A258490 Number of words of length 2n such that all letters of the ternary alphabet occur at least once and are introduced in ascending order and which can be built by repeatedly inserting doublets into the initially empty word.

5, 56, 465, 3509, 25571, 184232, 1325609, 9567545, 69387483, 505915981, 3708195075, 27314663271, 202116910415, 1501769001416, 11200258810265, 83815491037841, 629152465444715, 4735907436066401, 35740538971518155, 270356740041089471, 2049510329494271615

Offset

3,1

Links

Alois P. Heinz, Table of n, a(n) for n = 3..1000

Formula

a(n) ~ 8^n / (sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jun 01 2015

Conjecture: 4*n*(n-1)*(46829*n-161203)*a(n) -(n-1)*(4865671*n^2-22433759*n+19821114)*a(n-1) +6*(7756949*n^3-53792553*n^2+117956226*n-84118712)*a(n-2) +(-200071007*n^3+1677158106*n^2-4623144589*n+4201946850)*a(n-3) +2*(2*n-7)*(93171685*n^2-585009841*n+881711802)*a(n-4) -72*(2*n-7)*(2*n-9)*(744719*n-1901876)*a(n-5)=0. - R. J. Mathar, Aug 07 2015

Example

a(3) = 5: aabbcc, aabccb, abbacc, abbcca, abccba.

Maple

A:= proc(n, k) option remember; `if`(n=0, 1, k/n*
      add(binomial(2*n, j)*(n-j)*(k-1)^j, j=0..n-1))
    end:
T:= (n, k)-> add((-1)^i*A(n, k-i)/(i!*(k-i)!), i=0..k):
a:= n-> T(n, 3):
seq(a(n), n=3..25);

Mathematica

A[n_, k_] := A[n, k] = If[n == 0, 1, k/n*Sum[Binomial[2*n, j]*(n - j)*If[j == 0, 1, (k - 1)^j], {j, 0, n - 1}]];
T[n_, k_] := Sum[(-1)^i*A[n, k - i]/(i!*(k - i)!), {i, 0, k}];
a[n_] := T[n, 3];
Table[a[n], {n, 3, 25}] (* Jean-François Alcover, May 18 2018, translated from Maple *)

Crossrefs

Column k=3 of A256117.

Sequence in context: A030060 A247710 A247774 * A255953 A174249 A073563

Adjacent sequences: A258487 A258488 A258489 * A258491 A258492 A258493

Keyword

nonn

Author

Alois P. Heinz, May 31 2015

Status

approved