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
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. A249425 (record positions).
Cf. A249427 (record values).
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)))
Comments