OFFSET
1,4
COMMENTS
a(n) depends only on prime signature of n (cf. A025487). So a(24) = a(375) since 24 = 2^3*3 and 375 = 3*5^3 both have prime signature (3,1).
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..10000
FORMULA
Dirichlet g.f.: (2-zeta(s))/(3-2*zeta(s)).
Recurrence for number of ordered factorizations of n with k-1 levels of parentheses is a(n) = k*Sum_{d|n, d<n} a(d), n>1, a(1)= 1/k. - Vladeta Jovovic, May 25 2005
a(p^k) = 3^(k-1).
Sum_{k=1..n} a(k) ~ -n^r / (4*r*Zeta'(r)), where r = 2.185285451787482231198145140899733642292971552057774261555354324536... is the root of the equation Zeta(r) = 3/2. - Vaclav Kotesovec, Feb 02 2019
EXAMPLE
For n=6, we have (6) = (3*2) = (2*3) = (3)*(2) = (2)*(3), thus a(6) = 5.
MATHEMATICA
A[n_]:=If[n==1, n/2, 2*Sum[If[d<n, A[d], 0], {d, Divisors[n]}]]; Table[If[n==1, n, A[n]], {n, 1, 100}] (* Indranil Ghosh, May 19 2017 *)
PROG
(PARI)
A050354aux(n) = if(1==n, n/2, 2*sumdiv(n, d, if(d<n, A050354aux(d), 0)));
A050354(n) = if(1==n, n, A050354aux(n)); \\ Antti Karttunen, May 19 2017, after Jovovic's general recurrence.
(Sage)
def A(n): return 1/2 if n==1 else 2*sum(A(d) for d in divisors(n) if d<n)
def a(n): return 1 if n==1 else A(n)
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, May 19 2017, after Antti Karttunen's PARI program
CROSSREFS
KEYWORD
nonn
AUTHOR
Christian G. Bower, Oct 15 1999
EXTENSIONS
Duplicate comment removed by R. J. Mathar, Jul 15 2010
STATUS
approved