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Triangle read by rows. T(n, k) = k * binomial(n + 1, k + 1), for 1 <= k <= n.
4

%I #56 Jan 08 2023 05:27:01

%S 1,3,2,6,8,3,10,20,15,4,15,40,45,24,5,21,70,105,84,35,6,28,112,210,

%T 224,140,48,7,36,168,378,504,420,216,63,8,45,240,630,1008,1050,720,

%U 315,80,9,55,330,990,1848,2310,1980,1155,440,99,10

%N Triangle read by rows. T(n, k) = k * binomial(n + 1, k + 1), for 1 <= k <= n.

%C T(n,k) is the sum of the greatest element in each size k subset of {1,2,...,n}. - _Geoffrey Critzer_, Oct 17 2009

%C Reversed unsigned rows of A055137 with the diagonal and first subdiagonal removed. - _Tom Copeland_, Nov 04 2012

%C Consider the transformation 1 + 2x + 3x^2 + 4x^3 + ... + (n+1)*x^n = A_0*(x-1)^0 + A_1*(x-1)^1 + A_2*(x-1)^2 + ... + A_n*(x-1)^n. This sequence gives A_0, ..., A_n as the entries in the n-th row of this triangle, starting at n = 0. - _Derek Orr_, Oct 30 2014

%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.

%F A002260 * A007318 (Pascal's Triangle), where A002260 = the matrix [1; 1,2; 1,2,3,...].

%F T(n,k) = Sum_{i=k..n} binomial(i-1, k-1)*i. - _Geoffrey Critzer_, Oct 17 2009

%F Row sums = A000337: (1, 5, 17, 49, 129, ...) A007318 * A002260 = A127718.

%F From _Geoffrey Critzer_, Oct 18 2009: (Start)

%F T(n,k) = k*binomial(n+1, k+1).

%F Recurrence for column k: a(n) = a(n-1) + n*binomial(n-1, k-1) = a(n-1) + k*binomial(n, k).

%F O.g.f. for column k: k*x^k/(1-x)^(k+2). (End)

%F T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n+2-k, k+2-i). - _Mircea Merca_, Apr 11 2012

%F G.f.: 1/((1 - x)*(1 - x - x*y)^2), assuming the triangle (0,0)-based. - _Vladimir Kruchinin_, Jan 07 2023

%e First few rows of the triangle:

%e [1 2 3 4 5 6 7 8 9]

%e [1] 1;

%e [2] 3, 2;

%e [3] 6, 8, 3;

%e [4] 10, 20, 15, 4;

%e [5] 15, 40, 45, 24, 5;

%e [6] 21, 70, 105, 84, 35, 6;

%e [7] 28, 112, 210, 224, 140, 48, 7;

%e [8] 36, 168, 378, 504, 420, 216, 63, 8;

%e [9] 45, 240, 630, 1008, 1050, 720, 315, 80, 9;

%e ...

%e T(4, 3) = 15 because the size 3 subsets of {1, 2, 3, 4} are {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}. Adding the largest element from each subset we get 3 + 4 + 4 + 4 = 15. - _Geoffrey Critzer_, Oct 17 2009

%p # Assuming (1,1)-based triangle:

%p T := (n, k) -> k*binomial(n+1, k+1):

%p seq(seq(T(n, k), k = 1..n), n = 1..9);

%p # Assuming (0,0)-based triangle:

%p gf := 1/((1 - x)*(1 - x - x*y)^2): ser := series(gf, x, 11):

%p seq(seq(coeff(coeff(ser, x, n), y, k), k=0..n), n=0..9); # _Peter Luschny_, Jan 07 2023

%t Table[Table[Sum[Binomial[i - 1, k - 1]*i, {i, k, n}], {k, 1, n}], {n, 1, 10}] // Grid (* _Geoffrey Critzer_, Oct 17 2009 *)

%o (PARI) T(n,k) = k*sum(i=0,n-k,binomial(i+k,k))

%o for(n=1,15,for(k=1,n,print1(T(n,k),", "))) \\ _Derek Orr_, Oct 30 2014

%Y Cf. A002260, A007318, A000337, A127718.

%K nonn,tabl,easy

%O 1,2

%A _Gary W. Adamson_, Jan 25 2007

%E a(8) = 20, corrected by _Geoffrey Critzer_, Oct 17 2009

%E More terms from _Derek Orr_, Oct 30 2014

%E Offset set to 1 and new name using a formula of _Geoffrey Critzer_ by _Peter Luschny_, Jan 07 2023