A194459 -id:A194459 - cp's OEIS frontend

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

Jean-Marc Rebert, Aug 15 2025

nonn

			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.

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

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

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 *)

a(n) = if (n, sum(r=0, n, sum(k=0, r, !(binomial(r,k) % n))), 0); \\ Michel Marcus, Aug 15 2025

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

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

A387109 Number of entries in the n-th row of Pascal's triangle not divisible by 27.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 10, 20, 30, 19, 26, 33, 28, 32, 36, 25, 32, 39, 32, 37, 42, 39, 42, 45, 40, 44, 48, 45, 48, 51, 50, 52, 54, 19, 38, 57, 34, 47, 60, 49, 56, 63, 40, 53, 66, 51, 60

Offset: 0

Chai Wah Wu, Aug 16 2025

nonn

Eric Rowland, The number of nonzero binomial coefficients modulo p^alpha, arXiv:1001.1783 [math.NT]

Cf. A001316, A006047, A194459, A382720-A382725, A386952, A387050, A387051, A387108.

import re
from gmpy2 import digits
def A387109(n):
    s = digits(n,3)
    n1, n2, n10, n20, n21, n11 = s.count('1'), s.count('2'), s.count('10'), s.count('20'), s.count('21'), len(re.findall('(?=11)',s))
    n100, n110, n120, n101, n111, n121 = s.count('100'), s.count('110'), s.count('120'), len(re.findall('(?=101)',s)), len(re.findall('(?=111)',s)), len(re.findall('(?=121)',s))
    n200, n201, n210, n211, n220, n221 = s.count('200'), s.count('201'), s.count('210'), s.count('211'), s.count('220'), s.count('221')
    c = 144*n10+63*n11+128*(n20+n220)+80*n21+864*n100+216*(n101+n110)+54*n111+96*n120+24*n121+1152*n200+288*(n201+n210+1)+72*n211+32*n221
    c += (m:=4*n10+n11)*(96*n20+24*n21+9*m)+16*(4*n20+n21)**2
    return (c*3**n2<>5

A339146 a(n) = a(floor(n / 5)) * (n mod 5 + 1); initial terms are 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 2, 4, 6, 8, 10, 3, 6, 9, 12, 15, 4, 8, 12, 16, 20, 5, 10, 15, 20, 25, 1, 2, 3, 4, 5, 2, 4, 6, 8, 10, 3, 6, 9, 12, 15, 4, 8, 12, 16, 20, 5, 10, 15, 20, 25, 1, 2, 3, 4, 5, 2, 4, 6, 8, 10, 3, 6, 9, 12, 15, 4, 8, 12, 16, 20, 5, 10, 15, 20, 25

Offset: 0

Robert Dougherty-Bliss, Nov 25 2020

nonn

If a(n) is arranged in a table with row lengths 5, then the first column is the transpose of the first row, followed the transpose of the second row, followed by the transpose of the third row, and so on. The remainder of each row (except the first) is an arithmetic progression whose start and step size equals the first entry of the row.
a(n) = O(n).
limsup_n a(n) = +oo.

			a(10) = a(2) * 1 = 1.
a(13) = a(2) * 4 = 4.

Cf. A194459.

Cf. A048896 (with 2 instead of 5, but shifted).

a(n) = if (n < 5, 1, a(n\5)*(n % 5 + 1)); \\ Michel Marcus, Nov 26 2020

def a(n):
    if n < 5:
        return 1
    q, r = divmod(n, 5)
    return a(q) * (r + 1)

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

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A387109 Number of entries in the n-th row of Pascal's triangle not divisible by 27.

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A339146 a(n) = a(floor(n / 5)) * (n mod 5 + 1); initial terms are 1.

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