A374959 -id:A374959 - cp's OEIS frontend

A375082 a(n) = binomial(A349958(n),A374959(n))/n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 15, 1, 5, 1, 3, 1, 1, 1, 4, 1, 1, 5, 4, 1, 1, 1, 102, 2, 3, 1, 2, 1, 5, 1, 10659, 1, 91, 1, 6, 3, 33, 1, 325, 1, 1, 3, 7, 1, 2, 1, 8, 2, 1, 11, 5, 1, 2, 1463, 1, 1, 11, 1, 9, 4, 51, 6, 1, 1, 7, 1

Offset: 1

Pontus von Brömssen, Jul 29 2024

nonn

Pontus von Brömssen, Table of n, a(n) for n = 1..10000

Cf. A007318, A349958, A374959, A375083, A375084.

from math import comb
def A375082(n): return next(a for a, b in (divmod(comb(j,k),n) for j in range(n+1) for k in range(j+1)) if not b) # Chai Wah Wu, Jul 30 2024

A349958 a(n) is the index of the first row in Pascal's triangle that contains a multiple of n.

Original entry on oeis.org

0, 2, 3, 4, 5, 4, 7, 8, 9, 5, 11, 9, 13, 8, 6, 16, 17, 9, 19, 6, 7, 11, 23, 10, 25, 13, 27, 8, 29, 10, 31, 32, 11, 17, 7, 9, 37, 19, 13, 10, 41, 9, 43, 12, 10, 23, 47, 16, 49, 25, 18, 13, 53, 27, 11, 8, 19, 29, 59, 10, 61, 32, 9, 64, 13, 11, 67, 17, 23, 8, 71, 12, 73, 37, 25

Offset: 1

Nathan M Epstein, Dec 06 2021

nonn

a(n) is the minimum j such that binomial(j,k) is divisible by n for some k in 0..j.

a(n) is at most equal to A058084(n), the least m such that binomial(m,k) = n for some k.

			In the table below, the k value shown is the minimum k such that n divides binomial(a(n), k).
.
   n  a(n)  k  C(a(n), k)
  --  ----  -  ----------
   1    0   0       1
   2    2   1       2
   3    3   1       3
   4    4   1       4
   5    5   1       5
   6    4   2       6
   7    7   1       7
   8    8   1       8
   9    9   1       9
  10    5   2      10
  11   11   1      11
  12    9   2      36
.
The table below shows the left half (and middle column) of rows j = 0..12 of Pascal's triangle; each number in parentheses there is the first term encountered in Pascal's triangle (read by rows from left to right) that is a multiple of some number n in 1..12, and the corresponding term of {a(n)} whose value is j appears in the column at the right.
E.g., the first multiple of 12 encountered in Pascal's triangle is binomial(9,2) = 36; it appears in row 9, so a(12) = 9, and the column at the right includes a(12) in row 9.
                                               | terms in a(1)..a(12)
   j | left half of row j of Pascal's triangle | that are equal to j
  ---+-----------------------------------------+---------------------
   0 |                                    (1)  |        a(1)  =  0
   1 |                                  1      |
   2 |                               1    (2)  |        a(2)  =  2
   3 |                            1    (3)     |        a(3)  =  3
   4 |                         1    (4)   (6)  |  a(4), a(6)  =  4
   5 |                      1    (5)  (10)     |  a(5), a(10) =  5
   6 |                   1     6    15    20   |
   7 |                1    (7)   21    35      |        a(7)  =  7
   8 |             1    (8)   28    56    70   |        a(8)  =  8
   9 |          1    (9)  (36)   84   126      |  a(9), a(12) =  9
  10 |       1    10    45   120   210   252   |
  11 |    1   (11)   55   165   330   462      |        a(11) = 11
  12 | 1    12    66   210   496   792   924   |

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Cf. A007318, A058084, A374959.

a[n_] := Module[{k = 0}, While[!AnyTrue[Binomial[k, Range[0, Floor[k/2]]], Divisible[#, n] &], k++]; k]; Array[a, 75] (* Amiram Eldar, Dec 07 2021 *)

a(n) = my(k=0); while (!#select(x->(x==1), apply(denominator, vector((k+2)\2, i, binomial(k, i-1))/n)), k++); k; \\ Michel Marcus, Dec 07 2021

a(n) = { my (r = [1 % n]); for (i = 0, oo, if (vecmin(r)==0, return (i), r = (concat(0, r) + concat(r, 0)) % n;);); }

import numpy as np
def pascals(n):
  a = np.ones(1)
  f = np.ones(2)
  triangle = [a]
  for i in range(n):
    a = np.convolve(a,f)
    triangle.append(a)
  return triangle
def test(n,tri):
  for i, element in enumerate(tri):
    for sub_e in element:
      if sub_e % n == 0:
        return i
tri = pascals(500)
for i in range(1,50):
  print(test(i,tri),end=',')

from math import comb
def A349958(n):
    for j in range(n+1):
        for k in range(j+1):
            if comb(j,k) % n == 0: return j # Chai Wah Wu, Dec 10 2021

More terms from Michel Marcus, Dec 07 2021

A375083 Numbers k such that k is not the first multiple of k that appears in Pascal's triangle (read by rows), i.e., such that A375082(k) > 1.

Original entry on oeis.org

12, 14, 18, 22, 24, 26, 30, 33, 34, 38, 39, 40, 42, 44, 46, 48, 50, 51, 52, 54, 57, 58, 60, 62, 63, 65, 66, 68, 69, 72, 74, 75, 76, 77, 80, 82, 85, 86, 87, 88, 90, 92, 93, 94, 95, 96, 98, 99, 100, 102, 104, 105, 106, 108, 110, 111, 112, 114, 115, 116, 117, 118

Offset: 1

Pontus von Brömssen, Jul 29 2024

nonn

Cf. A007318, A349958, A374959, A375082.

isok(k) = for (i=1, k-1, for (j=1, i, my(b=binomial(i, j)); if (((b % k) == 0), return(b!=k)););); \\ Michel Marcus, Jul 30 2024

from itertools import count, islice
from math import comb
def A375083_gen(): # generator of terms
    return filter(lambda n:next(c for c in (comb(j,k) for j in range(n+1) for k in range(j+1)) if not c%n)>n, count(1))
A375083_list = list(islice(A375083_gen(),20)) # Chai Wah Wu, Jul 30 2024

cp's OEIS Frontend

A375082 a(n) = binomial(A349958(n),A374959(n))/n.

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A349958 a(n) is the index of the first row in Pascal's triangle that contains a multiple of n.

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A375083 Numbers k such that k is not the first multiple of k that appears in Pascal's triangle (read by rows), i.e., such that A375082(k) > 1.

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