login
A376886
The number of distinct factors of n of the form p^(k!), where p is a prime and k >= 1, when the factorization is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.
6
0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 2, 3, 2, 2, 1, 3, 1, 2, 2, 1, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 1, 3, 1, 3, 3
OFFSET
1,6
COMMENTS
See A376885 for details about this factorization.
First differs from A371090 at n = 2^18 = 262144.
Differs from A064547 at n = 64, 128, 192, 256, 320, 384, 448, 512, ... .
Differs from A058061 at n = 128, 384, 512, 640, 896, ... .
LINKS
FORMULA
Additive with a(p^e) = A060130(e).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) = 0.12589120926760155013..., where f(x) = -x + (1-x) * Sum_{k>=1} A060130(k) * x^k.
EXAMPLE
For n = 8 = 2^3, the representation of 3 in factorial base is 11, i.e., 3 = 1! + 2!, so 8 = (2^(1!))^1 * (2^(2!))^1 and a(8) = 1 + 1 = 2.
For n = 16 = 2^4, the representation of 4 in factorial base is 20, i.e., 4 = 2 * 2!, so 16 = (2^(2!))^2 and a(16) = 1.
MATHEMATICA
fdignum[n_] := Module[{k = n, m = 2, r, s = 0}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r > 0, s++]; m++]; s]; f[p_, e_] := fdignum[e]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
PROG
(PARI) fdignum(n) = {my(k = n, m = 2, r, s = 0); while([k, r] = divrem(k, m); k != 0 || r != 0, if(r > 0, s ++); m++); s; }
a(n) = {my(e = factor(n)[, 2]); sum(i = 1, #e, fdignum(e[i])); }
CROSSREFS
Similar sequences: A064547, A318464, A376885.
Sequence in context: A065031 A305832 A058061 * A371090 A064547 A318306
KEYWORD
nonn,easy,base
AUTHOR
Amiram Eldar, Oct 08 2024
STATUS
approved