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.

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A276482 a(n) = 5^n*Gamma(n+1/5)*Gamma(n+1)/Gamma(1/5).

Original entry on oeis.org

1, 1, 12, 396, 25344, 2661120, 415134720, 90084234240, 25944259461120, 9573431741153280, 4403778600930508800, 2470519795122015436800, 1660189302321994373529600, 1316530116741341538208972800, 1216473827868999581305090867200, 1295544626680484554089921773568000
Offset: 0

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Author

Ilya Gutkovskiy, Sep 05 2016

Keywords

Comments

12-gonal (or dodecagonal) factorial numbers, also polygorial(n, 12).
More generally, the m-gonal factorial numbers (or polygorial(n, m)) is 2^(-n)*(m - 2)^n*Gamma(n+2/(m-2))*Gamma(n+1)/Gamma(2/(m-2)), m>2.

Crossrefs

Cf. similar sequences of m-gonal factorial numbers (or polygorial(n, m)): A006472 (m=3), A001044 (m=4), A084939 (m=5), A000680 (m=6), A084940 (m=7), A084941 (m=8), A084942 (m=9), A084943 (m=10), A084944 (m=11).

Programs

  • Maple
    seq(mul(k*(5*k-4),k=1..n), n=0..20); # Robert Israel, Sep 18 2016
  • Mathematica
    FullSimplify[Table[5^n Gamma[n + 1/5] (Gamma[n + 1]/Gamma[1/5]), {n, 0, 15}]]
    polygorial[k_, n_] := FullSimplify[ n!/2^n (k -2)^n*Pochhammer[2/(k -2),n]]; Array[polygorial[12, #] &, 16, 0] (* Robert G. Wilson v, Dec 13 2016 *)
  • PARI
    a(n) = prod(k=1, n, k*(5*k - 4)); \\ Michel Marcus, Sep 06 2016

Formula

a(n) = Product_{k=1..n} k*(5*k - 4), a(0)=1.
a(n) = Product_{k=1..n} A051624(k), a(0)=1.
a(n) = A000142(n)*A008548(n).
a(n) ~ 2*Pi*5^n*n^(2*n+1/5)/(Gamma(1/5)*exp(2*n)).
Sum_{n>=0} 1/a(n) = BesselI(-4/5,2/sqrt(5))*Gamma(1/5)/5^(2/5) = Hypergeometric0F1(1/5, 1/5) = 2.085898421130914...

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).
		

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

A306390 Size of one subtree in the unlabeled binary rooted tree shape of size n whose leaf-labeled trees have the largest number of coalescence sequences.

Original entry on oeis.org

1, 2, 2, 2, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32
Offset: 3

Views

Author

Noah A Rosenberg, Feb 12 2019

Keywords

Comments

Consider the unlabeled, rooted, binary tree shapes (A001190). For each unlabeled shape, assign a labeling of the leaves: a labeled topology. This topology is a leaf-labeled, rooted, binary tree (chosen from among all such trees, A001147). From among the set of possible coalescence sequences (A006472), count all coalescence sequences, or labeled histories, that give rise to the specified labeled topology. For the unlabeled shape of size n whose labeled topologies have the largest number of coalescence sequences, the two subtrees immediately descended from the root have sizes a(n) and n-a(n) (Harding 1971, Eq. 5.7).
The decomposition of trees of n leaves into subtrees with sizes a(n) and n-a(n) also describes the set of unlabeled tree shapes whose labeled topologies have the largest number of root ancestral configurations (Disanto & Rosenberg 2017, Proposition 4).

Examples

			For n=5, the three unlabeled shapes are ((((.,.),.),.),.), (((.,.),(.,.)),.), and (((.,.),.),(.,.)). The formula gives a(5)=2, so that the shape with a subtree of size 2, (((.,.),.),(.,.)), is the one whose labeled topologies have the largest number of coalescence sequences. A labeled topology (((A,B),C),(D,E)) has 3 coalescence sequences, whereas ((((A,B),C),D),E) has 1 and (((A,B),(C,D)),E) has 2.
		

Crossrefs

Programs

  • Mathematica
    Table[2^(1 + Floor[Log2[(n - 1)/3]]), {n, 3, 100}]
  • PARI
    a(n) = 2^(1 + log((n-1)/3)\log(2)); \\ Michel Marcus, Mar 08 2019

Formula

a(n) = 2^(1 + floor(log_2((n-1)/3))).

A381536 Number of labeled histories for rooted 4-furcating trees with 3n+1 leaves if simultaneous 4-furcations are not allowed.

Original entry on oeis.org

1, 1, 35, 7350, 5255250, 9564555000, 37072215180000, 271183254041700000, 3430468163627505000000, 70238835650273164875000000, 2210064963735845132791875000000, 102493972758213553878355995000000000, 6769214430816214165896021689775000000000, 618638506832293812621237422228537250000000000
Offset: 0

Views

Author

Noah A Rosenberg, Feb 26 2025

Keywords

Crossrefs

Programs

  • Maple
    a:= n-> (3*n+1)!/24^n*mul(3*i-2, i=1..n):
    seq(a(n), n=0..13);  # Alois P. Heinz, Feb 26 2025

Formula

a(n) = ((3*n+1)!/24^n) * Product_{i=1..n} (3*i-2).

A120476 Triangle read by rows: a(n,m)=(2*n-1)*(n-m)*(n+m+1)/2, where n is the column and m the row index.

Original entry on oeis.org

1, 3, -2, 6, -5, -9, 10, -9, -21, -20, 15, -14, -36, -45, -35, 21, -20, -54, -75, -77, -54, 28, -27, -75, -110, -126, -117, -77, 36, -35, -99, -150, -182, -189, -165, -104, 45, -44, -126, -195, -245, -270, -264, -221, -135, 55, -54, -156, -245, -315, -360, -374, -351, -285, -170
Offset: 0

Views

Author

Roger L. Bagula, Jul 19 2006

Keywords

Comments

Triangular array based on recurrence in Laplace function in J. W. S. Rayleigh.

Examples

			1,
3, -2,
6, -5, -9,
10, -9, -21,-20,
15, -14,-36,-45, -35
		

References

  • J. W. S. Rayleigh, The Theory of Sound, volume 2, page 237,Dover, New York,1945

Crossrefs

Cf. A006472.

Programs

  • Mathematica
    a = Table[Table[(m + 1)*(2*n - 1)*(n - m)*(n + m + 1)/(2*(m + 1)), {n, 0, m - 1}], {m, 1, 10}] Flatten[a]

Formula

a(n,m) = (2n-1)*[A000217(n)-A000217(m)] = (1-2n)*A049777(n,m) . - R. J. Mathar, Dec 05 2007
Row sums: sum_{n=0..m-1} a(n,m) = -m(m+1)(3m^2-5m-4)/12. [From R. J. Mathar, Jan 15 2009]

Extensions

Edited by N. J. A. Sloane, Oct 01 2006

A233003 (n!)^2 mod Pt(n), where Pt(n) is product of first n positive triangular numbers (A000217).

Original entry on oeis.org

0, 1, 0, 36, 900, 8100, 0, 25401600, 514382400, 12859560000, 6224027040000, 56016243360000, 9466745127840000, 1855482045056640000, 0, 6679735362203904000000, 13513104637738497792000000, 156365925093831188736000000, 225792395835492236534784000000, 22579239583549223653478400000000
Offset: 1

Views

Author

Alex Ratushnyak, Dec 03 2013

Keywords

Comments

Indices of zeros appear to be 2^k-1.

Examples

			a(4) = 1*4*9*16 mod 1*3*6*10 = 576 mod 90 = 36.
		

Crossrefs

Cf. A006472 (triangular factorial, essentially equal to Pt(n)).
Cf. A006788 (floor(n!^2/Pt)).

Programs

  • Python
    s=t=1
    for n in range(1,33):
      s*=n*n
      t*=n*(n+1)//2
      print(s%t, end=', ')

A233004 Pt(n) mod n!, where Pt(n) is product of first n positive triangular numbers (A000217).

Original entry on oeis.org

0, 1, 0, 12, 60, 540, 0, 20160, 181440, 907200, 19958400, 359251200, 1556755200, 32691859200, 0, 10461394944000, 177843714048000, 1600593426432000, 60822550204416000, 608225502044160000, 38318206628782080000, 702500454861004800000, 12926008369442488320000
Offset: 1

Views

Author

Alex Ratushnyak, Dec 03 2013

Keywords

Comments

Pt(n) = n!*(n+1)! / 2^n.
Pt(n) mod n! = 0 if and only if 2^n divides (n+1)!, that is, n+1 is a power of 2. Thus indices of zeros are of the form 2^k-1.

Crossrefs

Cf. A006472 (triangular factorial, essentially equal to Pt(n)).
Cf. A067667 (Pt(n)/n! for n's of the form 2^k-1).

Programs

  • Python
    f=t=1
    for n in range(1,33):
      t*=n*(n+1)//2
      f*=n
      print(t%f, end=', ')

A259345 Number of labeled single-linkage dendrograms.

Original entry on oeis.org

1, 1, 1, 30, 750
Offset: 1

Views

Author

N. J. A. Sloane, Jun 27 2015, following a suggestion by Colin Mallows from 1981

Keywords

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)).
		

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.

Crossrefs

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).
Previous Showing 51-60 of 61 results. Next