A249151 Largest m such that m! divides the product of elements on row n of Pascal's triangle: a(n) = A055881(A001142(n)).

1, 1, 2, 1, 4, 2, 6, 1, 2, 4, 10, 7, 12, 6, 4, 1, 16, 2, 18, 4, 6, 10, 22, 11, 4, 12, 2, 6, 28, 25, 30, 1, 10, 16, 6, 36, 36, 18, 12, 40, 40, 6, 42, 10, 23, 22, 46, 19, 6, 4, 16, 12, 52, 2, 10, 35, 18, 28, 58, 47, 60, 30, 63, 1, 12, 10, 66, 16, 22, 49, 70, 41, 72, 36, 4, 18, 10, 12, 78, 80, 2

Offset: 0

Author

Antti Karttunen, Oct 25 2014

Keywords

• nonn
• changed

Comments

A000225 gives the positions of ones.

A006093 seems to give all such k, that a(k) = k.

Examples

			              Binomial coeff.   Their product  Largest k!
                 A007318          A001142(n)   which divides
Row 0                1                    1        1!
Row 1              1   1                  1        1!
Row 2            1   2   1                2        2!
Row 3          1   3   3   1              9        1!
Row 4        1   4   6   4   1           96        4! (96 = 4*24)
Row 5      1   5  10  10   5   1       2500        2! (2500 = 1250*2)
Row 6    1   6  15  20  15   6   1   162000        6! (162000 = 225*720)
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..10000 (terms 0..4096 from Antti Karttunen)

Crossrefs

One more than A249150.

Cf. A249423 (numbers k such that a(k) = k+1).

Cf. A249429 (numbers k such that a(k) > k).

Cf. A249433 (numbers k such that a(k) < k).

Cf. A249434 (numbers k such that a(k) >= k).

Cf. A249424 (numbers k such that a(k) = (k-1)/2).

Cf. A249428 (and the corresponding values, i.e. numbers n such that A249151(2n+1) = n).

Cf. A249425 (record positions).

Cf. A249427 (record values).

Cf. A001142, A006093, A000225, A007917, A055881, A187059, A249346, A249421, A249430, A249431, A249432.

Programs

PARI

A249151(n) = { my(uplim,padicvals,b); uplim = (n+3); padicvals = vector(uplim); for(k=0, n, b = binomial(n, k); for(i=1, uplim, padicvals[i] += valuation(b, prime(i)))); k = 1; while(k>0, for(i=1, uplim, if((padicvals[i] -= valuation(k, prime(i))) < 0, return(k-1))); k++); };
\\ Alternative implementation:
A001142(n) = prod(k=1, n, k^((k+k)-1-n));
A055881(n) = { my(i); i=2; while((0 == (n%i)), n = n/i; i++); return(i-1); }
A249151(n) = A055881(A001142(n));
for(n=0, 4096, write("b249151.txt", n, " ", A249151(n)));

Python

from itertools import count
from collections import Counter
from math import comb
from sympy import factorint
def A249151(n):
    p = sum((Counter(factorint(comb(n,i))) for i in range(n+1)),start=Counter())
    for m in count(1):
        f = Counter(factorint(m))
        if not f<=p:
            return m-1
        p -= f # Chai Wah Wu, Aug 19 2025

Scheme

(define (A249151 n) (A055881 (A001142 n)))

Formula

a(n) = A055881(A001142(n)).