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A136123
Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k maximal strings of increasing consecutive integers (0<=k<=floor(n/2)).
6
1, 1, 1, 1, 3, 3, 11, 12, 1, 53, 56, 11, 309, 321, 87, 3, 2119, 2175, 693, 53, 16687, 17008, 5934, 680, 11, 148329, 150504, 55674, 8064, 309, 1468457, 1485465, 572650, 96370, 5805, 53, 16019531, 16170035, 6429470, 1200070, 95575, 2119
OFFSET
0,5
COMMENTS
Row n has 1+floor(n/2) terms. Row sums are the factorials (A000142). Column 0 yields A000255. Column 1 yields A001277. Column 2 yields A001278. Column 3 yields A001279. Column 4 yields A001280. Sum(k*T(n,k),k>=0)=(n-2)!*(n^2 - 3n + 3)=A001564(n-2).
REFERENCES
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 264, Table 7.6.1.
FORMULA
G.f.: G(x,t) = Sum_{n>=0} n!*((1-t)*x^2 - x)/((1-t)*x^2-1))^n. - Vladeta Jovovic
EXAMPLE
T(3,0)=3 because we have 132, 213 and 321; T(6,3)=3 because we have 125634, 341256, 563412.
Triangle starts:
1;
1;
1, 1;
3, 3;
11, 12, 1;
53, 56, 11;
309, 321, 87, 3;
...
MAPLE
G:=Sum(factorial(n)*(((1-t)*x^2-x)/((1-t)*x^2-1))^n, n=0..infinity): Gser:= simplify(series(G, x=0, 13)): for n from 0 to 11 do P[n]:=sort(coeff(Gser, x, n)) end do: for n from 0 to 11 do seq(coeff(P[n], t, j), j=0..floor((1/2)*n)) end do; # yields sequence in triangular form
# alternative
A136123 := proc(n, k)
add( x^i*( ((1-y)*x-1)/((1-y)*x^2-1) )^i*i!, i=0..n+1) ;
coeftayl(%, x=0, n) ;
coeftayl(%, y=0, k) ;
end proc:
seq(seq( A136123(n, k), k=0..floor(n/2)), n=0..12) ; # R. J. Mathar, Jul 01 2022
MATHEMATICA
T[n_, k_] := Sum[x^i*(((1-y)*x-1)/((1-y)*x^2-1))^i*i!, {i, 0, n+1}] //
SeriesCoefficient[#, {x, 0, n}]& //
SeriesCoefficient[#, {y, 0, k}]&;
Table[Table[T[n, k], {k, 0, Floor[n/2]}], {n, 0, 12}] // Flatten (* Jean-François Alcover, May 09 2023, after R. J. Mathar *)
CROSSREFS
Sequence in context: A146458 A122573 A230011 * A045495 A045494 A217712
KEYWORD
nonn,tabf
AUTHOR
STATUS
approved