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

1,4

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.: (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

Example

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

Prog

(PARI)
A050356aux(n) = if(1==n, 1/3, 3*sumdiv(n, d, if(d<n, A050356aux(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.

Crossrefs

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

Sequence in context: A348981 A358077 A340073 * A245838 A158860 A335619

Adjacent sequences: A050353 A050354 A050355 * A050357 A050358 A050359

Keyword

nonn

Author

Christian G. Bower, Oct 15 1999

Status

approved