A384452 a(n) is the sum of squares of the unitary divisors of n!.

1, 5, 50, 650, 16900, 547924, 27396200, 1746641000, 139773881000, 13460683752200, 1642203417768400, 236441876606410000, 40195119023089700000, 7723888546922636420000, 1735183690969722609168800, 444206919394766468845892000, 128820006624482275965308680000, 41737604550102658693597600532800

Offset: 1

Author

Darío Clavijo, Jun 02 2025

Keywords

• nonn

Links

• Project Euler, Problem 429: Sum of Squares of Unitary Divisors.
• Wikipedia, Legendre's formula.

Crossrefs

Cf. A000142, A034676, A064028, A077610.

Programs

Mathematica

f[p_, e_] := p^(2*e)+1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 18] (* Amiram Eldar, Jun 02 2025 *)

PARI

row(n) = {my(d = divisors(n)); select(x->(gcd(x, n/x)==1), d); } \\ A077610
a(n) = norml2(row(n!)); \\ Michel Marcus, Jun 02 2025

Python

from sympy import nextprime
def f(n,p):
  if n==0: return 0
  return f(n//p,p) + n//p
def a(n):
  s,p = 1, 2
  while p<=n:
    s *= p**(f(n,p)<<1)+1
    p = nextprime(p)
  return s
print([a(n) for n in range(1, 19)])

Formula

a(n) = Sum_{d|n!} (d^2 if gcd(d,n!//d) = 1).

a(n) = Product_{p <= n, p prime} (p^(2*f(n,p)))+1 with f(n,p) = f(floor(n/p)) + floor(n/p) and f(0,p) = 0 where f(n,p) is equivalent to the Legendre formula.

a(n) = A034676(n!).