A384715 -id:A384715 - cp's OEIS frontend

A385285 a(n) = Sum_{k=0..n} (binomial(n, k) mod 8).

1, 2, 4, 8, 16, 16, 32, 32, 16, 32, 24, 48, 40, 64, 40, 64, 16, 32, 40, 64, 48, 48, 88, 96, 40, 80, 88, 112, 72, 112, 112, 128, 16, 32, 40, 64, 64, 96, 120, 160, 48, 96, 72, 144, 104, 176, 160, 192, 40, 80, 104, 176, 120, 176, 176, 224, 72, 144, 160, 224, 176

Offset: 0

Peter Luschny, Jun 26 2025

a(n) is a multiple of A001316(n). - Chai Wah Wu, Jun 26 2025

John Tyler Rascoe, Table of n, a(n) for n = 0..8192
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.
Chai Wah Wu, Sum of rows of Pascal's triangle modulo 8, blog post.

Cf. A034930 (row sums of), A001316 (mod 2), A384715 (mod 4).

A385285 := n -> local k; add(modp(binomial(n, k), 8), k = 0..n): seq(A385285(n), n = 0..60);

A385285[n_] := Sum[Mod[Binomial[n, k], 8], {k, 0, n}];
Array[A385285, 100, 0] (* Paolo Xausa, Jun 26 2025 *)

a(n) = sum(k=0, n, binomial(n, k) % 8); \\ Michel Marcus, Jun 26 2025

def A385285(n):
    if n==0: return 1
    s = '0'+bin(n)[2:]
    n1 = n>>1
    n2 = n1>>1
    n3 = n2>>1
    np = ~n
    n1111 = (n3&n2&n1&n).bit_count()
    n11 = (n1&n).bit_count()
    n101 = (n2&(~n1)&n).bit_count()
    n100 = (n2&(~n1)&np).bit_count()
    n110 = (n2&n1&np).bit_count()
    n10 = (n1&np).bit_count()
    n1100 = (n3&n2&(~n1)&np).bit_count()
    m11 = s.count('0110')
    m111 = s.endswith('0111')
    c = (n100<<2)+n110+(n10*(n10-1)>>1)
    if n11==0:
        c += n10+(n100<<1)
    elif n11==1:
        c += (n10-n1100<<1)+n110
    else:
        c += n10<<1
    if not (n1111 or n11 or n101):
        c += 1
    elif not (n1111 or n11) and n101:
        c += 3
    elif m111:
        c += 4
    elif not (n1111 or n101 or m11) and n11:
        c += 2
    else:
        c += 4
    return c<Chai Wah Wu, Jun 26 2025

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<

A385825 a(n) = Sum_{k=0..n} (binomial(n, k) mod 5).

Original entry on oeis.org

1, 2, 4, 8, 11, 2, 4, 8, 16, 22, 4, 8, 16, 22, 34, 8, 16, 22, 44, 48, 11, 22, 34, 48, 61, 2, 4, 8, 16, 22, 4, 8, 16, 32, 44, 8, 16, 32, 44, 68, 16, 32, 44, 88, 96, 22, 44, 68, 96, 122, 4, 8, 16, 22, 34, 8, 16, 32, 44, 68, 16, 32, 59, 68, 106, 22, 44, 68, 116, 142

Offset: 0

Chai Wah Wu, Jul 09 2025

nonn

Chai Wah Wu, Table of n, a(n) for n = 0..10000
Richard Garfield and Herbert S. Wilf, The distribution of the binomial coefficients modulo p, Journal of Number Theory, 41 (1) (1992), 1-5.

Cf. A001316, A051638, A384715, A385285, A385741.

a[n_]:=Sum[Mod[Binomial[n,k],5],{k,0,n}];Array[a,60,0] (* James C. McMahon, Jul 10 2025 *)

from gmpy2 import digits
from sympy.abc import x
from sympy import Poly, rem
def A385825(n):
    k = (1,2,4,3)
    s = digits(n,5)
    t = [s.count(str(i)) for i in range(1,5)]
    G = 2**t[0]*(x+2)**t[1]*(2*x**2+3)**t[3]*(2*x**3+2)**t[2]
    c = Poly(rem(G,x**4-1),x).all_coeffs()[::-1]
    return int(sum(k[i]*c[i] for i in range(len(c)) if c[i]))

Row sums of A095140. - R. J. Mathar, Jul 19 2025

A385393 a(n) = (Sum_{k=0..n} (binomial(n, k) mod 4)) / 2^bitcount(n).

Original entry on oeis.org

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

Offset: 0

Peter Luschny, Jun 27 2025

nonn

Apparently 2^bitcount(n) divides Sum_{k=0..n} (binomial(n, k) mod m) only if m is a power of 2.

Cf. A000120, A001316, A384715, A385394.

a := n -> local j; add(modp(binomial(n, j), 4), j=0..n) / 2^add(convert(n, base, 2)): seq(a(n), n = 0..86);

Table[Sum[Mod[Binomial[n, k], 4], {k, 0, n}]/2^DigitCount[n, 2, 1], {n, 0, 105}] (* Michael De Vlieger, Jun 27 2025 *)
A385393[n_] := StringCount[#, "10"] + Boole[StringContainsQ[#, "11"]] + 1 & [IntegerString[n, 2]]; Array[A385393,100,0] (* Paolo Xausa, Jun 28 2025 *)

a(n) = sum(k=0, n, binomial(n, k) % 4) / 2^hammingweight(n); \\ Michel Marcus, Jun 28 2025

def a(n: int) -> int:
    b = bin(n)[2:]
    return 1 + b.count("10") + int("11" in b)
print([a(n) for n in range(75)])

def A385393(n): return (((m:=n>>1)&~n).bit_count()+bool(n&m)+1) # Chai Wah Wu, Jun 28 2025

a(n) = A384715(n) / 2^A000120(n).

a(n) = A384715(n) / A001316(n).

A385826 a(n) = Sum_{k=0..n} (binomial(n, k) mod 7).

Original entry on oeis.org

1, 2, 4, 8, 16, 18, 22, 2, 4, 8, 16, 32, 36, 44, 4, 8, 16, 32, 43, 58, 67, 8, 16, 32, 36, 72, 60, 92, 16, 32, 43, 72, 81, 120, 121, 18, 36, 58, 60, 120, 100, 144, 22, 44, 67, 92, 121, 144, 169, 2, 4, 8, 16, 32, 36, 44, 4, 8, 16, 32, 64, 72, 88, 8, 16, 32, 64, 86

Offset: 0

Chai Wah Wu, Jul 09 2025

nonn

Sum of n-th row of Pascal's triangle mod 7, A095142.

Chai Wah Wu, Table of n, a(n) for n = 0..10000
Richard Garfield and Herbert S. Wilf, The distribution of the binomial coefficients modulo p, Journal of Number Theory, 41 (1) (1992), 1-5.

Cf. A001316, A051638, A384715, A385285, A385741, A385825.

from gmpy2 import digits
from sympy.abc import x
from sympy import Poly, rem
def A385826(n):
    k = (1,3,2,6,4,5)
    s = digits(n,7)
    t = [s.count(str(i)) for i in range(1,7)]
    G = 2**t[0]*(2*x+2)**t[2]*(x**2+2)**t[1]*(3*x**3+4)**t[5]*(2*x**4+x**3+2)**t[3]*(2*x**5+2*x+2)**t[4]
    c = Poly(rem(G,x**6-1),x).all_coeffs()[::-1]
    return int(sum(k[i]*c[i] for i in range(len(c)) if c[i]))

A385828 a(n) = Sum_{k=0..n} (binomial(n, k) mod 11).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 31, 40, 36, 50, 56, 2, 4, 8, 16, 32, 64, 62, 80, 72, 100, 112, 4, 8, 16, 32, 53, 106, 91, 116, 100, 156, 169, 8, 16, 32, 64, 62, 124, 116, 188, 134, 224, 228, 16, 32, 53, 62, 91, 182, 144, 222, 202, 272, 291, 32, 64, 106, 124, 182, 188, 244

Offset: 0

Chai Wah Wu, Jul 09 2025

nonn

Sum of n-th row of Pascal's triangle mod 11, A095144.

Chai Wah Wu, Table of n, a(n) for n = 0..10000
Richard Garfield and Herbert S. Wilf, The distribution of the binomial coefficients modulo p, Journal of Number Theory, 41 (1) (1992), 1-5.

Cf. A001316, A051638, A384715, A385285, A385741, A385825, A385826.

from gmpy2 import digits
from sympy.abc import x
from sympy import Poly, rem
def A385828(n):
    s = digits(n,11)
    t = tuple(s.count(digits(i,11)) for i in range(1,11))
    G = 2**t[0]*(x+2)**t[1]*(5*x**5+6)**t[9]*(2*x**8+2)**t[2]*(2*x**5+2*x**4+2)**t[4]*(x**9+2*x**2+2)**t[3]*(2*x**7+2*x**5+2*x+2)**t[6]*(2*x**9+2*x**3+x**2+4)**t[7]*(2*x**9+x**6+2*x**2+2)**t[5]*(2*x**8+2*x**7+2*x**6+2*x**4+2)**t[8]
    c = Poly(rem(G,x**10-1),x).all_coeffs()[::-1]
    return int(sum(pow(2,i,11)*c[i] for i in range(len(c)) if c[i]))

cp's OEIS Frontend

A385285 a(n) = Sum_{k=0..n} (binomial(n, k) mod 8).

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

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A385825 a(n) = Sum_{k=0..n} (binomial(n, k) mod 5).

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A385393 a(n) = (Sum_{k=0..n} (binomial(n, k) mod 4)) / 2^bitcount(n).

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A385826 a(n) = Sum_{k=0..n} (binomial(n, k) mod 7).

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A385828 a(n) = Sum_{k=0..n} (binomial(n, k) mod 11).

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