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A127717 Triangle read by rows. T(n, k) = k * binomial(n + 1, k + 1), for 1 <= k <= n.

%I A127717 #56 Jan 08 2023 05:27:01
%S A127717 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 A127717 224,140,48,7,36,168,378,504,420,216,63,8,45,240,630,1008,1050,720,
%U A127717 315,80,9,55,330,990,1848,2310,1980,1155,440,99,10
%N A127717 Triangle read by rows. T(n, k) = k * binomial(n + 1, k + 1), for 1 <= k <= n.
%C A127717 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 A127717 Reversed unsigned rows of A055137 with the diagonal and first subdiagonal removed. - _Tom Copeland_, Nov 04 2012
%C A127717 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 A127717 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 A127717 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 A127717 A002260 * A007318 (Pascal's Triangle), where A002260 = the matrix [1; 1,2; 1,2,3,...].
%F A127717 T(n,k) = Sum_{i=k..n} binomial(i-1, k-1)*i. - _Geoffrey Critzer_, Oct 17 2009
%F A127717 Row sums = A000337: (1, 5, 17, 49, 129, ...) A007318 * A002260 = A127718.
%F A127717 From _Geoffrey Critzer_, Oct 18 2009: (Start)
%F A127717 T(n,k) = k*binomial(n+1, k+1).
%F A127717 Recurrence for column k: a(n) = a(n-1) + n*binomial(n-1, k-1) = a(n-1) + k*binomial(n, k).
%F A127717 O.g.f. for column k: k*x^k/(1-x)^(k+2). (End)
%F A127717 T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n+2-k, k+2-i). - _Mircea Merca_, Apr 11 2012
%F A127717 G.f.: 1/((1 - x)*(1 - x - x*y)^2), assuming the triangle (0,0)-based. - _Vladimir Kruchinin_, Jan 07 2023
%e A127717 First few rows of the triangle:
%e A127717     [1    2    3     4     5    6    7     8    9]
%e A127717 [1]  1;
%e A127717 [2]  3,   2;
%e A127717 [3]  6,   8,   3;
%e A127717 [4] 10,  20,  15,    4;
%e A127717 [5] 15,  40,  45,   24,    5;
%e A127717 [6] 21,  70, 105,   84,   35,    6;
%e A127717 [7] 28, 112, 210,  224,  140,   48,    7;
%e A127717 [8] 36, 168, 378,  504,  420,  216,   63,   8;
%e A127717 [9] 45, 240, 630, 1008, 1050,  720,  315,  80,  9;
%e A127717   ...
%e A127717 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 A127717 # Assuming (1,1)-based triangle:
%p A127717 T := (n, k) -> k*binomial(n+1, k+1):
%p A127717 seq(seq(T(n, k), k = 1..n), n = 1..9);
%p A127717 # Assuming (0,0)-based triangle:
%p A127717 gf := 1/((1 - x)*(1 - x - x*y)^2): ser := series(gf, x, 11):
%p A127717 seq(seq(coeff(coeff(ser, x, n), y, k), k=0..n), n=0..9); # _Peter Luschny_, Jan 07 2023
%t A127717 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 A127717 (PARI) T(n,k) = k*sum(i=0,n-k,binomial(i+k,k))
%o A127717 for(n=1,15,for(k=1,n,print1(T(n,k),", "))) \\ _Derek Orr_, Oct 30 2014
%Y A127717 Cf. A002260, A007318, A000337, A127718.
%K A127717 nonn,tabl,easy
%O A127717 1,2
%A A127717 _Gary W. Adamson_, Jan 25 2007
%E A127717 a(8) = 20, corrected by _Geoffrey Critzer_, Oct 17 2009
%E A127717 More terms from _Derek Orr_, Oct 30 2014
%E A127717 Offset set to 1 and new name using a formula of _Geoffrey Critzer_ by _Peter Luschny_, Jan 07 2023