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A376887
The number of divisors of n that are products of factors of the form p^(k!) with multiplicities not larger than their multiplicity in n, where p is a prime and k >= 1, when the factorization of n is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.
3
1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 3, 2, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 6, 4, 4, 4, 4, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 6, 2, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 2, 4, 8, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 4, 8, 2, 6, 3, 4, 2, 8, 4, 4, 4
OFFSET
1,2
COMMENTS
See A376885 for details about this factorization.
If n = Product p_i^e_i is the canonical prime factorization of n, then the divisors that are counted by this function are d = Product p_i^s_i, where all the digits of s_i in factorial base are not larger than the corresponding digits of e_i.
The sum of these divisors is given by A376888(n).
LINKS
FORMULA
Multiplicative with a(p^e) = A227154(e).
EXAMPLE
For n = 12 = 2^2 * 3^1, the representation of 2 in factorial base is 10, i.e., 2 = 2!, so 12 = (2^(2!))^1 * (3^(1!))^1 and a(12) = (1+1) * (1+1) = 4, corresponding to the 4 divisors 1, 3, 4 and 12.
MATHEMATICA
fdigprod[n_] := Module[{k = n, m = 2, r, s = 1}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, s *= (r+1); m++]; s]; f[p_, e_] := fdigprod[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
PROG
(PARI) fdigprod(n) = {my(k = n, m = 2, r, s = 1); while([k, r] = divrem(k, m); k != 0 || r != 0, s *= (r+1); m++); s; }
a(n) = {my(e = factor(n)[, 2]); prod(i = 1, #e, fdigprod(e[i])); }
CROSSREFS
KEYWORD
nonn,easy,mult,base
AUTHOR
Amiram Eldar, Oct 08 2024
STATUS
approved