A083093 -id:A083093 - cp's OEIS frontend

A034932 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 16.

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 4, 15, 6, 1, 1, 7, 5, 3, 3, 5, 7, 1, 1, 8, 12, 8, 6, 8, 12, 8, 1, 1, 9, 4, 4, 14, 14, 4, 4, 9, 1, 1, 10, 13, 8, 2, 12, 2, 8, 13, 10, 1, 1, 11, 7, 5, 10, 14, 14, 10

Offset: 0

N. J. A. Sloane

T(n+1,k) = (T(n,k) + T(n,k-1)) mod 16. - Reinhard Zumkeller, Mar 14 2015

			Triangle begins:
                        1
                      1   1
                    1   2   1
                  1   3   3   1
                1   4   6   4   1
              1   5  10  10   5   1
            1   6  15   4  15   6   1
          1   7   5   3   3   5   7   1
        1   8  12   8   6   8  12   8   1
      1   9   4   4  14  14   4   4   9   1
    1  10  13   8   2  12   2   8  13  10   1
  1  11   7   5  10  14  14  10   5   7  11   1
.
Written in hexadecimal (with a=10, b=11, ..., f=15), rows 0..32 are
.
                                   1
                                  1 1
                                 1 2 1
                                1 3 3 1
                               1 4 6 4 1
                              1 5 a a 5 1
                             1 6 f 4 f 6 1
                            1 7 5 3 3 5 7 1
                           1 8 c 8 6 8 c 8 1
                          1 9 4 4 e e 4 4 9 1
                         1 a d 8 2 c 2 8 d a 1
                        1 b 7 5 a e e a 5 7 b 1
                       1 c 2 c f 8 c 8 f c 2 c 1
                      1 d e e b 7 4 4 7 b e e d 1
                     1 e b c 9 2 b 8 b 2 9 c b e 1
                    1 f 9 7 5 b d 3 3 d b 5 7 9 f 1
                   1 0 8 0 c 0 8 0 6 0 8 0 c 0 8 0 1
                  1 1 8 8 c c 8 8 6 6 8 8 c c 8 8 1 1
                 1 2 9 0 4 8 4 0 e c e 0 4 8 4 0 9 2 1
                1 3 b 9 4 c c 4 e a a e 4 c c 4 9 b 3 1
               1 4 e 4 d 0 8 0 2 8 4 8 2 0 8 0 d 4 e 4 1
              1 5 2 2 1 d 8 8 2 a c c a 2 8 8 d 1 2 2 5 1
             1 6 7 4 3 e 5 0 a c 6 8 6 c a 0 5 e 3 4 7 6 1
            1 7 d b 7 1 3 5 a 6 2 e e 2 6 a 5 3 1 7 b d 7 1
           1 8 4 8 2 8 4 8 f 0 8 0 c 0 8 0 f 8 4 8 2 8 4 8 1
          1 9 c c a a c c 7 f 8 8 c c 8 8 f 7 c c a a c c 9 1
         1 a 5 8 6 4 6 8 3 6 7 0 4 8 4 0 7 6 3 8 6 4 6 8 5 a 1
        1 b f d e a a e b 9 d 7 4 c c 4 7 d 9 b e a a e d f b 1
       1 c a c b 8 4 8 9 4 6 4 b 0 8 0 b 4 6 4 9 8 4 8 b c a c 1
      1 d 6 6 7 3 c c 1 d a a f b 8 8 b f a a d 1 c c 3 7 6 6 d 1
     1 e 3 c d a f 8 d e 7 4 9 a 3 0 3 a 9 4 7 e d 8 f a d c 3 e 1
    1 f 1 f 9 7 9 7 5 b 5 b d 3 d 3 3 d 3 d b 5 b 5 7 9 7 9 f 1 f 1
   1 0 0 0 8 0 0 0 c 0 0 0 8 0 0 0 6 0 0 0 8 0 0 0 c 0 0 0 8 0 0 0 1

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
James G. Huard, Blair K. Spearman, and Kenneth S. Williams, Pascal's triangle (mod 8), European Journal of Combinatorics 19:1 (1998), pp. 45-62.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A047999, A083093, A034931, A034930, A008975.

Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930 (m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), (this sequence) (m = 16).

a034932 n k = a034932_tabl !! n !! k
a034932_row n = a034932_tabl !! n
a034932_tabl = iterate
   (\ws -> zipWith ((flip mod 16 .) . (+)) ([0] ++ ws) (ws ++ [0])) [1]
-- Reinhard Zumkeller, Mar 14 2015

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 16] (* Robert G. Wilson v, May 26 2004 *)

from math import comb, isqrt
def A034932(n):
    g = (m:=isqrt(f:=n+1<<1))-(f<=m*(m+1))
    k = n-comb(g+1,2)
    if k.bit_count()+(g-k).bit_count()-g.bit_count()>3: return 0
    def g1(s,w,e):
        c, d = 1, 0
        if len(s) == 0: return c, d
        a, b = int(s,2), int(w,2)
        if a>=b:
            k = comb(a,b)&15
            j = (~k & k-1).bit_length()
            d += j*e
            k >>= j
            c = c*pow(k,e,16)&15
        else:
            if int(s[0:1],2)Chai Wah Wu, Jul 20 2025

T(i, j) = binomial(i, j) mod 16.

A051638 a(n) = Sum_{k=0..n} (C(n,k) mod 3).

Original entry on oeis.org

1, 2, 4, 2, 4, 8, 4, 8, 13, 2, 4, 8, 4, 8, 16, 8, 16, 26, 4, 8, 13, 8, 16, 26, 13, 26, 40, 2, 4, 8, 4, 8, 16, 8, 16, 26, 4, 8, 16, 8, 16, 32, 16, 32, 52, 8, 16, 26, 16, 32, 52, 26, 52, 80, 4, 8, 13, 8, 16, 26, 13, 26, 40, 8, 16, 26, 16, 32, 52, 26, 52, 80, 13

Offset: 0

N. J. A. Sloane

Row sums of the triangle in A083093. - Reinhard Zumkeller, Jul 11 2013

Reinhard Zumkeller, Table of n, a(n) for n = 0..6561=3^8
Michael Gilleland, Some Self-Similar Integer Sequences
A. Granville, Binomials modulo a prime

Cf. A001316.

a051638 = sum . a083093_row  -- Reinhard Zumkeller, Jul 11 2013

Table[2^(DigitCount[n,3,1]-1) (3^(DigitCount[n,3,2]+1)-1),{n,0,80}] (* Harvey P. Dale, Jun 20 2019 *)

from gmpy2 import digits
def A051638(n): return 3*3**(s:=digits(n,3)).count('2')-1<>1 # Chai Wah Wu, Jun 25 2025

Write n in base 3; if the representation contains r 1's and s 2's then a(n) = 2^{r-1} * (3^(s+1) - 1) = 1/2 * (3*A006047(n) - 2^(A062756(n))). - Robin Chapman, Ahmed Fares (ahmedfares(AT)my-deja.com) and others, Jul 16 2001

a(3n) = a(n), a(3n+1) = 2a(n), a(9n+2) = a(3n+2), a(9n+5) = 2a(3n+2), a(9n+8) = 4a(3n+2) - 3a(n). - David Radcliffe, Jun 25 2025

A095140 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 5.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 1, 4, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 2, 1, 0, 0, 1, 2, 1, 1, 3, 3, 1, 0, 1, 3, 3, 1, 1, 4, 1, 4, 1, 1, 4, 1, 4, 1, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 2, 2, 0, 0, 0, 1, 1, 1, 2, 1, 0, 0, 2, 4, 2, 0, 0, 1, 2, 1, 1, 3, 3, 1, 0, 2, 1, 1, 2, 0, 1, 3, 3, 1

Offset: 0

Robert G. Wilson v, May 29 2004

{T(n,k)} is a fractal gasket with fractal (Hausdorff) dimension log(A000217(5))/log(5) = log(15)/log(5) = 1.68260... (see Reiter reference). Replacing values greater than 1 with 1 produces a binary gasket with the same dimension (see Bondarenko reference). - Richard L. Ollerton, Dec 14 2021

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see pp. 130-132.
A. M. Reiter, Determining the dimension of fractals generated by Pascal's triangle, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A047999, A083093, A034931, A095141, A095142, A034930, A095143, A008975, A095144, A095145, A034932.

Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), (this sequence) (m = 5), A095141 (m = 6), A095142 (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 5]

from math import isqrt, comb
def A095140(n):
    def f(m,k):
        if m<5 and k<5: return comb(m,k)%5
        c,a = divmod(m,5)
        d,b = divmod(k,5)
        return f(c,d)*f(a,b)%5
    return f(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Apr 30 2025

T(i, j) = binomial(i, j) mod 5.

A095142 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 7.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 3, 3, 5, 1, 1, 6, 1, 6, 1, 6, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 2, 1, 0, 0, 0, 0, 1, 2, 1, 1, 3, 3, 1, 0, 0, 0, 1, 3, 3, 1, 1, 4, 6, 4, 1, 0, 0, 1, 4, 6, 4, 1, 1, 5, 3, 3, 5, 1, 0, 1, 5, 3, 3, 5, 1, 1, 6, 1, 6, 1, 6, 1, 1, 6, 1, 6, 1, 6, 1

Offset: 0

Robert G. Wilson v, May 29 2004

{T(n,k)} is a fractal gasket with fractal (Hausdorff) dimension log(A000217(7))/log(7) = log(28)/log(7) = 1.71241... (see Reiter reference). Replacing values greater than 1 with 1 produces a binary gasket with the same dimension (see Bondarenko reference). - Richard L. Ollerton, Dec 14 2021

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see pp. 130-132.
A. M. Reiter, Determining the dimension of fractals generated by Pascal's triangle, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A047999, A083093, A034931, A095140, A095141, A034930, A095143, A008975, A095144, A095145, A034932.

Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), (this sequence) (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 7]

from math import comb, isqrt
def A095142(n):
    def f(m,k):
        if m<7 and k<7: return comb(m,k)%7
        c,a = divmod(m,7)
        d,b = divmod(k,7)
        return f(c,d)*f(a,b)%7
    return f(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Apr 30 2025

T(i, j) = binomial(i, j) mod 7.

A095144 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 11.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 4, 9, 4, 6, 1, 1, 7, 10, 2, 2, 10, 7, 1, 1, 8, 6, 1, 4, 1, 6, 8, 1, 1, 9, 3, 7, 5, 5, 7, 3, 9, 1, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 1

Offset: 0

Robert G. Wilson v, May 29 2004

{T(n,k)} is a fractal gasket with fractal (Hausdorff) dimension log(A000217(11))/log(11) = log(66)/log(11) = 1.74722... (see Reiter reference). Replacing values greater than 1 with 1 produces a binary gasket with the same dimension (see Bondarenko reference). - Richard L. Ollerton, Dec 14 2021

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see pp. 130-132.
Zubeyir Cinkir and Aysegul Ozturkalan, An extension of Lucas Theorem, arXiv:2309.00109 [math.NT], 2023. See Figures 3 and 4 p. 6.
Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
A. M. Reiter, Determining the dimension of fractals generated by Pascal's triangle, Fibonacci Quarterly, 31(2), 1993, pp. 112-120.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A047999, A083093, A034931, A095140, A095141, A095142, A034930, A095143, A008975, A095145, A034932.

Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930 (m = 8), A095143 (m = 9), A008975 (m = 10), (this sequence) (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).

R[0]:= 1:
for  n from 1 to 20 do
  R[n]:= op([R[n-1],0] + [0,R[n-1]] mod 11);
od:
for n from 0 to 20 do R[n] od; # Robert Israel, Jan 02 2019

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 11]

from math import isqrt, comb
def A095144(n):
    def f(m,k):
        if m<11 and k<11: return comb(m,k)%11
        c,a = divmod(m,11)
        d,b = divmod(k,11)
        return f(c,d)*f(a,b)%11
    return f(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Apr 30 2025

T(i, j) = binomial(i, j) mod 11.
From Robert Israel, Jan 02 2019: (Start)
T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod 11 with T(n,0) = 1.
T(n,k) = (Product_i binomial(n_i, k_i)) mod 11, where n_i and k_i are the base-11 digits of n and k. (End)

A095141 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 6.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 0, 4, 1, 1, 5, 4, 4, 5, 1, 1, 0, 3, 2, 3, 0, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 2, 4, 2, 4, 2, 4, 2, 1, 1, 3, 0, 0, 0, 0, 0, 0, 3, 1, 1, 4, 3, 0, 0, 0, 0, 0, 3, 4, 1, 1, 5, 1, 3, 0, 0, 0, 0, 3, 1, 5, 1, 1, 0, 0, 4, 3, 0, 0, 0, 3, 4, 0, 0, 1, 1, 1, 0, 4, 1, 3, 0, 0, 3, 1, 4, 0, 1, 1

Offset: 0

Robert G. Wilson v, May 29 2004

Bill Gosper, Pastel-colored illustration of triangle
Ilya Gutkovskiy, Illustrations (triangle formed by reading Pascal's triangle mod m)
Ivan Korec, Definability of Pascal's Triangles Modulo 4 and 6 and Some Other Binary Operations from Their Associated Equivalence Relations, Acta Univ. M. Belii Ser. Math. 4 (1996), pp. 53-66.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A047999, A083093, A034931, A095140, A095142, A034930, A095143, A008975, A095144, A095145, A034932.

Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), (this sequence) (m = 6), A095142 (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 6]
Graphics[Table[{%[Mod[Binomial[n, k], 6]/5], RegularPolygon[{4√3 (k - n/2), -6 n}, {4,π/6}, 6]}, {n, 0, 105}, {k, 0, n}]] (* Mma code for illustration, Bill Gosper, Aug 05 2017 *)

from math import isqrt, comb
from sympy.ntheory.modular import crt
def A095141(n):
    w, c = n-((r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)))*(r+1)>>1), 1
    d = int(not ~r & w)
    while True:
        r, a = divmod(r,3)
        w, b = divmod(w,3)
        c = c*comb(a,b)%3
        if r<3 and w<3:
            c = c*comb(r,w)%3
            break
    return crt([3,2],[c,d])[0] # Chai Wah Wu, May 01 2025

T(i, j) = binomial(i, j) mod 6.

A095145 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 12.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 3, 8, 3, 6, 1, 1, 7, 9, 11, 11, 9, 7, 1, 1, 8, 4, 8, 10, 8, 4, 8, 1, 1, 9, 0, 0, 6, 6, 0, 0, 9, 1, 1, 10, 9, 0, 6, 0, 6, 0, 9, 10, 1, 1, 11, 7, 9, 6, 6, 6, 6, 9, 7, 11, 1, 1, 0, 6, 4, 3, 0, 0, 0, 3, 4, 6, 0, 1, 1, 1, 6, 10, 7, 3, 0, 0, 3, 7, 10, 6, 1, 1

Offset: 0

Robert G. Wilson v, May 29 2004

Cf. A007318, A047999, A083093, A034931, A095140, A095141, A095142, A034930, A095143, A008975, A095144, A034932.

Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930 (m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), (this sequence) (m = 12), A275198 (m = 14), A034932 (m = 16).

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 12]

# uses python code from A034931 and A083093
from sympy.ntheory.modular import crt
def A095145(n): return crt([4,3],[A034931(n),A083093(n)])[0] # Chai Wah Wu, Jul 19 2025

T(i, j) = binomial(i, j) mod 12.

A275198 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 14.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 1, 6, 1, 6, 1, 1, 7, 7, 7, 7, 7, 7, 1, 1, 8, 0, 0, 0, 0, 0, 8, 1, 1, 9, 8, 0, 0, 0, 0, 8, 9, 1, 1, 10, 3, 8, 0, 0, 0, 8, 3, 10, 1, 1, 11, 13, 11, 8, 0, 0, 8, 11, 13, 11, 1, 1, 12, 10, 10, 5, 8, 0, 8, 5, 10, 10, 12, 1, 1, 13, 8, 6, 1, 13, 8, 8, 13, 1, 6, 8, 13, 1, 1, 0, 7, 0, 7, 0, 7, 2, 7, 0, 7, 0, 7, 0, 1

Offset: 0

Ilya Gutkovskiy, Aug 11 2016

			Triangle begins:
                      1,
                    1,  1,
                  1,  2,  1,
                1,  3,  3,  1,
              1,  4,  6,  4,  1,
            1,  5, 10, 10,  5,  1,
          1,  6,  1,  6,  1,  6,  1,
        1,  7,  7,  7,  7,  7,  7,  1,
      1,  8,  0,  0,  0,  0,  0,  8,  1,
    1,  9,  8,  0,  0,  0,  0,  8,  9,  1,
  1, 10,  3,  8,  0,  0,  0,  8,  3, 10,  1,
  ...

Cf. A007318, A070696.

Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930 (m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), (this sequence) (m = 14), A034932 (m = 16).

Mod[Flatten[Table[Binomial[n, k], {n, 0, 14}, {k, 0, n}]], 14]

from math import comb, isqrt
from sympy.ntheory.modular import crt
def A275198(n):
    w, c = n-((r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)))*(r+1)>>1), 1
    d = int(not ~r & w)
    while True:
        r, a = divmod(r,7)
        w, b = divmod(w,7)
        c = c*comb(a,b)%7
        if r<7 and w<7:
            c = c*comb(r,w)%7
            break
    return crt([7,2],[c,d])[0] # Chai Wah Wu, May 01 2025

T(n, k) = binomial(n, k) mod 14.

a(n) = A070696(A007318(n)).

A006996 a(n) = C(2n,n) mod 3.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, James Propp

Removing 0's from the sequence gives Thue-Morse sequence A001285 : 1,2,0,2,1,0,0,0,0,2,1,0,1,2,..->1,2,2,1,2,1,1,2,... - Benoit Cloitre, Jan 04 2004
a(n) = 0 if n in A074940, a(n) = 1 if n in A074939, a(n) = 2 if n in A074938.
Central terms of the triangle in A083093. - Reinhard Zumkeller, Jul 11 2013

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..2187=3^7
Michael Gilleland, Some Self-Similar Integer Sequences
Index entries for sequences that are fixed points of mappings

Cf. A001285.
Cf. A074938, A074939, A074940.
Cf. A083093.
Cf. A000984, A005704.

a006996 n = a083093 (2 * n) n  -- Reinhard Zumkeller, Jul 11 2013

Table[ Mod[ Binomial[2n, n], 3], {n, 0, 104}] (* Or *)
Nest[ Function[ l, {Flatten[(l /. {0 -> {0, 0, 0}, 1 -> {1, 2, 0}, 2 -> {2, 1, 0}})]}], {1}, 7] (* Robert G. Wilson v, Mar 28 2005 *)

a(n)=if(n==0, return(1)); if(vecmax(Set(digits(n,3)))>1, 0, 1 + n%2) \\ Charles R Greathouse IV, May 09 2016

from gmpy2 import digits
def A006996(n): return 0 if '2' in digits(n,3) else 1+(n&1) # Chai Wah Wu, Jun 26 2025

a(n) = A000984(n) mod 3.
a(n) = A005704(n) mod 3. - Benoit Cloitre, Jan 04 2004
A fixed point of the morphism : 1 -> 120, 2 -> 210, 0 -> 000. - Philippe Deléham, Jan 08 2004

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

Original entry on oeis.org

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

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

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

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

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.

Cf. A006047, A083093, A206424, A227428.

Cf. A062756, A081603.

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

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

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 *)

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

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

More terms from Emeric Deutsch, Feb 03 2005

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