A034931 -id:A034931 - 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

A384715 a(n) = Sum_{k=0..n} (binomial(n, k) mod 4).

Original entry on oeis.org

1, 2, 4, 8, 4, 8, 12, 16, 4, 8, 12, 24, 12, 24, 24, 32, 4, 8, 12, 24, 12, 24, 32, 48, 12, 24, 32, 48, 24, 48, 48, 64, 4, 8, 12, 24, 12, 24, 32, 48, 12, 24, 32, 64, 32, 64, 64, 96, 12, 24, 32, 48, 32, 64, 64, 96, 24, 48, 64, 96, 48, 96, 96, 128, 4, 8, 12, 24

Offset: 0

David Radcliffe, Jun 23 2025

This is a 2-automatic sequence.

			Let b(n) be the binary expansion of n. Then a(n) = (1 + p10 + p11) * 2^c, where c is the number of set bits in b(n), p10 is the number of '10' patterns in b(n), and p11 is 1 or 0 depending on whether the pattern '11' is occurring in b(n) or not. This formula is used by _Chai Wah Wu_ in the Python function below. For instance:
  n = 25 -> b(n) = 11001 -> a(n) = (1+1+1) * 2^3 = 24.
  n = 26 -> b(n) = 11010 -> a(n) = (1+2+1) * 2^3 = 32.
  n = 27 -> b(n) = 11011 -> a(n) = (1+1+1) * 2^4 = 48.
- _Peter Luschny_, Jun 25 2025

David Radcliffe, Table of n, a(n) for n = 0..10000
Kenneth S. Davis and William A. Webb, Pascal's triangle modulo 4, Fib. Quart., 29 (1991), 79-83.

Cf. A001316, A014081, A033264, A051638 (mod 3 analog), A085357. Row sums of triangle A034931.

A384715[n_] := 2^DigitSum[n, 2]*(StringCount[IntegerString[n, 2], "10"] - Boole[BitAnd[n,2*n] == 0] + 2);
Array[A384715, 100, 0] (* Paolo Xausa, Jun 26 2025 *)

a(n) = sum(k=0, n, binomial(n,k)%4); \\ Michel Marcus, Jun 25 2025

def A001316(n): return (1 + (n % 2)) * A001316(n // 2) if n else 1
def A033264(n): return (n % 4 == 2) + A033264(n // 2) if n else 0
def A085357(n): return int(n & (n<<1) == 0)
def A384715(n): return A001316(n) * (A033264(n) - A085357(n) + 2)

def A384715(n): return (((n>>1)&~n).bit_count()+bool(n&(n<<1))+1)<Chai Wah Wu, Jun 25 2025

def a(n: int) -> int:  # after Chai Wah Wu
    b = bin(n)[2:]; p = b.count("10"); q = b.count("11")
    return (p + (2 if q else 1)) * 2**n.bit_count()  # Peter Luschny, Jun 25 2025

a(n) = A001316(n) * (A033264(n) - A085357(n) + 2) for n > 0.
Recurrences:
a(4n) = a(2n),
a(4n+1) = 2a(2n),
a(8n+2) = a(4n+2) + 2a(2n) - a(2n+1),
a(8n+3) = a(4n+3) + 4a(2n) - 4a(n),
a(8n+6) = a(4n+3) + 2a(4n+2) - 2a(2n+1),
a(8n+7) = 2a(4n+3).

A163000 Count of integers x in [0,n] satisfying A000120(x) + A000120(n-x) = A000120(n) + 1.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 2, 0, 1, 2, 4, 4, 2, 4, 4, 0, 1, 2, 4, 4, 4, 8, 8, 8, 2, 4, 8, 8, 4, 8, 8, 0, 1, 2, 4, 4, 4, 8, 8, 8, 4, 8, 12, 16, 8, 16, 16, 16, 2, 4, 8, 8, 8, 16, 16, 16, 4, 8, 16, 16, 8, 16, 16, 0, 1, 2, 4, 4, 4, 8, 8, 8, 4, 8, 12, 16, 8, 16, 16, 16, 4, 8, 12, 16, 12, 24, 24, 32, 8, 16, 24, 32, 16

Offset: 0

Vladimir Shevelev, Jul 20 2009

For every solution x, binomial(n,x) is 2 times an odd integer.
A generalization: for every solution 0 <= x <= n of the equation A000120(x) + A000120(n-x) = A000120(n) + r, binomial(n,x) is 2^r times an odd integer.
Apparently this is also the number of 2's in the n-th row of A034931. - R. J. Mathar, Jul 28 2017

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Kenneth S. Davis and William A. Webb, Pascal's triangle modulo 4, Fib. Quart., 29 (1991), 79-83.
Vladimir Shevelev, Binomial predictors, arXiv:0907.3302 [math.NT], 2009.
L. Spiegelhofer and M. Wallner, Divisibility of binomial coefficients by powers of two, arXiv:1710.10884 [math.NT], 2017.

Cf. A000120, A007814.

A001316 and A163577 count binomial coefficients with 2-adic valuation 0 and 2. A275012 gives a measure of complexity of these sequences. - Eric Rowland, Mar 15 2017

A163000 := proc(n) local a,x; a := 0 ; for x from 0 to n do if A000120(x)+A000120(n-x) = A000120(n)+1 then a := a+1; fi; od: a; end:
seq(A163000(n),n=0..100) ; # R. J. Mathar, Jul 21 2009

okQ[x_, n_] := DigitCount[x, 2, 1] + DigitCount[n - x, 2, 1] == DigitCount[n, 2, 1] + 1; a[n_] := Count[Range[0, n], x_ /; okQ[x, n]]; Table[a[n], {n, 0, 92}] (* Jean-François Alcover, Jul 13 2017 *)

a(n) = my(z=hammingweight(n)+1); sum(x=0, n, hammingweight(x) + hammingweight(n-x) == z); \\ Michel Marcus, Jun 06 2021

a(n)=0 iff n=2^k-1, k>=0. a(n)=1 iff n=2^k, k>=1.

Conjecture: a(n) = A033264(n)* 2^(A000120(n)-1); from [Davis & Webb]. - R. J. Mathar, Jul 28 2017

Extended beyond a(22) by R. J. Mathar, Jul 21 2009

A249732 Number of (not necessarily distinct) multiples of 4 on row n of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 1, 0, 6, 4, 3, 0, 7, 2, 3, 0, 14, 12, 11, 8, 13, 6, 7, 0, 19, 14, 11, 4, 17, 6, 7, 0, 30, 28, 27, 24, 29, 22, 23, 16, 33, 26, 23, 12, 29, 14, 15, 0, 43, 38, 35, 28, 37, 22, 23, 8, 45, 34, 27, 12, 37, 14, 15, 0, 62, 60, 59, 56, 61, 54, 55, 48, 65, 58, 55, 44, 61, 46, 47, 32

Offset: 0

Antti Karttunen, Nov 04 2014

nonn

a(n) = Number of zeros on row n of A034931 (Pascal's triangle reduced modulo 4).

This should have a formula (see A048967).

			Row 9 of Pascal's triangle is: {1,9,36,84,126,126,84,36,9,1}. The terms 36 and 84 are only multiples of four, and both of them occur two times on that row, thus a(9) = 2*2 = 4.
Row 10 of Pascal's triangle is: {1,10,45,120,210,252,210,120,45,10,1}. The terms 120 (= 4*30) and 252 (= 4*63) are only multiples of four, 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..8192
Kenneth S. Davis and William A. Webb, Pascal's triangle modulo 4, Fib. Quart., 29 (1991), 79-83.

Cf. A007318, A034931, A048277, A048645, A072823, A048967, A062296, A187059, A249733.

A249732(n) = { my(c=0); for(k=0,n\2,if(!(binomial(n,k)%4),c += (if(k<(n/2),2,1)))); return(c); } \\ Slow...
for(n=0, 8192, write("b249732.txt", n, " ", A249732(n)));

def A249732(n): return n+1-(2+((n>>1)&~n).bit_count()<>1) # Chai Wah Wu, Jul 24 2025

Other identities:
a(n) <= A048277(n) for all n.
a(n) <= A048967(n) for all n.

A046097 Values of n for which binomial(2n-1, n) is squarefree.

Original entry on oeis.org

1, 2, 3, 4, 6, 9, 10, 12, 36

Offset: 1

Eric W. Weisstein

No more terms up to 2^300. The sequence is finite by results of Sander and of Granville and Ramaré (see links). - Robert Israel, Dec 10 2015

Eric Weisstein's World of Mathematics, Binomial Coefficient.
A. Granville and O. Ramaré, Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients, Mathematika 43 (1996), 73-107.
J. W. Sander, Prime power divisors of binomial coefficients, Journal für die reine und angewandte Mathematik 430 (1992), 1-20.

Cf. A001700.

For a term to be here, it needs to be at least in the intersection of A048645, A051382, A050607, A050608 and an infinitude of similar sequences. The corresponding location in next-to-center column should be nonzero in A034931 (Pascal's triangle mod 4) and all similarly constructed fractal triangles (Pascal's triangle mod p^2).

[n: n in [1..150] | IsSquarefree(Binomial(2*n-1,n))]; // Vincenzo Librandi, Dec 10 2015

select(n -> numtheory:-issqrfree(binomial(2*n-1,n)), [$1..2000]); # Robert Israel, Dec 09 2015
N:= 300: # to find all terms <= 2^N
carries:= proc(n,m,p)
# number of carries when adding n + m in base p.
local A,B,C,j,nc, t;
   A:= convert(m,base,p);
   B:= convert(n,base,p);
C:= 0; nc:= 0;
   if nops(A) < nops(B) then A = [op(A),0$(nops(B)-nops(A))]
   elif nops(A) > nops(B) then B:= [op(B), 0$(nops(A)-nops(B))]
   fi;
for j from 1 to nops(A) do
    t:= C + A[j] + B[j];
    if t >= p then
       nc:= nc+1;
       C:= 1;
    else
       C:= 0
    fi
od:
nc;
end proc:
Cands:=  {seq(2^j,j=0..N), seq(seq(2^j + 2^k, k=0..j-1),j=1..N-1)}:
for i from 2 to 10 do
  Cands:= select(n -> carries(n-1,n,ithprime(i)) <= 1, Cands)
od:
select(n -> numtheory:-issqrfree(binomial(2*n-1,n)),Cands); # Robert Israel, Dec 10 2015

Select[ Range[1500], SquareFreeQ[ Binomial[ 2#-1, #]] &] (* Jean-François Alcover, Oct 25 2012 *)

is(n)=issquarefree(binomial(2*n-1,n)) \\ Anders Hellström, Dec 09 2015

James Sellers reports no further terms below 1500.
Michael Somos checked to 99999. Probably there are no more terms.
Mauro Fiorentini checked up to 2^64, as for n = 545259520, the binomial coefficient is a multiple of 5^4 and other possible exceptions have been checked (see Weisstein page for details).

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

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).

cp's OEIS Frontend

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

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

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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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A384715 a(n) = Sum_{k=0..n} (binomial(n, k) mod 4).

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A163000 Count of integers x in [0,n] satisfying A000120(x) + A000120(n-x) = A000120(n) + 1.

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A249732 Number of (not necessarily distinct) multiples of 4 on row n of Pascal's triangle.

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A046097 Values of n for which binomial(2n-1, n) is squarefree.

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