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.

A007123 Number of connected unit interval graphs with n nodes; also number of bracelets (turnover necklaces) with n black beads and n-1 white beads.

Original entry on oeis.org

1, 1, 2, 4, 10, 26, 76, 232, 750, 2494, 8524, 29624, 104468, 372308, 1338936, 4850640, 17685270, 64834550, 238843660, 883677784, 3282152588, 12233309868, 45741634536, 171530482864, 644953425740, 2430975800876, 9183681736376, 34766785487152, 131873995933480
Offset: 1

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Comments

Also number of rooted planar general trees (of n vertices or n-1 edges) up to reflection. - Antti Karttunen, Aug 09 2002 (For the correspondence with bracelets, start by considering Raney's lemma as explained by Graham, Knuth & Patashnik.)
Number of connected lattice path matroids on n elements up to isomorphism.
a(n) = number of noncrossing set partitions of [n] up to reflection (i<->n+1-i). Example: a(4) counts 123, 1-23, 13-2, 1-2-3 but not 12-3 because it is the reflection of 1-23. - David Callan, Oct 08 2005
From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of n beads, each of which is painted by one of 2*n-1 colors.
The sequence solves the so-called Reis problem about convex k-gons in case N=2*n-1, k=n. H. Gupta (1979) gave a full solution; I gave a short proof of Gupta's result and showed an equivalence of this problem and each of the following problems: the problem of enumerating the bracelets of n beads of 2 colors, k of them black, and the problem of enumerating the necklaces of k beads, each painted by one of n colors.
a(n) is an essentially unimprovable upper estimate for the number of distinct values of the permanent in (0,1)-circulants of order 2*n-1 with n 1's in every row. (End)
The number of Dyck paths of semilength n-1 up to reversal; that is, the number of Dyck paths of semilength n-1, treating as identical a path and that path when traveled in reverse order. - Noah A Rosenberg, Jan 28 2019

Examples

			x + x^2 + 2*x^3 + 4*x^4 + 10*x^5 + 26*x^6 + 76*x^7 + 232*x^8 + 750*x^9 + ...
		

References

  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 5.6.7.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 345 & 346.
  • R. W. Robinson, personal communication.
  • R. W. Robinson, Numerical implementation of graph counting algorithms, AGRC Grant, Math. Dept., Univ. Newcastle, Australia, 1980.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Occurs as row 164 in A073201.
Next-to-center columns of triangle A052307.
Equal to A001405 plus A006079.

Programs

  • Mathematica
    f[k_Integer, n_] := (Plus @@ (EulerPhi[ # ]Binomial[n/#, k/# ] & /@ Divisors[GCD[n, k]])/n + Binomial[(n - If[OddQ@n, 1, If[OddQ@k, 2, 0]])/2, (k - If[OddQ@k, 1, 0])/2])/2 (* Robert A. Russell, Sep 27 2004 *)
    Table[ f[n, 2n - 1], {n, 10}]
    (* Comment from Wouter Meeussen, Feb 02 2013, added by N. J. A. Sloane, Feb 02 2013: To get lists of the necklaces in Mathematica, use (if n=4, say):
    <
    				
  • PARI
    {a(n) = if( n<1, 0, (2 * binomial(n-1, (n-1)\2) + binomial(2*n, n) / (2*n - 1)) / 4)} /* Michael Somos, Apr 16 2012 */
    
  • Python
    from sympy import catalan, binomial, floor
    def a(n): return 1 if n==1 else (catalan(n - 1) + binomial(n - 1, floor((n - 1)/2)))/2 # Indranil Ghosh, Jun 03 2017

Formula

a(n+1) = (Catalan(n) + binomial(n, floor(n/2)))/2 = (A000108(n) + A001405(n))/2. - Antti Karttunen, Aug 09 2002
G.f.: (1 + 2*x - sqrt(1 - 4*x)*sqrt(1 - 4*x^2))/(4*sqrt(1 - 4*x^2)).
G.f.: (sqrt((1 + 2*x) / (1 - 2*x)) - sqrt(1 - 4*x)) / 4. - Michael Somos, Apr 16 2012
a(n) = (A063886(n) - A002420(n)) / 4. - Michael Somos, Apr 16 2012
D-finite with recurrence n*(n-1)*(n-4)*a(n) - 4*(n-1)*(n^2-5*n+5)*a(n-1) - 4*(n-2)*(n^2-7*n+11)*a(n-2) + 8*(2*n-7)*(n-2)*(n-3)*a(n-3)=0. - R. J. Mathar, Aug 22 2018

Extensions

Extended by Christian G. Bower
Edited by Jon E. Schoenfield, Feb 14 2015