A050356 Number of ordered factorizations of n with 2 levels of parentheses.

1, 1, 1, 4, 1, 7, 1, 16, 4, 7, 1, 40, 1, 7, 7, 64, 1, 40, 1, 40, 7, 7, 1, 208, 4, 7, 16, 40, 1, 73, 1, 256, 7, 7, 7, 292, 1, 7, 7, 208, 1, 73, 1, 40, 40, 7, 1, 1024, 4, 40, 7, 40, 1, 208, 7, 208, 7, 7, 1, 544, 1, 7, 40, 1024, 7, 73, 1, 40, 7, 73, 1, 1840, 1, 7, 40, 40, 7, 73, 1

Offset: 1

Author

Christian G. Bower, Oct 15 1999

Keywords

• nonn

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).

Examples

			For n=6 we have ((6)) = ((3*2)) = ((2*3)) = ((3)*(2)) = ((2)*(3)) = ((3))*((2)) = ((2))*((3)), thus a(6) = 7.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002033, A050351, A050352, A050353, A050354, A050355, A050357, A050358, A050359.

Programs

PARI

A050356aux(n) = if(1==n,1/3, 3*sumdiv(n,d, if(dA050356aux(d), 0)));
A050356(n) = if(1==n,n,A050356aux(n)); \\ Antti Karttunen, May 19 2017, after the general recurrence given by Vladeta Jovovic May 25 2005 in A050354.

Formula

Dirichlet g.f.: (3-2*zeta(s))/(4-3*zeta(s)).

a(p^k) = 4^(k-1).

a(A002110(n)) = A050352(n).

Sum_{k=1..n} a(k) ~ -n^r / (9*r*Zeta'(r)), where r = 2.52138975790328306967497455387140053675965539610041801606891036... is the root of the equation Zeta(r) = 4/3. - Vaclav Kotesovec, Feb 02 2019