A059797 - OEIS

%I #54 Mar 26 2021 19:26:27

%S 2,5,5,9,16,9,14,35,35,14,20,64,90,64,20,27,105,189,189,105,27,35,160,

%T 350,448,350,160,35,44,231,594,924,924,594,231,44,54,320,945,1728,

%U 2100,1728,945,320,54,65,429,1430,3003,4290,4290,3003,1430,429,65

%N Second in a series of arrays counting standard tableaux by partition type.

%C The first array in the series is Pascal's triangle, A007318. The initial partition for each subsequent array in the series is chosen as described in A053445. When cells are squared, as in A008459, row sums yield 1, 2, 6, 24, ... (A000142). E.g., (1 + 16 + 36 + 16 + 1) + (25 + 25) = 70 + 50 = 120 using row five from A007318 and row two from this array.

%C Number of lattice paths from (0,0) to (n,k) in exactly n+k+2 steps. Moves allowed: (0,1), (0,-1), (-1,0) and (1,0). Paths must stay in Quadrant I but may touch the axes. - _Juan Luis Vargas-Molina_, Mar 03 2014

%D Stanton and White, Constructive Combinatorics, 1986, pp. 84, 91.

%H Cyann Donnot, Antoine Genitrini, Yassine Herida, <a href="https://hal.sorbonne-universite.fr/hal-02462764">Unranking Combinations Lexicographically: an efficient new strategy compared with others</a>, hal-02462764 [cs] / [cs.DS] / [math] / [math.CO], 2020.

%H Antoine Genitrini and Martin Pépin, <a href="https://hal.sorbonne-universite.fr/hal-03040740v2">Lexicographic unranking of combinations revisited</a>, hal-03040740v2 [cs.DM] [cs.DS] [math.CO], 2020.

%H Juan Luis Vargas-Molina, <a href="/A059797/a059797.cpp.txt">C++ Algorithm</a>

%F T(row, col) = T(row-1, col-1) + T(row-1, col) + A007318(row+2, col+1). - Corrected by _R. J. Mathar_, Jan 27 2011

%F T(n,k) = (k+1)(n+k+4)FallingFactorial(n+k+2,n)/((n+1)!+n!) n,k >= 0. - _Juan Luis Vargas-Molina_, Mar 03 2014

%e a(5) = 16 because we can write T(2,2) = T(1,2) + T(1,1) + A007318(4,2) = 5 + 5 + 6.

%e    2;

%e    5,   5;

%e    9,  16,   9;

%e   14,  35,  35,  14;

%e   20,  64,  90,  64,  20;

%e   27, 105, 189, 189, 105,  27;

%e T(n,k) as a grid for the first few lattice points. Where T(0,0)=2 since there are two ways to get to (0,0) with "0+0+2" steps under the move restrictions.

%e      k = 0  1   2   3    4    5

%e T(0,k) = 2  5   9   14   20   27

%e T(1,k) = 5  16  35  64   105  160

%e T(2,k) = 9  35  90  189  350  594

%e T(3,k) = 14 64  189 448  924  1728

%e T(4,k) = 20 105 350 924  2100 4290

%e T(5,k) = 27 160 594 1728 4290 9504

%e - _Juan Luis Vargas-Molina_, Mar 03 2014

%p A059797 := proc(n,k) option remember; if n <0 or k<0 or k>n then 0; else procname(n-1,k-1)+procname(n-1,k)+binomial(n+2,k+1) ; end if; end proc:

%p seq(seq( A059797(n,k),k=0..n),n=0..15) ; # _R. J. Mathar_, Jan 27 2011

%t t[n_, k_] /; (n < 0 || k < 0 || k > n) = 0; t[n_, k_] := t[n, k] = t[n - 1, k - 1] + t[n - 1, k] + Binomial[n + 2, k + 1]; Flatten[ Table[ t[n, k], {n, 0, 9}, {k, 0, n}]] (* _Jean-François Alcover_, Dec 20 2011, after _R. J. Mathar_ *)

%t T[n_, k_] := (k + 1)(n + k +4) FactorialPower[k + n + 2, n]/(n! + (n + 1)!)

%t MatrixForm[Table[T[n, k], {n, 0, 10}, {k, 0, 10}]] (* _Juan Luis Vargas-Molina_, Mar 03 2014 *)

%Y Cf. A000041, A000142, A007318, A008459, A053445.

%K nice,nonn,tabl

%O 0,1

%A _Alford Arnold_, Feb 22 2001