A387064 Total number of entries in rows 0 to n of Pascal's triangle multiple of n.

0, 3, 1, 2, 2, 4, 3, 6, 4, 6, 10, 10, 12, 12, 21, 22, 8, 16, 18, 18, 30, 42, 47, 22, 38, 20, 74, 18, 65, 28, 81, 30, 16, 113, 136, 132, 94, 36, 147, 195, 140, 40, 162, 42, 199, 210, 217, 46, 126, 42, 146, 302, 261, 52, 110, 335, 243, 374, 394, 58, 363, 60, 465, 416

Offset

0,2

Links

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Formula

a(p) = p-1, a(p^2) = p*(p-1) for p prime. Conjecture: a(p^k) = (p-1)*p^(k-1) for p prime. - Chai Wah Wu, Aug 21 2025

Example

The first two rows of Pascal's triangle are [1] and [1, 1]. Since all elements are divisible by 1, a(1) equals the total number of such divisible terms: 1 + 2 = 3.

Mathematica

a[n_] := Sum[Boole[Divisible[Binomial[k, i], n]], {k, 0, n}, {i, 0, k}]; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Aug 17 2025 *)

Prog

(PARI) a(n) = if (n, sum(r=0, n, sum(k=0, r, !(binomial(r, k) % n))), 0); \\ Michel Marcus, Aug 15 2025
(Python)
from sympy import isprime, integer_nthroot
def A387064(n):
    if isprime(n): return n-1
    a, b = integer_nthroot(n, 2)
    if b and isprime(a): return n-a
    r, c = [1], n==1
    for m in range(n):
        s = [1]
        for i in range(m):
            s.append((r[i]+r[i+1])%n)
            c += s[-1]==0
        r = s+[1]
        c += (n==1)<<1
    return int(c) # Chai Wah Wu, Aug 21 2025

Crossrefs

Cf. A001316, A006046, A006047, A006048, A007318, A194458, A194459, A382720-A382731, A386953.

Sequence in context: A002016 A282016 A238848 * A117494 A342913 A221490

Adjacent sequences: A387061 A387062 A387063 * A387065 A387066 A387067

Keyword

nonn

Author

Jean-Marc Rebert, Aug 15 2025

Status

approved