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 1-2 of 2 results.

A135805 Eighth column (k=7) of triangle A134832 (circular succession numbers).

Original entry on oeis.org

1, 0, 0, 120, 330, 6336, 61776, 785928, 10456875, 151099520, 2339361024, 38655753552, 678721170036, 12615988058880, 247449420044640, 5106608041235184, 110596074738524661, 2507849090860975488
Offset: 0

Views

Author

Wolfdieter Lang, Jan 21 2008

Keywords

Comments

a(n) enumerates circular permutations of {1,2,...,n+7} with exactly seven successor pairs (i,i+1). Due to cyclicity also (n+7,1) is a successor pair.

Examples

			a(0)=1 because from the 7!/7 = 720 circular permutations of n=7 elements only one, namely (1,2,3,4,5,6,7), has seven successors.
		

References

  • Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 183, eq. (5.15), for k=7.

Crossrefs

Cf. A135804 (column k=6), A135806 (column k=8).

Programs

  • Mathematica
    f[n_] := (-1)^n + Sum[(-1)^k*n!/((n - k)*k!), {k, 0, n - 1}]; a[n_, n_] = 1; a[n_, 0] := f[n]; a[n_, k_] := a[n, k] = n/k*a[n - 1, k - 1]; Table[a[n, 7], {n, 7, 25}] (* G. C. Greubel, Nov 10 2016 *)

Formula

a(n) = binomial(n+7,7)*A000757(n), n>=0.
E.g.f.: (d^7/dx^7) (x^7/7!)*(1-log(1-x))/e^x.

A135807 Tenth column (k=9) of triangle A134832 (circular succession numbers).

Original entry on oeis.org

1, 0, 0, 220, 715, 16016, 180180, 2619760, 39503750, 642172960, 11111964864, 204016477080, 3959206825210, 80952590044480, 1739019535313720, 39150661649469744, 921633956154372175, 22640304292494917600
Offset: 0

Views

Author

Wolfdieter Lang, Jan 21 2008, Feb 22 2008

Keywords

Comments

a(n) enumerates circular permutations of {1,2,...,n+9} with exactly nine successor pairs (i,i+1). Due to cyclicity also (n+9,1) is a successor pair.

Examples

			a(0)=1 because from the 9!/9 = 40320 circular permutations of n=9 elements only one, namely (1,2,3,4,5,6,7,8,9), has nine successors.
		

References

  • Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 183, eq. (5.15), for k=9.

Crossrefs

Cf. A135806 (column k=8).

Programs

  • Mathematica
    f[n_] := (-1)^n + Sum[(-1)^k*n!/((n - k)*k!), {k, 0, n - 1}]; a[n_, n_] = 1; a[n_, 0] := f[n]; a[n_, k_] := a[n, k] = n/k*a[n - 1, k - 1]; Table[a[n, 9], {n, 9, 25}] (* G. C. Greubel, Nov 10 2016 *)
  • PARI
    a(n)=((-1)^n + sum( k=0, n-1, (-1)^k * binomial( n, k) * (n - k - 1)!))*binomial(n+9,9) \\ Charles R Greathouse IV, Nov 10 2016

Formula

a(n) = binomial(n+9,9)*A000757(n), n>=0.
E.g.f.: (d^9/dx^9) (x^9/9!)*(1-log(1-x))/e^x.
Showing 1-2 of 2 results.