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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.
2
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
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
KEYWORD
nonn
AUTHOR
Alois P. Heinz, May 31 2015
STATUS
approved