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.

A306350 Number of paraphyletic coalescence sequences for 2n lineages, n each in 2 species.

Original entry on oeis.org

0, 4, 162, 23328, 9072000, 7873200000, 13367512620000, 40367907740160000, 201793403949096960000, 1578804075215377920000000, 18484433452834116768000000000, 312162837144268369009766400000000, 7374810540967959718955457331200000000
Offset: 1

Views

Author

Noah A Rosenberg, Feb 09 2019

Keywords

Comments

Consider a binary tree evolving in time from a single node until the tree has 2n labeled leaves. Color the 2n leaves in 2 colors, red and blue, assigning n leaves to each color. Suppose coalescences of pairs of leaves happen at distinct times (i.e., no simultaneous mergers). A coalescence sequence is a sequence of coalescence events backward in time, tracing the reduction of the 2n leaves to the single ancestral node. A paraphyletic coalescence sequence is a sequence in which (1) all n red leaves have a common ancestor node that is not the ancestor of any blue leaves; or (2) all n blue leaves have a common ancestor node that is not the ancestor of any red leaves; but not both (1) and (2).

Examples

			For n=2, consider two red leaves R1 and R2 and two blue leaves B1 and B2. The a(2)=4 paraphyletic coalescence sequences, separated by semicolons, are (R1,R2), ((R1,R2),B1), (((R1,R2),B1),B2); (R1,R2), ((R1,R2),B2), (((R1,R2),B2),B1); (B1,B2), ((B1,B2),R1), (((B1,B2),R1),R2); and (B1,B2), ((B1,B2),R2), (((B1,B2),R2),R1).

Links

Crossrefs

The total number of coalescence sequences for n leaves, from among which the paraphyletic coalescence sequences are identified, follows A006472. Reciprocally monophyletic coalescence sequences, in which conditions (1) and (2) above both hold, follow A306266.

Programs

  • Mathematica
    Table[3*(n!)^2*(2n - 2)!*(n - 1)/((n + 1)(2^(2 n - 3))), {n, 1, 30}]

Formula

a(n) = 3*(n!)^2*(2*n-2)!*(n-1)/((n+1)*2^(2*n-3)).
a(n) ~ 24*exp(-4*n)*n^(4*n-1/2)*Pi^(3/2). - Stefano Spezia, Apr 30 2024

A306391 Number of polyphyletic coalescence sequences for 2n lineages, n each in 2 species.

Original entry on oeis.org

0, 12, 2484, 1557792, 2560572000, 9326330280000, 66250877823900000, 834917902101803520000, 17373747843395915811840000, 564479089176417832085760000000, 27382950623629177584815808000000000, 1912097851374544604017590025267200000000, 186429568131038636125345650494922854400000000
Offset: 1

Views

Author

Noah A Rosenberg, Feb 12 2019

Keywords

Comments

Consider a binary tree evolving in time from a single node until the tree has 2n labeled leaves. Color the 2n leaves in 2 colors, red and blue, assigning n leaves to each color. Suppose coalescences of pairs of leaves happen at distinct times (i.e., no simultaneous mergers). A coalescence sequence is a sequence of coalescence events backward in time, tracing the reduction of the 2n leaves to the single ancestral node. Consider two possible features of the resulting tree: (1) all n red leaves have a common ancestor node that is not the ancestor of any blue leaves; (2) all n blue leaves have a common ancestor node that is not the ancestor of any red leaves. A reciprocally monophyletic sequence satisfies both (1) and (2). A paraphyletic coalescence sequence satisfies (1) or (2) but not both. A polyphyletic coalescence sequence does not satisfy either (1) or (2).

Examples

			For n=2, consider two red leaves R1 and R2 and two blue leaves B1 and B2. The a(2)=12 polyphyletic coalescence sequences, separated by semicolons, are (B1,R1), ((B1,R1),B2), (((B1,R1),B2),R2); (B1,R1), ((B1,R1),R2), (((B1,R1),R2),B2); (B1,R2), ((B1,R2),B2), (((B1,R2),B2),R1); (B1,R2), ((B1,R2),R1), (((B1,R2),R1),B2); (B2,R1), ((B2,R1),B1), (((B2,R1),B1),R2); (B2,R1), ((B2,R1),R2), (((B2,R1),R2),B1); (B2,R2), ((B2,R2),B1), (((B2,R2),B1),R1); (B2,R2), ((B2,R2),R1), (((B2,R2),R1),B1); (B1,R1), (B2,R2), ((B1,R1),(B2,R2)); (B1,R2), (B2,R1), ((B1,R2),(B2,R1)); (B2,R1), (B1,R2), ((B2,R1),(B1,R2)); (B2,R2), (B1,R1), ((B2,R2),(B1,R1)).

Links

Crossrefs

The total number of coalescence sequences for n leaves, from among which the polyphyletic coalescence sequences are identified, follows A006472. The sum of a(n), A306266(n), and A306350(n) is A006472(2n).

Programs

  • Mathematica
    Table[(1 - (2 n! n!/(2 n)!) (7 n - 5)/((n + 1) (2 n - 1))) (2 n)! (2 n - 1)!/2^(2 n - 1), {n, 1, 30}]
  • PARI
    a(n) = (1 - (2*n!*n!/(2*n)!)*(7*n-5)/((n+1)*(2*n-1)))*(2*n)!*(2*n-1)!/2^(2*n-1); \\ Michel Marcus, Feb 12 2019

Formula

a(n) = (1 - (2 n! n!/(2n)!)(7n-5)/((n+1)(2n-1))) (2n)! (2n-1)!/2^(2n-1).
a(n) ~ exp(-4*n)*n^(4*n-1)*(4^n + 3*4^(1+n)*n - 84*sqrt(n*Pi))*Pi/3. - Stefano Spezia, Apr 30 2024

Extensions

a(12)-a(13) from Stefano Spezia, Apr 30 2024

A377461 Number of ranked labeled trees compatible with the 2-leaf perfect phylogeny of sample size n that possesses the largest number of compatible ranked labeled trees.

Original entry on oeis.org

1, 1, 2, 9, 54, 540, 6480, 113400, 2268000, 61236000, 1837080000, 70727580000, 2970558360000, 154469034720000, 8650265944320000, 583892951241600000, 42040292489395200000, 3573424861598592000000, 321608237543873280000000, 33608060823334757760000000
Offset: 2

Views

Author

Noah A Rosenberg, Jan 03 2025

Keywords

Comments

The 2-leaf perfect phylogeny of sample size n that possesses the largest number of compatible ranked labeled trees is (floor(n/2), ceiling(n/2)); a(n) is the number of ranked labeled trees for this perfect phylogeny.

Links

Crossrefs

Cf. A001405, A006472, A306266.

Programs

  • Mathematica
    a[n_] := ((n - 2)!/((Floor[n/2] - 1)! (n - 1 - Floor[n/2])!)) Product[Binomial[i, 2], {i, 2, Floor[n/2]}] Product[Binomial[i, 2], {i, 2, Ceiling[n/2]}]
a[n_] := ((n - 2)!/((Floor[n/2] - 1)! (n - 1 - Floor[n/2])!)) Floor[n/2]! (Floor[n/2] - 1)! Ceiling[n/2]! (Ceiling[n/2] - 1)! /(2^(Floor[n/2] - 1) 2^(Ceiling[n/2] - 1))

Formula

a(n) = ((n-2)! / ((floor(n/2)-1)! (n-1-floor(n/2))!)) * (floor(n/2))! (floor(n/2)-1)! (ceiling(n/2))! (ceiling(n/2)-1)! / (2^(floor(n/2)-1) 2^(ceiling(n/2)-1)).
a(n) = A001405(n-2)*A006472(floor(n/2))*A006472(ceiling(n/2)).
a(2n) = A306266(n).
Showing 1-3 of 3 results.

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