A050358 Number of ordered factorizations of n with 3 levels of parentheses.

1, 1, 1, 5, 1, 9, 1, 25, 5, 9, 1, 65, 1, 9, 9, 125, 1, 65, 1, 65, 9, 9, 1, 425, 5, 9, 25, 65, 1, 121, 1, 625, 9, 9, 9, 605, 1, 9, 9, 425, 1, 121, 1, 65, 65, 9, 1, 2625, 5, 65, 9, 65, 1, 425, 9, 425, 9, 9, 1, 1145, 1, 9, 65, 3125, 9, 121, 1, 65, 9, 121, 1, 4825, 1, 9, 65, 65, 9, 121

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).

The Dirichlet inverse is given by A050356, turning all but the first element of A050356 negative. - R. J. Mathar, Jul 15 2010

Examples

			6 = (((6))) = (((3*2))) = (((2*3))) = (((3)*(2))) = (((2)*(3))) = (((3))*((2))) = (((2))*((3))) = (((3)))*(((2))) = (((2)))*(((3))).
		

Links

• R. J. Mathar, Table of n, a(n) for n = 1..899

Crossrefs

Cf. A002033, A050351-A050359. a(p^k)=5^(k-1). a(A002110)=A050353.

Formula

Dirichlet g.f.: (4-3*zeta(s))/(5-4*zeta(s)).

a(n) = A050359(A101296(n)). - R. J. Mathar, May 26 2017

Sum_{k=1..n} a(k) ~ -n^r / (16*r*Zeta'(r)), where r = 2.7884327053324956670606046076818023223650950899573090550836329583345... is the root of the equation Zeta(r) = 5/4. - Vaclav Kotesovec, Feb 02 2019