0,5

a(p) = 0 for every odd prime p. Proof: Since, according to the divisor function, tau(k) = product (e_i + 1) = p, we must have k = q^(p - 1) for some prime q. Thus we would need a prime q whose exponent in the factorization of p! equals p - 1. The largest exponent occurs for q = 2. By Legendre's formula v_2(p!) = p - s_2(p), where v_2(p!) is the exponent of 2 in the prime factorization of p!, and s_2(p) is the sum of the binary digits of p. For any odd prime p, the binary expansion contains at least two 1s, so s_2(p) >= 2. Hence v_2(p!) = p - s_2(p) < p - 1. Therefore no prime can occur with exponent p - 1 in p!, so such a k does not exist and a(p) = 0.

Eric Weisstein's World of Mathematics, Divisor Function.

Wikipedia, Legendre's formula.

a(p) = 0 for odd primes p.

The a(6) = 4 divisors of 6! = 720 with exactly 6 divisors are 12 (1, 2, 3, 4, 6, 12), 18 (1, 2, 3, 6, 9, 18), 20 (1, 2, 4, 5, 10, 20) and 45 (1, 3, 5, 9, 15, 45).

A393242 := proc(N)

local a, i, k, l, n, p, t, b, w;

w := proc(f, g)

local c, e, m, r;

if f = 1 then return 1; end if;

if nops(g) = 0 then return 0; end if;

c := 0;

m := g[1];

r := g[2..-1];

for e from 0 to m do

if f mod (e + 1) = 0 then

c := c + w(f / (e + 1), r);

end if;

end do;

return c;

end proc;

a := [0];

l := [];

t := table();

for n from 1 to N do

p := ifactors(n)[2];

for i in p do

if member(i[1], l) then

t[i[1]] := t[i[1]] + i[2];

else

l := [op(l), i[1]];

t[i[1]] := i[2];

end if;

end do;

b := [seq(t[k], k in l)];

a := [op(a), w(n, b)];

end do;

return a;

end proc:

A393242(66);

a[n_]:=Count[Length/@Divisors[ Divisors[n!]], n]; Array[a, 20, 0] (* James C. McMahon, Feb 24 2026 *)

Cf. A000005, A000142, A005179, A027423, A099311, A393241.

Sequence in context: A077957 A077966 A275670 * A021102 A021053 A182443

Adjacent sequences: A393239 A393240 A393241 * A393243 A393244 A393245

Felix Huber, Feb 18 2026

approved