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.

A185369 Number of simple labeled graphs on n nodes of degree 1 or 2 without cycles.

Original entry on oeis.org

1, 0, 1, 3, 15, 90, 645, 5355, 50505, 532980, 6219045, 79469775, 1103335695, 16533226710, 265888247625, 4566885297975, 83422361847825, 1614626682669000, 33003508539026025, 710350201433547675, 16057073233633006575
Offset: 0

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Author

Geoffrey Critzer, Feb 20 2011

Keywords

Examples

			a(4) = 15 because there are 15 simple labeled graphs on 4 nodes of degree 1 or 2 without cycles: 1-2 3-4, 1-3 2-4, 1-4 2-3, 1-2-3-4, 1-2-4-3, 1-3-2-4, 1-3-4-2, 1-4-2-3, 1-4-3-2, 2-1-3-4, 2-1-4-3, 3-1-2-4, 3-1-4-2, 4-1-2-3, 4-1-3-2.
		

References

  • Herbert S. Wilf, Generatingfunctionology, p. 104.

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember;
           `if`(n<2, 1-n, add(binomial(n-1, k-1) *k!/2 *a(n-k), k=2..n))
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, Feb 24 2011
  • Mathematica
    a=1/(2(1-x))-1/2-x/2;
    Range[0,20]! CoefficientList[Series[Exp[a],{x,0,20}],x]

Formula

E.g.f.: exp(1/(2*(1-x))-x/2-1/2).
a(n) = 1-n if n<2, else a(n) = Sum_{k=2..n} C(n-1,k-1) * k!/2 * a(n-k).
a(n) ~ 2^(-3/4)*n^(n-1/4)*exp(-3/4+sqrt(2*n)-n). - Vaclav Kotesovec, Sep 25 2013
Conjecture: +2*a(n) +4*(-n+1)*a(n-1) +2*(n-1)*(n-3)*a(n-2) +(n-1)*(n-2)*a(n-3)=0. - R. J. Mathar, Jun 14 2016
a(n) = n! * Sum_{k=0..floor(n/2)} binomial(n-k-1,n-2*k)/(2^k * k!). - Seiichi Manyama, Jun 17 2024

A335344 Expansion of e.g.f. exp(x^2/(2*(1 - x)^2)).

Original entry on oeis.org

1, 0, 1, 6, 39, 300, 2715, 28350, 335265, 4422600, 64298745, 1020477150, 17542820295, 324552648420, 6426708843555, 135573281994150, 3034400481137025, 71801374285040400, 1790523094644709425, 46923435009924823350, 1289032229351717425575
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 02 2020

Keywords

Comments

For n>0, a(n) is also the number of ways to split n people into nonempty groups, have each group sit around a circular table, and select 2 people from each table (where two seating arrangements are considered identical if each person has the same left neighbors in both of them). See example below. - Enrique Navarrete, Oct 01 2023

Examples

			For n = 5, using one table, there are 4! circular seatings and binomial(5,2) ways to select 2 persons, hence 240 ways. Using two tables, the only way we can select 2 persons from each one is seating 3 persons in one table and 2 in the other, which can be done in 20 ways; then choosing 2 persons from each table can be done in 3 ways, hence giving another 60 ways for a total of 300. - _Enrique Navarrete_, Oct 01 2023
		

Crossrefs

Programs

  • Maple
    f:= gfun:-rectoproc({a(n) = -(n-1)*(3*n-7)*a(n-2) + 3*(n-1)*a(n-1) + (n - 1)*(n - 2)*(n - 3)*a(n-3),a(0)=1,a(1)=0,a(2)=1},a(n),remember):
    map(f, [$0..30]); # Robert Israel, Jun 04 2020
  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[x^2/(2 (1 - x)^2)], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = (1/2) Sum[Binomial[n - 1, k - 1] (k - 1) k! a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 20}]
  • PARI
    seq(n)=Vec(serlaplace(exp(x^2/(2*(1 - x)^2) + O(x*x^n)))) \\ Andrew Howroyd, Jun 02 2020

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * A001286(k) * a(n-k).
D-finite with recurrence a(n) = -(n - 1)*(3*n - 7)*a(n - 2) + 3*(n - 1)*a(n - 1) + (n - 1)*(n - 2)*(n - 3)*a(n - 3). - Robert Israel, Jun 04 2020
a(n) ~ n^(n - 1/6) * exp(1/6 - n^(1/3)/2 + 3*n^(2/3)/2 - n) / sqrt(3). - Vaclav Kotesovec, Jun 11 2020
a(n) = n! * Sum_{k=0..floor(n/2)} binomial(n-1,n-2*k)/(2^k * k!). - Seiichi Manyama, Mar 16 2023

A373771 Expansion of e.g.f. exp(x^2 / (2 * (1 - x)^3)) / (1 - x).

Original entry on oeis.org

1, 1, 3, 18, 147, 1425, 15855, 200130, 2838465, 44767485, 777046095, 14705245170, 301014595035, 6621102973485, 155640761791515, 3891902825660850, 103115436832433025, 2884715829245475225, 84950805438277854075, 2626194012669689512050
Offset: 0

Views

Author

Seiichi Manyama, Jun 18 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = n!*sum(k=0, n\2, binomial(n+k, n-2*k)/(2^k*k!));

Formula

a(n) = n! * Sum_{k=0..floor(n/2)} binomial(n+k,n-2*k)/(2^k * k!).
Showing 1-3 of 3 results.