cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A050359 Number of ordered factorizations with 3 levels of parentheses indexed by prime signatures: A050358(A025487(n)).

Original entry on oeis.org

1, 1, 5, 9, 25, 65, 125, 425, 121, 625, 605, 2625, 1145, 3125, 4825, 15625, 9225, 15625, 35125, 13265, 90625, 2169, 45225, 67625, 78125, 240625, 126025, 515625, 25385, 377625, 465625, 390625, 1578125, 1062625, 2890625, 242825, 2915625
Offset: 1

Views

Author

Christian G. Bower, Oct 15 1999

Keywords

Extensions

Link in title corrected by Matthew Vandermast, Jul 03 2012

A050354 Number of ordered factorizations of n with one level of parentheses.

Original entry on oeis.org

1, 1, 1, 3, 1, 5, 1, 9, 3, 5, 1, 21, 1, 5, 5, 27, 1, 21, 1, 21, 5, 5, 1, 81, 3, 5, 9, 21, 1, 37, 1, 81, 5, 5, 5, 111, 1, 5, 5, 81, 1, 37, 1, 21, 21, 5, 1, 297, 3, 21, 5, 21, 1, 81, 5, 81, 5, 5, 1, 201, 1, 5, 21, 243, 5, 37, 1, 21, 5, 37, 1, 513, 1, 5, 21, 21, 5, 37, 1, 297, 27, 5, 1, 201
Offset: 1

Views

Author

Christian G. Bower, Oct 15 1999

Keywords

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).
Dirichlet inverse of (A074206*A153881). - Mats Granvik, Jan 12 2009

Examples

			For n=6, we have (6) = (3*2) = (2*3) = (3)*(2) = (2)*(3), thus a(6) = 5.
		

Crossrefs

Programs

  • Mathematica
    A[n_]:=If[n==1, n/2, 2*Sum[If[dIndranil Ghosh, May 19 2017 *)
  • PARI
    A050354aux(n) = if(1==n,n/2, 2*sumdiv(n,d, if(dA050354aux(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 dIndranil Ghosh, May 19 2017, after Antti Karttunen's PARI program

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, d1, a(1)= 1/k. - Vladeta Jovovic, May 25 2005
a(p^k) = 3^(k-1).
a(A002110(n)) = A050351(n).
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

Extensions

Duplicate comment removed by R. J. Mathar, Jul 15 2010

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

Original entry on oeis.org

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

Views

Author

Christian G. Bower, Oct 15 1999

Keywords

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.
		

Crossrefs

Programs

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
Showing 1-3 of 3 results.