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

Author

Peter Luschny, Jun 26 2025

Keywords

• nonn
• easy

Comments

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

Links

• 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.

Crossrefs

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

Programs

Maple

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

Mathematica

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

PARI

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

Python

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