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 1-5 of 5 results.

A110707 Number of linear arrangements of n blue, n red and n green items such that there are no adjacent items of the same color (first and last elements considered as adjacent).

Original entry on oeis.org

6, 24, 132, 804, 5196, 34872, 240288, 1688244, 12040188, 86892384, 633162360, 4650680640, 34390540320, 255773538240, 1911730760832, 14350853162676, 108139250403804, 817629606524112, 6200696697358344, 47152195812692664
Offset: 1

Views

Author

Max Alekseyev, Aug 04 2005

Keywords

Comments

The number of linear arrangements is given by A110706 and the number of circular arrangements counted up to rotations is given by A110710.

Crossrefs

Programs

  • Mathematica
    b = Binomial; a[n_] := 2*Sum[b[n-1, k]*(b[n-1, k]*(b[2*n+1-2*k, n+1] - 3* b[2*n-1-2*k, n+1]) + b[n-1, k+1]*(b[2*n-2*k, n+1] - 3*b[2*n-2*k-2, n+1]) ), {k, 0, n/2}]; Array[a, 20] (* Jean-François Alcover, Dec 04 2015, adapted from PARI *)
  • PARI
    a(n) = 2 * sum(k=0,n\2, binomial(n-1,k) * ( binomial(n-1,k)*(binomial(2*n+1-2*k,n+1)-3*binomial(2*n-1-2*k,n+1)) + binomial(n-1,k+1)*(binomial(2*n-2*k,n+1)-3*binomial(2*n-2*k-2,n+1)) ))

Formula

a(n) = 2 * Sum[k=0..[n/2]] binomial(n-1, k) * ( binomial(n-1, k)*(binomial(2n+1-2k, n+1)-3*binomial(2n-1-2k, n+1)) + binomial(n-1, k+1)*(binomial(2n-2k, n+1)-3*binomial(2n-2k-2, n+1)) )
a(n) = A110706(n) - A110711(n)
a(n) = 2*A000172(n-1)+2*A000172(n) - Mark van Hoeij, Jul 14 2010
Conjecture: n^2*a(n) -3*n*(2*n-1)*a(n-1) -3*(n-1)*(5*n-12)*a(n-2) -8*(n-3)^2*a(n-3)=0. - R. J. Mathar, Jul 26 2014
a(n) ~ 3^(3/2) * 2^(3*n - 1) / (Pi*n). - Vaclav Kotesovec, Nov 09 2024

A110710 Number of ternary necklaces with n beads of each color and no adjacent beads of the same color (i.e., no substrings 00, 11, 22).

Original entry on oeis.org

1, 2, 5, 16, 70, 348, 1948, 11444, 70380, 445944, 2896590, 19186740, 129186596, 881808728, 6089851874, 42482906040, 298976142764, 2120377458900, 15141289233972, 108784152585236, 785869931659980, 5705406374249272
Offset: 0

Views

Author

Max Alekseyev, Aug 05 2005

Keywords

Comments

The number of circular arrangements (counted up to rotations) of n blue, n red and n green items such that there are no adjacent items of the same color. The number of various linear arrangements is given by A110706, A110707 and A110711.

Examples

			For n=2 there are 5 necklaces: 010212, 012012, 012021, 012102, 021021.
		

Crossrefs

Programs

  • Mathematica
    b = Binomial; A110707[n_] := 2*Sum[b[n - 1, k]*(b[n - 1, k]*(b[2*n + 1 - 2*k, n + 1] - 3*b[2*n - 1 - 2*k, n + 1]) + b[n - 1, k + 1]*(b[2*n - 2*k, n + 1] - 3*b[2*n - 2*k - 2, n + 1])), {k,0, n/2}]; a[n_] := DivisorSum[n, A110707[n/#]*EulerPhi[#]&]/(3n); a[0]=1; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Dec 04 2015, adapted from PARI *)
  • PARI
    { A110707(n) = 2 * sum(k=0,n\2, binomial(n-1,k) * (binomial(n-1,k)*(binomial(2*n+1-2*k,n+1)-3*binomial(2*n-1-2*k,n+1)) + binomial(n-1,k+1)*(binomial(2*n-2*k,n+1)-3*binomial(2*n-2*k-2,n+1)) )); A110710(n) = sumdiv(n,d,A110707(n\d)*eulerphi(d))\(3*n); }

Formula

a(n) = Sum_{d|n} A110707(n/d)*eulerphi(d) / (3n) for n>0, a(0)=1.
a(n) ~ sqrt(3) * 2^(3*n - 1) / (Pi * n^2). - Vaclav Kotesovec, Mar 20 2023

Extensions

a(0)=1 prepended by Alois P. Heinz, Dec 04 2015

A141147 Number of linear arrangements of n blue, n red and n green items such that the first item is blue and there are no adjacent items of the same color (first and last elements considered as adjacent).

Original entry on oeis.org

2, 8, 44, 268, 1732, 11624, 80096, 562748, 4013396, 28964128, 211054120, 1550226880, 11463513440, 85257846080, 637243586944, 4783617720892, 36046416801268, 272543202174704, 2066898899119448, 15717398604230888
Offset: 1

Views

Author

Max Alekseyev, Jun 10 2008

Keywords

Crossrefs

Programs

  • Maple
    A141147 := n -> 2^n*hypergeom([n, (1-n)/2, -n/2],[1, 1],1);
    seq(simplify(A141147(i)),i=1..20); # Peter Luschny, Jan 15 2012
  • PARI
    { a(n) = sum(k=0,n\2, binomial(n,2*k) * binomial(2*k,k) * binomial(n-1+k,k) * 2^(n-2*k) ) }

Formula

a(n) = A110707(n) / 3 = (A000172(n) + A000172(n-1)) * 2 / 3.
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2k) * binomial(2k,k) * binomial(n-1+k,k) * 2^(n-2k).
a(n) = 2^n*Hypergeometric([n,(1-n)/2,-n/2],[1, 1],1). - Peter Luschny, Jan 15 2012
Recurrence: (3*n^3 + 13*n^2 + 16*n + 4)*a(n+2) = (21*n^3 + 73*n^2 + 74*n + 16)*a(n+1) + (24*n^3 + 32*n^2)*a(n). - Ralf Stephan, Feb 11 2014
a(n) = (1/n) * Sum_{k = floor(n/2)..n} k * binomial(n,k)^2 * binomial(2*k,n). - Peter Bala, Mar 19 2023

A141146 Number of linear arrangements of n blue, n red and n green items such that first and last elements are blue but there are no adjacent items of the same color.

Original entry on oeis.org

0, 2, 14, 96, 664, 4660, 33144, 238448, 1732112, 12685428, 93552700, 694072720, 5176136640, 38777105120, 291661779920, 2201518518240, 16670124621472, 126586920736564, 963723103197516, 7354034055776864, 56236603567496720
Offset: 1

Views

Author

Max Alekseyev, Jun 10 2008

Keywords

Crossrefs

Programs

  • PARI
    { a(n) = sum(k=0,n\2, binomial(n-1,2*k) * binomial(2*k,k) * binomial(n-1+k,k+1) * 2^(n-1-2*k) ) }

Formula

a(n) = A110711(n) / 3.
a(n) = Sum[k=0..[n/2]] binomial(n-1,2k) * binomial(2k,k) * binomial(n-1+k,k+1) * 2^(n-1-2k).
G.f.: (2*x-1)^2*(1-8*x)^(-4/3)*(x+1)^(-8/3)*hypergeom([4/3, 4/3],[2],27*x^2/((8*x-1)*(x+1)^2))-(1-8*x)^(-1/3)*(x+1)^(-2/3)*hypergeom([1/3, 1/3],[1],27*x^2/((8*x-1)*(x+1)^2)). - Mark van Hoeij, May 14 2013
Conjecture: -(n+1)*(n-2)*a(n) +(7*n^2-13*n+4)*a(n-1) +8*(n-2)^2*a(n-2)=0. - R. J. Mathar, Jul 23 2014

A141148 Number of aperiodic ternary necklaces with n beads of each color and no adjacent beads of the same color.

Original entry on oeis.org

2, 3, 14, 65, 346, 1929, 11442, 70310, 445928, 2896239, 19186738, 129184583, 881808726, 6089840427, 42482905678, 298976072384, 2120377458898, 15141288786096, 108784152585234, 785869928763325, 5705406374237814
Offset: 1

Views

Author

Max Alekseyev, Jun 10 2008

Keywords

Crossrefs

Formula

a(n) = Sum_{d|n} moebius(n/d) * A141147(d) / n.
Showing 1-5 of 5 results.