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
Examples
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.
Links
- Chai Wah Wu, Table of n, a(n) for n = 0..10000
Crossrefs
Programs
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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 *) -
PARI
a(n) = if (n, sum(r=0, n, sum(k=0, r, !(binomial(r,k) % n))), 0); \\ Michel Marcus, Aug 15 2025
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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
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