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A344687 a(n) is the lowest nonnegative exponent k such that n!^k is the product of the divisors of n!. 1
0, 1, 2, 4, 8, 15, 30, 48, 80, 135, 270, 396, 792, 1296, 2016, 2688, 5376, 7344, 14688, 20520, 30400, 48000, 96000, 121440, 170016, 266112, 338688, 458640, 917280, 1166400, 2332800, 2764800, 3932160, 6082560, 8211456, 9797760, 19595520, 30233088, 42550272 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
This sequence is a subsequence of A001222, because the product of divisors of n! is n^(d(n)/2) (where d(n) is the number of divisors of n), so a(n) = d(n!)/2.
For prime p, d(p!) = 2*d((p-1)!), so a(p) = 2*a(p-1).
LINKS
FORMULA
a(n) = d(n!)/2 = A000005(A000142(n))/2 = A027423(n)/2 for n > 1.
a(n) = A157672(n-1) + 1 for all n >= 2.
EXAMPLE
For n = 4, n! = 24 = 2^3 * 3, which has (3+1)*(1+1) = 8 divisors: {1,2,3,4,6,8,12,24} whose product is 331776 = (24)^4 = (4!)^4. So a(4) = 4.
MATHEMATICA
Join[{0}, Table[DivisorSigma[0, n!]/2, {n, 2, 39}]] (* Stefano Spezia, Aug 18 2021 *)
PROG
(Python)
def a(n):
d = {}
for i in range(2, n+1):
tmp = i
j = 2
while(tmp != 1):
if(tmp % j == 0):
d.setdefault(j, 0)
tmp //= j
d[j] += 1
else:
j += 1
res = 1
for i in d.values():
res *= (i+1)
return res // 2
(Python)
from math import prod
from collections import Counter
from sympy import factorint
def A344687(n): return prod(e+1 for e in sum((Counter(factorint(i)) for i in range(2, n+1)), start=Counter()).values())//2 # Chai Wah Wu, Jun 25 2022
(PARI) a(n) = if (n==1, 0, numdiv(n!)/2); \\ Michel Marcus, Aug 18 2021
CROSSREFS
Sequence in context: A176503 A262333 A293335 * A301480 A217777 A034338
KEYWORD
nonn
AUTHOR
Alex Sokolov, Aug 17 2021
STATUS
approved

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Last modified June 15 18:09 EDT 2024. Contains 373410 sequences. (Running on oeis4.)