A157672 -id:A157672 - cp's OEIS frontend

A157836 Triangle read by rows where T(n,k) is the number of factorizations of (n+1)! into k distinct factors.

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 14, 28, 13, 1, 1, 29, 103, 95, 24, 1, 1, 47, 273, 448, 249, 41, 1, 1, 79, 725, 1897, 1837, 671, 74, 1, 1, 134, 1876, 7301, 10856, 6780, 1686, 127, 1, 1, 269, 5791, 31811, 65782, 59434, 24017, 3960, 197, 1, 1, 395, 12061, 92987, 272932, 362956, 232152, 69765, 8703, 323, 1

Offset: 1

Ray Chandler, Mar 07 2009

n-th row has n terms; first and last term in each row = 1.

			Triangle begins:
2! 1
3! 1 1
4! 1 3 1
5! 1 7 7 1
6! 1 14 28 13 1
7! 1 29 103 95 24 1
8! 1 47 273 448 249 41 1
9! 1 79 725 1897 1837 671 74 1
10! 1 134 1876 7301 10856 6780 1686 127 1
11! 1 269 5791 31811 65782 59434 24017 3960 197 1
12! 1 395 12061 92987 272932 362956 232152 69765 8703 323 1
...

Andrew Howroyd, Table of n, a(n) for n = 1..465 (first 30 rows)

A157612 gives row sums. A157672 gives 2nd column.

EulerT(v)={Vec(exp(x*Ser(dirmul(v, vector(#v, n, 1/n))))-1, -#v)}
D(p, n, sig)={my(v=vector(n)); for(i=1, #p, v[p[i]]++); my(r=EulerT(v)); prod(i=1, #sig, r[sig[i]])/prod(i=1, #v, i^v[i]*v[i]!)}
detail(sig)={my(m=vecsum(sig)+1,n=vecmax(sig), q=Vec(exp(intformal(O(x^m) - x^n/(1-x)))/(y+x))); if(n==0, 1, (-1)^m*sum(j=0, m, my(s=0); forpart(p=j, s+=(-1)^#p*D(p, n, sig), [1, n]); s*q[#q-j]*y^m)/(1+y))}
row(n)={if(n<=1, [], Vecrev(detail(factor(n!)[,2])))}
{ for(n=1, 10, print(row(n+1))) } \\ Andrew Howroyd, Feb 01 2020

A344687 a(n) is the lowest nonnegative exponent k such that n!^k is the product of the divisors of n!.

Original entry on oeis.org

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

Offset: 1

Alex Sokolov, Aug 17 2021

nonn

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).

			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.

Cf. A000005, A000142, A027423, A280420, A157672.

Join[{0},Table[DivisorSigma[0,n!]/2,{n,2,39}]] (* Stefano Spezia, Aug 18 2021 *)

a(n) = if (n==1, 0, numdiv(n!)/2); \\ Michel Marcus, Aug 18 2021

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

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

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.

cp's OEIS Frontend

A157836 Triangle read by rows where T(n,k) is the number of factorizations of (n+1)! into k distinct factors.

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