in recurrence double sequences, J. Number Theory, 103 (2003), 109-121; see Fig. 2.

Row sums: A091563. Cf. A007318, A083093, A090171-A090174.

a(n, k) = A091562(n, k) mod 2.

Edited and extended by Christian G. Bower, Jan 20 2004

A083094 Numbers k such that Sum_{j=0..k} (binomial(k,j) mod 3) is odd.

Original entry on oeis.org

0, 8, 20, 24, 56, 60, 72, 80, 164, 168, 180, 188, 216, 224, 236, 240, 488, 492, 504, 512, 540, 548, 560, 564, 648, 656, 668, 672, 704, 708, 720, 728, 1460, 1464, 1476, 1484, 1512, 1520, 1532, 1536, 1620, 1628, 1640, 1644, 1676, 1680, 1692, 1700, 1944, 1952

Offset: 1

Benoit Cloitre, Apr 22 2003

Apparently a(n)/2 (mod 3) = A010060(n), the Thue-Morse sequence.

Cf. A010060, A051638, A074939, A083093, A083095 (gives a b-file of 16384 terms).

Select[Range[0, 2000],OddQ[Sum[Mod[Binomial[#, j], 3], {j, 0, #}]] &] (* Paul F. Marrero Romero, Dec 28 2024 *)

isok(n) = sum(k=0, n, binomial(n,k) % 3) % 2; \\ Michel Marcus, Dec 05 2013

def A083094(n): return int(bin(((m:=n-1).bit_count()&1)+(m<<1))[2:],3)<<1 # Chai Wah Wu, Jun 26 2025

a(n) = 4*A083095(n). - Hugo Pfoertner, Jan 12 2025

Numbers that are multiples of 4 and such that base-3 digits contain no 1's, or equivalently, numbers such that base-3 digits contains an even number of 2's and no 1's, i.e. a(n) = 2*A074939(n-1). This characterization can be derived from the formula in A051638. - Chai Wah Wu, Jun 26 2025

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 29 2003

A090173 Triangle read by rows, related to Pascal's triangle read mod 2, starting with 0, 0, 1.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1

Offset: 0

N. J. A. Sloane, Jan 19 2004

			Triangle begins
  0;
  0,1;
  0,1,1;
  0,0,1,0;
  0,1,1,1,1;
  ...

Y. Moshe, The density of 0's in recurrence double sequences, J. Number Theory, 103 (2003), 109-121; see Fig. 2.

Cf. A007318, A083093.
Cf. A090171, A090172, A090174, A091533, A091562, A205575 (same recurrence).
a(n, k) = A090174(n-1, k-1), k>0, 0 otherwise.

T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k) + T(n-2, k-1) + T(n-2, k-2) for n >= 2, k >= 0, with initial conditions specified by first two rows.

Edited and extended by Christian G. Bower, Jan 20 2004

A006940 Rows of Pascal's triangle mod 3.

Original entry on oeis.org

1, 11, 121, 1001, 11011, 121121, 1002001, 11022011, 121212121, 1000000001, 11000000011, 121000000121, 1001000001001, 11011000011011, 121121000121121, 1002001001002001, 11022011011022011, 121212121121212121, 1000000002000000001, 11000000022000000011, 121000000212000000121

Offset: 0

N. J. A. Sloane

nonn

Subsequence of A118594. - Chai Wah Wu, Jul 30 2025

C. Pickover, Mazes for the Mind, St. Martin's Press, NY, 1992, p. 353.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Cf. A007318, A083093, A118594.

a[n_] := FromDigits[Table[Mod[Binomial[n, k], 3], {k, 0, n}]]; Array[a, 25, 0] (* Amiram Eldar, Nov 22 2018 *)

a(n)=fromdigits(apply(x->x%3, binomial(n))); \\ Michel Marcus, Nov 21 2018

from math import prod, comb
from gmpy2 import digits
def A006940(n):
    if n==0: return 1
    c, l = '', len(s:=digits(n,3))
    for k in range(m:=n+2>>1):
        t = digits(k,3).zfill(l)
        c += str(prod(comb(int(s[i]),int(t[i]))%3 for i in range(l))%3)
    return int(c+c[m-2+(n&1)::-1]) # Chai Wah Wu, Jul 30 2025

More terms from Michel Marcus, Nov 21 2018

A243759 Triangle T(m,k): exponent of the highest power of 3 dividing the binomial coefficient binomial(m,k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 1, 2, 2, 1, 2, 2, 0, 0, 0, 2, 1, 1, 2, 1, 1, 2, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 2, 2, 1, 2, 2

Offset: 0

Tom Edgar, Jun 10 2014

T(m,k) is the number of 'carries' that occur when adding k and n-k in base 3 using the traditional addition algorithm.

			The triangle begins:
0,
0, 0,
0, 0, 0,
0, 1, 1, 0;
0, 0, 1, 0, 0;
0, 0, 0, 0, 0, 0;
0, 1, 1, 0, 1, 1, 0;
0, 0, 1, 0, 0, 1, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0, 0;
0, 2, 2, 1, 2, 2, 1, 2, 2, 0;

Robert Israel, Table of n, a(n) for n = 0..10010
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.

Row sums: A249343.

Cf. A007318, A007949, A083093, A065040, A242849, A060828, A038500.

A243759:= (m,k) -> padic[ordp](binomial(m,k),3);
for m from 0 to 50 do
  seq(A243759(m,k),k=0..m)
od;   # Robert Israel, Jun 15 2014

T[m_, k_] := IntegerExponent[Binomial[m, k], 3];
Table[T[m, k], {m, 0, 12}, {k, 0, m}] // Flatten (* Jean-François Alcover, Jun 05 2022 *)

m=50
T=[0]+[3^valuation(i, 3) for i in [1..m]]
Table=[[prod(T[1:i+1])/(prod(T[1:j+1])*prod(T[1:i-j+1])) for j in [0..i]] for i in [0..m-1]]
[log(Integer(x),base=3) for sublist in Table for x in sublist]

(define (A243759 n) (A007949 (A007318 n))) ;; Antti Karttunen, Oct 28 2014

T(m,k) = log_3(A242849(m,k)).
From Antti Karttunen, Oct 28 2014: (Start)
a(n) = A007949(A007318(n)).
a(n) * A083093(n) = 0 and a(n) + A083093(n) > 0 for all n.
(End)

Name clarified by Antti Karttunen, Oct 28 2014

A153459 Decimal expansion of log_3 (6).

Original entry on oeis.org

1, 6, 3, 0, 9, 2, 9, 7, 5, 3, 5, 7, 1, 4, 5, 7, 4, 3, 7, 0, 9, 9, 5, 2, 7, 1, 1, 4, 3, 4, 2, 7, 6, 0, 8, 5, 4, 2, 9, 9, 5, 8, 5, 6, 4, 0, 1, 3, 1, 8, 8, 0, 4, 2, 7, 8, 7, 0, 6, 5, 4, 9, 4, 3, 8, 3, 8, 6, 8, 5, 2, 0, 1, 3, 8, 0, 9, 1, 4, 8, 0, 5, 0, 6, 1, 1, 7, 2, 6, 8, 8, 5, 4, 9, 4, 5, 1, 7, 4

Offset: 1

N. J. A. Sloane, Oct 30 2009

Equals the Hausdorff dimension of Pascal's triangle modulo 3 (A083093). In general, the dimension of Pascal's triangle modulo a prime p is log(p*(p+1)/2) / log(p) (see Reiter link, theorem 2 page 117). - Bernard Schott, Dec 01 2022

			1.6309297535714574370995271143427608542995856401318804278706...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000.
A. M. Reiter, Determining the dimension of fractals generated by Pascal's triangle, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
Wikipedia, List of fractals by Hausdorff dimension (see Pascal triangle modulo 3).
Index entries for transcendental numbers

cf. A002391, A016629, A083093, A102525.

evalf(log(6)/log(3),80); # Bernard Schott, Dec 01 2022

RealDigits[Log[3, 6], 10, 120][[1]] (* Vincenzo Librandi, Aug 29 2013 *)

Equals A016629 / A002391 = 1 + A102525. - Bernard Schott, Dec 01 2022

Previous Showing 21-30 of 36 results. Next

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Python

Scheme

Formula

A090174 Triangle read by rows, related to Pascal's triangle read mod 2.

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

Extensions

A083095 a(n) = A083094(n)/4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Python

Formula

Extensions

A090171 Triangle read by rows, related to Pascal's triangle read mod 2, starting with 0, 1, 0.

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

Extensions

A090172 Triangle read by rows, related to Pascal's triangle read mod 2.

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

Extensions

A083094 Numbers k such that Sum_{j=0..k} (binomial(k,j) mod 3) is odd.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

Python

Formula

Extensions

A090173 Triangle read by rows, related to Pascal's triangle read mod 2, starting with 0, 0, 1.

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

Extensions

A006940 Rows of Pascal's triangle mod 3.

Views

Author

Keywords