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

1, 3, 2, 6, 8, 3, 10, 20, 15, 4, 15, 40, 45, 24, 5, 21, 70, 105, 84, 35, 6, 28, 112, 210, 224, 140, 48, 7, 36, 168, 378, 504, 420, 216, 63, 8, 45, 240, 630, 1008, 1050, 720, 315, 80, 9, 55, 330, 990, 1848, 2310, 1980, 1155, 440, 99, 10

Offset: 1

Author

Gary W. Adamson, Jan 25 2007

Keywords

• nonn
• tabl
• easy

Comments

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

Reversed unsigned rows of A055137 with the diagonal and first subdiagonal removed. - Tom Copeland, Nov 04 2012

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

Examples

			First few rows of the triangle:
    [1    2    3     4     5    6    7     8    9]
[1]  1;
[2]  3,   2;
[3]  6,   8,   3;
[4] 10,  20,  15,    4;
[5] 15,  40,  45,   24,    5;
[6] 21,  70, 105,   84,   35,    6;
[7] 28, 112, 210,  224,  140,   48,    7;
[8] 36, 168, 378,  504,  420,  216,   63,   8;
[9] 45, 240, 630, 1008, 1050,  720,  315,  80,  9;
  ...
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
		

Links

• Cyann Donnot, Antoine Genitrini, Yassine Herida, Unranking Combinations Lexicographically: an efficient new strategy compared with others, hal-02462764 [cs] / [cs.DS] / [math] / [math.CO], 2020.
• Antoine Genitrini and Martin Pépin, Lexicographic unranking of combinations revisited, hal-03040740v2 [cs.DM] [cs.DS] [math.CO], 2020.

Crossrefs

Cf. A002260, A007318, A000337, A127718.

Programs

Maple

# Assuming (1,1)-based triangle:
T := (n, k) -> k*binomial(n+1, k+1):
seq(seq(T(n, k), k = 1..n), n = 1..9);
# Assuming (0,0)-based triangle:
gf := 1/((1 - x)*(1 - x - x*y)^2): ser := series(gf, x, 11):
seq(seq(coeff(coeff(ser, x, n), y, k), k=0..n), n=0..9); # Peter Luschny, Jan 07 2023

Mathematica

Table[Table[Sum[Binomial[i - 1, k - 1]*i, {i, k, n}], {k, 1, n}], {n, 1, 10}] // Grid (* Geoffrey Critzer, Oct 17 2009 *)

PARI

T(n,k) = k*sum(i=0,n-k,binomial(i+k,k))
for(n=1,15,for(k=1,n,print1(T(n,k),", "))) \\ Derek Orr, Oct 30 2014

Formula

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

T(n,k) = Sum_{i=k..n} binomial(i-1, k-1)*i. - Geoffrey Critzer, Oct 17 2009

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

From Geoffrey Critzer, Oct 18 2009: (Start)

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

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

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

T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n+2-k, k+2-i). - Mircea Merca, Apr 11 2012

G.f.: 1/((1 - x)*(1 - x - x*y)^2), assuming the triangle (0,0)-based. - Vladimir Kruchinin, Jan 07 2023

Extensions

a(8) = 20, corrected by Geoffrey Critzer, Oct 17 2009

More terms from Derek Orr, Oct 30 2014

Offset set to 1 and new name using a formula of Geoffrey Critzer by Peter Luschny, Jan 07 2023