A095143 -id:A095143 - cp's OEIS frontend

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

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)).

A249723 Numbers n such that there is a multiple of 9 on row n of Pascal's triangle with property that all multiples of 4 on the same row (if they exist) are larger than it.

Original entry on oeis.org

9, 10, 13, 15, 18, 19, 21, 27, 29, 31, 37, 39, 43, 45, 46, 47, 54, 55, 59, 63, 75, 79, 81, 82, 83, 85, 87, 90, 91, 93, 95, 99, 103, 109, 111, 117, 118, 119, 123, 126, 127, 135, 139, 151, 153, 154, 157, 159, 162, 163, 165, 167, 171, 175, 181, 183, 187, 189, 190, 191, 198, 199, 207, 219, 223, 225, 226, 229, 231, 234, 235, 237, 239, 243, 245, 247, 251, 253, 255

Offset: 1

Antti Karttunen, Nov 04 2014

nonn

All n such that on row n of A095143 (Pascal's triangle reduced modulo 9) there is at least one zero and the distance from the edge to the nearest zero is shorter than the distance from the edge to the nearest zero on row n of A034931 (Pascal's triangle reduced modulo 4), the latter distance taken to be infinite if there are no zeros on that row in the latter triangle.

A052955 from its eight term onward, 31, 47, 63, 95, 127, ... seems to be a subsequence. See also the comments at A249441.

			Row 13 of Pascal's triangle (A007318) is: {1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1} and the term binomial(13, 5) = 1287 = 9*11*13 occurs before any term which is a multiple of 4. Note that one such term occurs right next to it, as binomial(13, 6) = 1716 = 4*3*11*13, but 1287 < 1716, thus 13 is included.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Complement: A249724.
Natural numbers (A000027) is a disjoint union of the sequences A048278, A249722, A249723 and A249726.
Cf. A007318, A034931, A048645, A051382, A095143, A052955, A249441, A249695.
Cf. A048278, A249722, A249726, A249731, A249732, A249733.

A249723list(upto_n) = { my(i=0, n=0); while(i

A249733 Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 3, 0, 4, 2, 0, 2, 1, 0, 12, 6, 0, 8, 4, 0, 4, 2, 0, 24, 21, 18, 19, 14, 9, 14, 7, 0, 28, 20, 12, 20, 13, 6, 12, 6, 0, 32, 19, 6, 21, 12, 3, 10, 5, 0, 48, 42, 36, 38, 28, 18, 28, 14, 0, 50, 37, 24, 36, 24, 12, 22, 11, 0, 52, 32, 12, 34, 20, 6, 16, 8, 0

Offset: 0

Antti Karttunen, Nov 04 2014

nonn

Number of zeros on row n of A095143 (Pascal's triangle reduced modulo 9).

This should have a formula. See for example A062296, A006047 and A048967.

			Row 9 of Pascal's triangle is {1, 9, 36, 84, 126, 126, 84, 36, 9, 1}. The terms 9, 36, and 126 are the only multiples of nine, and each of them occurs two times on that row, thus a(9) = 2*3 = 6.
Row 10 of Pascal's triangle is {1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1}. The terms 45 (= 9*5) and 252 (= 9*28) are the only multiples of nine, and the former occurs twice, while the latter is alone at the center, thus a(10) = 2+1 = 3.

Antti Karttunen, Table of n, a(n) for n = 0..6561
James G. Huard, Blair K. Spearman and Kenneth S. Williams, Pascal's triangle (mod 9), Acta Arithmetica (1997), Volume: 78, Issue: 4, page 331-349.

Cf. A007318, A048277, A048967, A062296, A095143, A249343, A249723, A249731, A249732, A051382, A249719, A249720, A006047.

Total/@Table[If[Mod[Binomial[n,k],9]==0,1,0],{n,0,80},{k,0,n}] (* Harvey P. Dale, Feb 12 2020 *)

A249733(n) = { my(c=0); for(k=0,n\2,if(!(binomial(n,k)%9),c += (if(k<(n/2),2,1)))); return(c); } \\ Unoptimized.
for(n=0, 6561, write("b249733.txt", n, " ", A249733(n)));

import re
from gmpy2 import digits
def A249733(n):
    s = digits(n,3)
    n1 = s.count('1')
    n2 = s.count('2')
    n01 = s.count('10')
    n02 = s.count('20')
    n11 = len(re.findall('(?=11)',s))
    n12 = s.count('21')
    return n+1-(((3*(n01+1)+(n02<<2)+n12<<2)+3*n11)*(3**n2<Chai Wah Wu, Jul 24 2025

For all n >= 0, the following holds:
a(n) <= A048277(n).
a(n) <= A062296(n).
a(2*A249719(n)) > 0 and a((2*A249719(n))-1) > 0.
a(n) is odd if and only if n is one of the terms of A249720.

A249722 Numbers n such that there is a multiple of 4 on row n of Pascal's triangle with property that all multiples of 9 on the same row (if they exist) are larger than it.

Original entry on oeis.org

4, 6, 8, 12, 14, 16, 17, 20, 22, 24, 25, 26, 28, 30, 32, 33, 34, 35, 38, 40, 41, 42, 44, 48, 49, 50, 51, 52, 53, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 74, 76, 77, 78, 80, 84, 86, 88, 89, 92, 94, 96, 97, 98, 100, 101, 102, 104, 105, 106, 107, 112, 113, 114, 115, 116, 120, 121, 122, 124, 125

Offset: 1

Antti Karttunen, Nov 04 2014

nonn

All n such that on row n of A034931 (Pascal's triangle reduced modulo 4) there is at least one zero and the distance from the edge to the nearest zero is shorter than the distance from the edge to the nearest zero on row n of A095143 (Pascal's triangle reduced modulo 9), the latter distance taken to be infinite if there are no zeros on that row in the latter triangle.

			Row 4 of Pascal's triangle (A007318) is {1,4,6,4,1}. The least multiple of 4 occurs as C(4,1) = 4, and there are no multiples of 9 present, thus 4 is included among the terms.
Row 12 of Pascal's triangle is {1,12,66,220,495,792,924,792,495,220,66,12,1}. The least multiple of 4 occurs as C(12,1) = 12, which is less than the least multiple of 9 present at C(12,4) = 495 = 9*55, thus 12 is included among the terms.

Antti Karttunen, Table of n, a(n) for n = 1..10000

A subsequence of A249724.
Natural numbers (A000027) is a disjoint union of the sequences A048278, A249722, A249723 and A249726.
Cf. A007318, A034931, A095143, A048645, A051382.

A249722list(upto_n) = { my(i=0, n=0); while(i

A249726 Numbers n such that there is a multiple of 36 on row n of Pascal's triangle with property that it is also the least multiple of 4 and the least multiple of 9 on the same row.

Original entry on oeis.org

36, 72, 73, 108, 110, 144, 145, 147, 180, 216, 217, 218, 221, 252, 288, 289, 291, 295, 324, 326, 360, 361, 396, 432, 433, 434, 435, 437, 443, 468, 504, 505, 540, 542, 576, 577, 579, 583, 612, 648, 649, 650, 653, 684, 720, 721, 723, 756, 758, 792, 793, 828, 864, 865, 866, 867, 869, 871, 875, 887, 900, 936, 937, 972, 974, 1008, 1009, 1011, 1044, 1080

Offset: 1

Antti Karttunen, Nov 04 2014

nonn

All n such that both on row n of A034931 (Pascal's triangle reduced modulo 4) and on row n of A095143 (Pascal's triangle reduced modulo 9) there is at least one zero and the distance from the edge to the nearest zero is same on both rows.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Subsequence of A249724.
A044102 is a subsequence (after zero).
Natural numbers (A000027) is a disjoint union of the sequences A048278, A249722, A249723 and A249726.
Cf. A007318, A034931, A095143, A048645, A051382.

A249726list(upto_n) = { my(i=0, n=0); while(i

A385741 a(n) = Sum_{k=0..n} (binomial(n, k) mod 9).

Original entry on oeis.org

1, 2, 4, 8, 16, 14, 28, 38, 31, 8, 16, 32, 28, 56, 49, 62, 52, 68, 28, 56, 76, 62, 79, 122, 91, 92, 112, 8, 16, 32, 28, 56, 76, 80, 124, 140, 28, 56, 103, 80, 142, 158, 145, 146, 184, 62, 124, 158, 100, 146, 184, 188, 232, 230, 28, 56, 76, 80, 151, 158, 136, 236

Offset: 0

Chai Wah Wu, Jul 09 2025

nonn

Sum of n-th row of Pascal's triangle mod 9, A095143.

Chai Wah Wu, Table of n, a(n) for n = 0..10000
James G. Huard, Blair K. Spearman, and Kenneth S. Williams, Pascal's triangle (mod 9), Acta Arithmetica, 78 (1997), 331-349.

Cf. A001316, A051638, A384715, A385285.

a[n_]:=Sum[Mod[Binomial[n,k],9],{k,0,n}];Table[a[n],{n,0,61}] (* James C. McMahon, Jul 10 2025 *)

a(n) = sum(k=0, n, binomial(n, k) % 9); \\ Michel Marcus, Jul 10 2025

from gmpy2 import digits
import re, sympy
from sympy import S, I, sqrt, simplify, Rational
def A385741(n):
    s = digits(n,3)
    n1 = s.count('1')
    n2 = s.count('2')
    n01 = s.count('10')
    n02 = s.count('20')
    n11 = len(re.findall('(?=11)',s))
    n12 = s.count('21')
    n121 = len(re.findall('(?=121)',s))
    n122 = s.count('221')
    n21 = s.count('12')
    n22 = len(re.findall('(?=22)',s))
    x1 = (3*(3**n2*(12*n01+(n02<<4)+3*n11+(n12<<2))-(n01+n12<<2)+(n02<<4)+n11)<>3
    beta = S.Half*(I*sqrt(3)-1)
    def ind2(t): return (0,0,1,0,2,5,0,4,3)[t]
    def X(t): return beta**(ind2(t)-n11-n12+n121-n122)*(2-beta)**(n21-n121)*(3+beta)**(n2-n12-n21-n22+n121+n122)
    def Y(t): return beta**(n11-ind2(t))*(1-beta)**(n21-n121)*(2+beta)**(n2-n21-n22)*(1+2*beta)**n121
    def f(t): return ((3**n2<

A249731 Number of multiples of 4 on row n of Pascal's triangle minus the number of multiples of 9 on the same row: a(n) = A249732(n) - A249733(n).

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 1, 0, 6, -2, 0, 0, 3, 0, 3, -2, 13, 12, -1, 2, 13, -2, 3, 0, 15, 12, 11, -20, -4, -12, -12, -14, 21, 14, 20, 24, 1, 2, 11, -4, 20, 20, 11, 6, 29, -18, -4, -6, 22, 26, 32, 18, 32, 22, -25, -34, 9, -4, -1, -6, 9, 0, 15, -50, 25, 36, 23, 32, 49, 32, 44, 48, 13, 26, 43, 10, 41, 40, 31, 24, 73, -12

Offset: 0

Antti Karttunen, Nov 05 2014

sign

Antti Karttunen, Table of n, a(n) for n = 0..6561

Cf. A007318, A034931, A095143, A249723, A249732, A249733.

import re
from gmpy2 import digits
def A249731(n):
    s = digits(n,3)
    n1 = s.count('1')
    n2 = s.count('2')
    n01 = s.count('10')
    n02 = s.count('20')
    n11 = len(re.findall('(?=11)',s))
    n12 = s.count('21')
    return (((3*(n01+1)+(n02<<2)+n12<<2)+3*n11)*(3**n2<>1)&~n).bit_count()<>1) # Chai Wah Wu, Jul 24 2025

(define (A249731 n) (- (A249732 n) (A249733 n)))

a(n) = A249732(n) - A249733(n).

A386441 Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 27.

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, 15, 20, 15, 6, 1, 1, 7, 21, 8, 8, 21, 7, 1, 1, 8, 1, 2, 16, 2, 1, 8, 1, 1, 9, 9, 3, 18, 18, 3, 9, 9, 1, 1, 10, 18, 12, 21, 9, 21, 12, 18, 10, 1, 1, 11, 1, 3, 6, 3, 3, 6, 3, 1, 11, 1, 1, 12, 12, 4, 9, 9, 6, 9, 9, 4, 12, 12, 1, 1, 13, 24, 16, 13, 18, 15, 15, 18, 13, 16, 24, 13, 1

Offset: 0

Chai Wah Wu, Jul 21 2025

			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,  20,  15,  6,  1;
 1,  7,  21,  8,   8,  21,  7,  1;
  ...

Chai Wah Wu, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)
Kenneth S. Davis and William A. Webb, Lucas' Theorem for Prime Powers, Europ. J. Combinatorics, Vol. 11, No. 3 (1990), 229-233.

Cf. A007318, A047999, A083093, A034931, A095140, A095141, A095142, A034930, A008975, A095143, 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), A095142 (m = 7), A034930 (m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).

T[i_,j_]:=Mod[Binomial[i,j],27]; Table[T[n,k],{n,0,13},{k,0,n}]//Flatten (* Stefano Spezia, Jul 22 2025 *)

from math import isqrt, comb
from sympy import multiplicity
from gmpy2 import digits
def A386441(n):
    def g1(s,w,e):
        c, d = 1, 0
        if len(s) == 0: return c, d
        a, b = int(s,3), int(w,3)
        if a>=b:
            k = comb(a,b)%27
            j = multiplicity(3,k)
            d += j*e
            k = k//3**j
            c = c*pow(k,e,27)%27
        else:
            if int(s[0:1],3)4: return 0
    s = s.zfill(3)
    w = w.zfill(l:=len(s))
    c, d = g1(s[:3],w[:3],1)
    for i in range(1,l-2):
        c0, d0 = g1(s[i:i+3],w[i:i+3],1)
        c1, d1 = g1(s[i:i+2],w[i:i+2],-1)
        c = c*c0*c1%27
        d += d0+d1
    return c*3**d%27

T(i, j) = binomial(i, j) mod 27.

cp's OEIS Frontend

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

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Python

Formula

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Python

Formula

A249723 Numbers n such that there is a multiple of 9 on row n of Pascal's triangle with property that all multiples of 4 on the same row (if they exist) are larger than it.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A249733 Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A249722 Numbers n such that there is a multiple of 4 on row n of Pascal's triangle with property that all multiples of 9 on the same row (if they exist) are larger than it.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A249726 Numbers n such that there is a multiple of 36 on row n of Pascal's triangle with property that it is also the least multiple of 4 and the least multiple of 9 on the same row.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

A385741 a(n) = Sum_{k=0..n} (binomial(n, k) mod 9).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Python

A249731 Number of multiples of 4 on row n of Pascal's triangle minus the number of multiples of 9 on the same row: a(n) = A249732(n) - A249733(n).

Views

Author

Keywords

Links