|
%I #60 Sep 08 2022 08:45:53
%S 1,1,3,1,4,10,1,5,15,35,1,6,21,56,126,1,7,28,84,210,462,1,8,36,120,
%T 330,792,1716,1,9,45,165,495,1287,3003,6435,1,10,55,220,715,2002,5005,
%U 11440,24310,1,11,66,286,1001,3003,8008,19448,43758,92378,1,12,78,364,1365,4368,12376,31824,75582,167960,352716,1,13,91,455,1820,6188,18564,50388,125970,293930,646646,1352078
%N Triangle T(n,k) = binomial(n+k-1,n) read by rows, 1 <= k <= n.
%C Obtained from A176992 by reversing entries in each row, from A092392 by removing the left column and reversing entries in each row, or from A100100 by removing the first two columns and reversing entries in each row.
%C Also T(n,k) = count of degree k monomials in the Monomial symmetric polynomials m(mu,k) summed over all partitions mu of n.
%C T(n,k) is the number of ways to put n indistinguishable balls into k distinguishable boxes. - _Dennis P. Walsh_, Apr 11 2012
%C T(n,k) is the number of compositions of n into k parts if zeros are allowed as parts. - _L. Edson Jeffery_, Jul 23 2014
%C T(n,k) is the number of compositions (ordered partitions) of n+k into exactly k parts. - _Juergen Will_, Jan 23 2016
%C T(n,k) is the number of binary strings with exactly n zeros and k-1 ones. - _Dennis P. Walsh_, Apr 09 2016
%C T(n,k) is the number of functions f:[k-1]->[n+1] that are nondecreasing. There is a unique correspondence between such a function and a binary string with exactly n zeros and k-1 ones. Given a string, let the corresponding function f be defined by f(i)=1 + (the number of zeros in the string that precede the i-th one in the string) for i=1,..,k-1. - _Dennis P. Walsh_, Apr 09 2016
%H P. A. MacMahon, <a href="http://www.jstor.org/stable/90632?seq=1#page_scan_tab_contents">Memoir on the Theory of the Compositions of Numbers</a>, Phil. Trans. Royal Soc. London A, 184 (1893), 835-901.
%H Ch. Stover and E. W. Weisstein, <a href="http://mathworld.wolfram.com/Composition.html">Composition</a>. From MathWorld - A Wolfram Web Resource.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Symmetric_polynomials">Symmetric Polynomials</a>
%F T(n,k) = A046899(n,k-1) = A038675(n,k)/A008292(n,k).
%F T(n,1) = 1.
%F T(n,2) = n+1.
%F T(n,3) = A000217(n+1).
%F T(n,4) = A000292(n+1).
%F T(n,5) = A000332(n+4).
%F T(n,n) = A001700(n-1) = A088218(n). - _Dennis P. Walsh_, Apr 10 2012
%e Triangle begins
%e 1;
%e 1, 3;
%e 1, 4, 10;
%e 1, 5, 15, 35;
%e 1, 6, 21, 56, 126;
%e 1, 7, 28, 84, 210, 462;
%e 1, 8, 36, 120, 330, 792, 1716;
%e T(3,3)=10 since there are 10 ways to put 3 identical balls into 3 distinguishable boxes, namely, (OOO)()(), ()(OOO)(), ()()(OOO), (OO)(O)(), (OO)()(O), (O)(OO)(), ()(OO)(O), (O)()(OO), ()(O)(OO), and (O)(O)(O). - _Dennis P. Walsh_, Apr 11 2012
%e For example, T(3,3)=10 since there are ten functions f:[2]->[4] that are nondecreasing, namely, <f(1),f(2)> = <1,1> or <1,2> or <1,3> or <1,4> or <2,2> or <2,3> or <2,4> or <3,3> or <3,4> or <4,4>. - _Dennis P. Walsh_, Apr 09 2016
%p seq(seq(binomial(n+k-1,n),k=1..n),n=1..15); # _Dennis P. Walsh_, Apr 11 2012
%t m[par_?PartitionQ, v_] := Block[{le = Length[par], it }, If[le > v, Return[0]]; it = Permutations[PadRight[par, v]]; Tr[ Apply[Times, Table[Subscript[x, j], {j, v}]^# & /@ it, {1}]]];
%t Table[Tr[(m[#, k] & /@ Partitions[l]) /. Subscript[x, _] -> 1], {l, 11}, {k, l}](* _Wouter Meeussen_, Mar 11 2012 *)
%t Quiet[Needs["Combinatorica`"], All]; Grid[Table[Length[Combinatorica`Compositions[n, k]], {n, 10}, {k, n}]] (* _L. Edson Jeffery_, Jul 24 2014 *)
%t t[n_, k_] := Binomial[n + k - 1, n]; Table[ t[n, k], {n, 10}, {k, n}] // Flatten (* _Robert G. Wilson v_, Jul 24 2014 *)
%o (Magma) (* As triangle *) [[Binomial(n+k-1,n): k in [1..n]]: n in [1.. 15]]; // _Vincenzo Librandi_, Jan 24 2016
%Y Cf. A000142, A000312, A007318, A001791 (row sums), A209664-A209673.
%Y Cf. A000217, A000292, A000332, A001700, A008292, A038675, A046899, A088218.
%Y Cf. A176992, A092392, A100100.
%K easy,nonn,tabl
%O 1,3
%A _Alford Arnold_, May 24 2010
|
