A249732 -id:A249732 - cp's OEIS frontend

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).

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).

A034931 Triangle read by rows: Pascal's triangle (A007318) mod 4.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

The number of 3's in row n is given by 2^(A000120(n)-1) if A014081(n) is nonzero, else by 0 [Davis & Webb]. - R. J. Mathar, Jul 28 2017

			Triangle begins:
                 1
                1 1
               1 2 1
              1 3 3 1
             1 0 2 0 1
            1 1 2 2 1 1
           1 2 3 0 3 2 1
          1 3 1 3 3 1 3 1
         1 0 0 0 2 0 0 0 1
        1 1 0 0 2 2 0 0 1 1
       1 2 1 0 2 0 2 0 1 2 1
      1 3 3 1 2 2 2 2 1 3 3 1
  ...

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
Kenneth S. Davis and William A. Webb, Lucas' theorem for prime powers, European Journal of Combinatorics 11:3 (1990), pp. 229-233.
Kenneth S. Davis and William A. Webb, Pascal's triangle modulo 4, Fib. Quart., 29 (1991), 79-83.
Marc Evanstein, Hearing Pascal's Triangle Mod 4, YouTube video.
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.
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, A034930, A008975, A034932, A163000 (# 2's), A270438 (# 1's), A249732 (# 0's).
Cf. A000120, A014081.
Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), (this sequence) (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).

a034931 n k = a034931_tabl !! n !! k
a034931_row n = a034931_tabl !! n
a034931_tabl = iterate
   (\ws -> zipWith ((flip mod 4 .) . (+)) ([0] ++ ws) (ws ++ [0])) [1]
-- Reinhard Zumkeller, Mar 14 2015

A034931 := proc(n,k)
    modp(binomial(n,k),4) ;
end proc:
seq(seq(A034931(n,k),k=0..n),n=0..10); # R. J. Mathar, Jul 28 2017

Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 4] (* Robert G. Wilson v, May 26 2004 *)

C(n, k)=binomial(n, k)%4 \\ Charles R Greathouse IV, Aug 09 2016

f(n,k)=2*(bitand(n-k, k)==0);
T(n,j)=if(j==0,return(1)); my(k=logint(n,2),K=2^k,K1=K/2,L=n-K); if(LCharles R Greathouse IV, Aug 11 2016

from math import isqrt, comb
def A034931(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()>1: return 0
    s, c, d = bin(g)[2:], 1, 0
    w = (bin(k)[2:]).zfill(l:=len(s))
    for i in range(0,l-1):
        r, t = s[i:i+2], w[i:i+2]
        if (x:=int(r,2)) < (y:=int(t,2)):
            d += (t[0]>r[0])+(t[1]>r[1])
        else:
            c = c*comb(x,y)&3
    d -= sum(1 for i in range(1,l-1) if w[i]>s[i])
    return (c<Chai Wah Wu, Jul 19 2025

T(n+1,k) = (T(n,k) + T(n,k-1)) mod 4. - Reinhard Zumkeller, Mar 14 2015

A048277 Number of (not necessarily distinct) nonsquarefree numbers among C(n,k), k=0..n.

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 1, 0, 6, 8, 5, 0, 9, 4, 3, 2, 15, 12, 17, 12, 13, 12, 11, 0, 21, 22, 19, 26, 25, 18, 25, 20, 31, 30, 27, 28, 35, 30, 25, 28, 37, 30, 29, 18, 29, 38, 27, 6, 47, 48, 49, 48, 47, 36, 51, 50, 55, 52, 49, 38, 53, 36, 23, 56, 63, 62, 61, 60, 61, 54, 59, 54, 71, 66, 57

Offset: 0

Labos Elemer

nonn

Number of nonsquarefree numbers (A013929) on row n of Pascal's triangle (A007318). - Antti Karttunen, Nov 05 2014

			a(13) = 4 because C(13,5) = C(13,8) = 3^2*11*13 and C(13,6) = C(13,7) = 2^2*3*11*13.
If n=20, then C[ 20, k ] is divisible by a square for 13 values of k, i.e. for k = 1, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 19, so a[ 20 ] = 13.

Antti Karttunen, Table of n, a(n) for n = 0..8192

Cf. A005117, A007318, A013929, A046098, A048276, A064460, A249732, A249733, A249442, A249716.

seq(nops(remove(numtheory:-issqrfree,[seq(binomial(n,k),k=0..n)])),n=0..100); # Robert Israel, Nov 05 2014

f[ n_ ] := (c = 0; k = 1; While[ k < n, If[ Union[ Transpose[ FactorInteger[ Binomial[ n, k ] ] ] [ [ 2 ] ] ] [ [ -1 ] ] > 1, c++ ]; k++ ]; c); Table[ f[ n ], {n, 0, 75} ]
Table[(1 + n) - Length[Select[Binomial[n, Range[0, n]], SquareFreeQ[#] &]], {n, 0, 100}] (* Vincenzo Librandi, Nov 06 2014 *)

a(n) = sum(k=0, n, !issquarefree(binomial(n, k))); \\ Michel Marcus, Mar 05 2014

A048277(n) = sum(k=0,n\2,((0==moebius(binomial(n,k)))*(if(k<(n/2),2,1))));
for(n=0, 8192, write("b048277.txt", n, " ", A048277(n))); \\ b-file was computed with this program. - Antti Karttunen, Nov 05 2014

From Antti Karttunen, Nov 05 2014: (Start)
a(n) = 1 + n - A048276(n).
Also, for all n >= 0:
a(n) >= A249732(n).
a(n) >= A249733(n).
(End)

Definition corrected by Michel Marcus, Mar 05 2014

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.

cp's OEIS Frontend

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).

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A034931 Triangle read by rows: Pascal's triangle (A007318) mod 4.

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A048277 Number of (not necessarily distinct) nonsquarefree numbers among C(n,k), k=0..n.

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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.

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

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