A158498 -id:A158498 - cp's OEIS frontend

A059481 Triangle read by rows. T(n, k) = binomial(n+k-1, k) for 0 <= k <= n.

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1716, 1, 8, 36, 120, 330, 792, 1716, 3432, 6435, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 24310, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 92378

Offset: 0

Fabian Rothelius, Feb 04 2001

T(n,k) is the number of ways to distribute k identical objects in n distinct containers; containers may be left empty.
T(n,k) is the number of nondecreasing functions f from {1,...,k} to {1,...,n}. - Dennis P. Walsh, Apr 07 2011
Coefficients of Faber polynomials for function x^2/(x-1). - Michael Somos, Sep 09 2003
Consider k-fold Cartesian products CP(n,k) of the vector A(n)=[1,2,3,...,n].
An element of CP(n,k) is a n-tuple T_t of the form T_t=[i_1,i_2,i_3,...,i_k] with t=1,...,n^k.
We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.
For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j > i_(j+1), T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j > i_(j+1)).
The indicator function which tests on i_j = i_(j+1) generates A158497, which contains further examples of this type of counting.
Triangle of the numbers of combinations of k elements with repetitions from n elements {1,2,...,n} (when every element i, i=1,...,n, appears in a k-combination either 0, or 1, or 2, ..., or k times). - Vladimir Shevelev, Jun 19 2012
G.f. for Faber polynomials is -log(-t*x-(1-sqrt(1-4*t))/2+1)=sum(n>0, T(n,k)*t^k/n). - Vladimir Kruchinin, Jul 04 2013
Values of complete homogeneous symmetric polynomials with all arguments equal to 1, or, equivalently, the number of monomials of degree k in n variables. - Tom Copeland, Apr 07 2014
Row k >= 0 of the infinite square array A[k,n] = C(n,k), n >= 0, would start with k zeros in front of the first nonzero element C(k,k) = 1; this here is the triangle obtained by taking the first k+1 nonzero terms C(k .. 2k, k) of rows k = 0, 1, 2, ... of that array. - M. F. Hasler, Mar 05 2017

			The triangle T(n,k), n >= 0, 0 <= k <= n, begins
  1      A000217
  1 1   /     A000292
  1 2  3    /    A000332
  1 3  6  10    /    A000389
  1 4 10  20  35    /     A000579
  1 5 15  35  70 126     /
  1 6 21  56 126 252  462
  1 7 28  84 210 462  924 1716
  1 8 36 120 330 792 1716 3432 6435
.
T(3,2)=6 considers the CP with the 3^2=9 elements (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3), and does not count the 3 of them which are (2,1),(3,1) and (3,2).
T(3,3) = 10 because the ways to distribute the 3 objects into the three containers are: (3,0,0) (0,3,0) (0,0,3) (2,1,0) (1,2,0) (2,0,1) (1,0,2) (0,1,2) (0,2,1) (1,1,1), for a total of 10 possibilities.
T(3,3)=10 since (x^2/(x-1))^3 = (x+1+1/x+O(1/x^2))^3 = x^3+3x^2+6x+10+O(x).
T(4,2)=10 since there are 10 nondecreasing functions f from {1,2} to {1,2,3,4}. Using <f(1),f(2)> to denote such a function, the ten functions are <1,1>, <1,2>, <1,3>, <1,4>, <2,2>, <2,3>, <2,4>, <3,3>, <3,4>, and <4,4>. - _Dennis P. Walsh_, Apr 07 2011
T(4,0) + T(4,1) + T(4,2) + T(4,3) = 1 + 4 + 10 + 20 = 35 = T(4,4). - _Jonathan Sondow_, Jun 28 2014
From _Paul Curtz_, Jun 18 2018: (Start)
Consider the array
2,    1,    1,    1,    1,    1,     ... = A054977(n)
1,    1/2,  1/3,  1/4,  1/5,  1/6,   ... = 1/(n+1) = 1/A000027(n)
1/3,  1/6,  1/10, 1/15, 1/21, 1/28,  ... = 2/((n+2)*(n+3)) = 1/A000217(n+2)
1/10, 1/20, 1/35, 1/56, 1/84, 1/120, ... = 6/((n+3)*(n+4)*(n+5)) =1/A000292(n+2) (see the triangle T(n,k)).
Every row is an autosequence of the second kind. (See OEIS Wiki, Autosequence.)
By decreasing antidiagonals the denominator of the array is a(n).
Successive vertical denominators: A088218(n), A000984(n), A001700(n), A001791(n+1), A002054(n), A002694(n).
Successive diagonal denominators: A165817(n), A005809(n), A045721(n), A025174(n+1), A004319(n). (End)
Without the first row (2, 1, 1, 1, ... ), the array leads to A165257(n) instead of a(n). - _Paul Curtz_, Jun 19 2018

R. Grimaldi, Discrete and Combinatorial Mathematics, Addison-Wesley, 4th edition, chapter 1.4.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
J. Abate and W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011) # 11.2.6, (referring to A158498 before recycling)
M. A. A. Obaid, S. K. Nauman, W. M. Fakieh, and C. M. Ringel, The numbers of support-tilting modules for a Dynkin algebra, 2014 and J. Int. Seq. 18 (2015) 15.10.6.
C. M. Ringel, The Catalan combinatorics of the hereditary artin algebras, arXiv:1502.06553 [math.RT], 2015.
Dennis Walsh, Notes on finite monotonic and non-monotonic functions

Columns: T(n,1) = A000027(n), n >= 1. T(n,2) = A000217(n) = A161680(n+1), n >= 2. T(n,3) = A000292(n), n >= 3. T(n,4) = A000332(n+3), n >= 4. T(n,5) = A000389(n+4), n >= 5. T(n,6) = A000579(n+5), n >=6. T(n,k) = A001405(n+k-1) for k <= n <= k+2. [Corrected and extended by M. F. Hasler, Mar 05 2017]
Rows: T(5,k) = A000332(k+4). T(6,k) = A000389(k+5). T(7,k) = A000579(k+6).
Diagonals: T(n,n) = A001700(n-1). T(n,n-1) = A000984(n-1).
T(n,k) = A046899(n-1,k). - R. J. Mathar, Mar 26 2009
Take Pascal's triangle A007318, delete entries to the right of a vertical line just right of center, then scan the diagonals.
For a signed version of this triangle see A027555.
Row sums give A000984.
Cf. A007318, A158497, A100100 (mirrored), A009766.
A000984, A001700, A001791, A002054, A002694, A004319, A005809, A025174, A045721, A088218, A165817, A054977, A165257, A000027, A000217, A006134 (trace of the symmetric Pascal matrix).

Flat(List([0..10], n->List([0..n], k->Binomial(n+k-1, k)))); # Stefano Spezia, Oct 30 2018

a059481 n k = a059481_tabl !! n !! n
a059481_row n = a059481_tabl !! n
a059481_tabl = map reverse a100100_tabl
-- Reinhard Zumkeller, Jan 15 2014

&cat [[&*[ Binomial(n+k-1,k)]: k in [0..n]]: n in [0..30] ]; // Vincenzo Librandi, Apr 08 2011

for n from 0 to 10 do for k from 0 to n do print(binomial(n+k-1,k)) ; od: od: # R. J. Mathar, Mar 31 2009

t[n_, k_] := Binomial[n+k-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 09 2013 *)
(* The combinatorial objects defined in the first comment can, for n >= 1, be generated by: *) r[n_, k_] := FrobeniusSolve[ConstantArray[1,n],k]; (* Peter Luschny, Jan 24 2019 *)

sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$ display_triangle(n) := for i from 0 thru n do disp(sjoin(makelist(binomial(i+j-1, j), j, 0, i), " ")); display_triangle(10); /* triangle output */ /* Stefano Spezia, Oct 30 2018 */

{T(n, k) = binomial( n+k-1, k)}; \\ Michael Somos, Sep 09 2003, edited by M. F. Hasler, Mar 05 2017

{T(n, k) = if( n<0, 0, polcoeff( Pol(((1 / (x - x^2) + x * O(x^n))^n + O(x)) * x^n), k))}; /* Michael Somos, Sep 09 2003 */

[[binomial(n+k-1,k) for k in range(n+1)] for n in range(11)] # G. C. Greubel, Nov 21 2018

T(n,0) + T(n,1) + . . . + T(n,n-1) = T(n,n). - Jonathan Sondow, Jun 28 2014
From Peter Bala, Jul 21 2015: (Start)
T(n, k) = Sum_{j = k..n} (-1)^(k+j)*binomial(2*n,n+j)*binomial(n+j-1,j)* binomial(j,k) (gives the correct value T(n,k) = 0 for k > n).
O.g.f.: 1/2*( x*(2*x - 1)/(sqrt(1 - 4*t*x)*(1 - x - t)) + (1 + 2*x)/sqrt(1 - 4*t*x) + (1 - t)/(1 - x - t) ) = 1 + (1 + t)*x + (1 + 2*t + 3*t^2)*x^2 + (1 + 3*t + 6*t^2 + 10*t^3)*x^3 + ....
n-th row polynomial R(n,t) = [x^n] ( (1 + x)^2/(1 + x(1 - t)) )^n.
exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + t)*x + (1 + 2*t + 2*t^2)*x^2 + (1 + 3*t + 5*t^2 + 5*t^3)*x^3 + ... is the o.g.f for A009766. (End)
a(n) = abs(A027555(n)). - M. F. Hasler, Mar 05 2017
For n >= k > 0, T(n, k) = Sum_{j=1..n} binomial(k + j - 2, k - 1) = Sum_{j=1..n} A007318(k + j - 2, k - 1). - Stefano Spezia, Oct 30 2018
T(n, k) = RisingFactorial(n, k) / k!. - Peter Luschny, Nov 24 2023

Offset changed from 1 to 0 by R. J. Mathar, Jan 15 2013

Edited by M. F. Hasler, Mar 05 2017

A158497 Triangle T(n,k) formed by the coordination sequences and the number of leaves for trees.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 6, 12, 1, 4, 12, 36, 108, 1, 5, 20, 80, 320, 1280, 1, 6, 30, 150, 750, 3750, 18750, 1, 7, 42, 252, 1512, 9072, 54432, 326592, 1, 8, 56, 392, 2744, 19208, 134456, 941192, 6588344, 1, 9, 72, 576, 4608, 36864, 294912, 2359296, 18874368, 150994944, 1, 10, 90, 810, 7290, 65610, 590490, 5314410, 47829690, 430467210, 3874204890

Offset: 0

Thomas Wieder, Mar 20 2009

Consider the k-fold Cartesian products CP(n,k) of the vector A(n) = [1, 2, 3, ..., n].
An element of CP(n,k) is a n-tuple T_t of the form T_t = [i_1, i_2, i_3, ..., i_k] with t=1, .., n^k.
We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.
For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j = i_(j+1): T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j = i_(j+1)).
The test on i_j > i_(j+1) generates A158498. One gets the Pascal triangle A007318 if the indicator function tests whether for any two i_j, i_(j+1) in T_t one has i_j >= i_(j+1).
Use of other indicator functions can also calculate the Bell numbers A000110, A000045 or A000108.

			Array, A(n, k) = n*(n-1)^(k-1) for n > 1, A(n, k) = 1 otherwise, begins as:
  1,  1,   1,    1,     1,      1,       1,        1,        1, ... A000012;
  1,  1,   1,    1,     1,      1,       1,        1,        1, ... A000012;
  1,  2,   2,    2,     2,      2,       2,        2,        2, ... A040000;
  1,  3,   6,   12,    24,     48,      96,      192,      384, ... A003945;
  1,  4,  12,   36,   108,    324,     972,     2916,     8748, ... A003946;
  1,  5,  20,   80,   320,   1280,    5120,    20480,    81920, ... A003947;
  1,  6,  30,  150,   750,   3750,   18750,    93750,   468750, ... A003948;
  1,  7,  42,  252,  1512,   9072,   54432,   326592,  1959552, ... A003949;
  1,  8,  56,  392,  2744,  19208,  134456,   941192,  6588344, ... A003950;
  1,  9,  72,  576,  4608,  36864,  294912,  2359296, 18874368, ... A003951;
  1, 10,  90,  810,  7290,  65610,  590490,  5314410, 47829690, ... A003952;
  1, 11, 110, 1100, 11000, 110000, 1100000, 11000000, ............. A003953;
  1, 12, 132, 1452, 15972, 175692, 1932612, 21258732, ............. A003954;
  1, 13, 156, 1872, 22464, 269568, 3234816, 38817792, ............. A170732;
  ... ;
The triangle begins as:
  1
  1, 1;
  1, 2,  2;
  1, 3,  6,  12;
  1, 4, 12,  36,  108;
  1, 5, 20,  80,  320,  1280;
  1, 6, 30, 150,  750,  3750,  18750;
  1, 7, 42, 252, 1512,  9072,  54432, 326592;
  1, 8, 56, 392, 2744, 19208, 134456, 941192, 6588344;
  ...;
T(3,3) = 12 counts the triples (1,2,1), (1,2,3), (1,3,1), (1,3,2), (2,1,2), (2,1,3), (2,3,1), (2,3,2), (3,1,2), (3,1,3), (3,2,1), (3,2,3) out of a total of 3^3 = 27 triples in the CP(3,3).

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Array rows n: A170733 (n=14), ..., A170769 (n=50).
Columns k: A000012(n) (k=0), A000027(n) (k=1), A002378(n-1) (k=2), A011379(n-1) (k=3), A179824(n) (k=4), A101362(n-1) (k=5), 2*A168351(n-1) (k=6), 2*A168526(n-1) (k=7), 2*A168635(n-1) (k=8), 2*A168675(n-1) (k=9), 2*A170783(n-1) (k=10), 2*A170793(n-1) (k=11).
Diagonals k: A055897 (k=n), A055541 (k=n-1), A373395 (k=n-2), A379612 (k=n-3).
Sums: (-1)^n*A065440(n) (signed row).
Cf. A000045, A000108, A000110, A007318, A079901, A158498.

A158497:= func< n,k | k le 1 select n^k else n*(n-1)^(k-1) >;
[A158497(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 18 2025

A158497[n_, k_]:= If[n<2 || k==0, 1, n*(n-1)^(k-1)];
Table[A158497[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 18 2025 *)

def A158497(n,k): return n^k if k<2 else n*(n-1)^(k-1)
print(flatten([[A158497(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Mar 18 2025

T(n, k) = (n-1)^(k-1) + (n-1)^k = n*A079901(n-1,k-1), k > 0.

Sum_{k=0..n} T(n,k) = (n*(n-1)^n - 2)/(n-2), n > 2.

Edited by R. J. Mathar, Mar 31 2009

More terms added by G. C. Greubel, Mar 18 2025

A174952 Triangle t(n,m)= binomial(n+m-1,n-1) + binomial(2n-m-1,n-1) -binomial(2n-1,n-1) read by rows, 0<=m<=n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, -1, -1, 1, 1, -11, -15, -11, 1, 1, -51, -76, -76, -51, 1, 1, -204, -315, -350, -315, -204, 1, 1, -785, -1226, -1422, -1422, -1226, -785, 1, 1, -2995, -4683, -5523, -5775, -5523, -4683, -2995, 1, 1, -11431, -17830, -21142, -22528, -22528

Offset: 0

Roger L. Bagula, Apr 02 2010

Row sums are 1, 2, 3, 0, -35, -252, -1386, -6864, -32175, -145860, -646646,....

			1;
1, 1;
1, 1, 1;
1, -1, -1, 1;
1, -11, -15, -11, 1;
1, -51, -76, -76, -51, 1;
1 -204, -315, -350, -315, -204, 1;
1, -785, -1226, -1422, -1422, -1226, -785, 1;
1, -2995, -4683, -5523, -5775, -5523, -4683, -2995, 1;
1, -11431, -17830, -21142, -22528, -22528, -21142, -17830, -11431, 1;
1, -43748, -68013, -80718, -86658, -88374, -86658, -80718, -68013, -43748, 1;

Cf. A158498, A059481

A174952 := proc(n,m)
        binomial(n+m-1,n-1)+binomial(2*n-m-1,n-1) -binomial(2*n-1,n-1);
end proc: # R. J. Mathar, Jan 15 2013

t(n,m) = t(n,n-m).

Definition corrected and deobfuscated. R. J. Mathar, Jan 15 2013

A176566 Triangle T(n, k) = binomial(n*(n+1)/2 + k, k), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 4, 10, 20, 1, 7, 28, 84, 210, 1, 11, 66, 286, 1001, 3003, 1, 16, 136, 816, 3876, 15504, 54264, 1, 22, 253, 2024, 12650, 65780, 296010, 1184040, 1, 29, 435, 4495, 35960, 237336, 1344904, 6724520, 30260340, 1, 37, 703, 9139, 91390, 749398, 5245786, 32224114, 177232627, 886163135

Offset: 0

Roger L. Bagula, Apr 20 2010

			Square array of T(n, k):
  1,  1,   1,    1,     1,     1,      1 ...
  1,  1,   1,    1,     1,     1,      1 ... A000012;
  1,  2,   3,    4,     5,     6,      7 ... A000027;
  1,  4,  10,   20,    35,    56,     84 ... A000292;
  1,  7,  28,   84,   210,   462,    924 ... A000579;
  1, 11,  66,  286,  1001,  3003,   8008 ... A001287;
  1, 16, 136,  816,  3876, 15504,  54264 ... A010968;
  1, 22, 253, 2024, 12650, 65780, 296010 ... A010974;
Triangle begins as:
  1;
  1,  1;
  1,  2,   3;
  1,  4,  10,   20;
  1,  7,  28,   84,   210;
  1, 11,  66,  286,  1001,   3003;
  1, 16, 136,  816,  3876,  15504,   54264;
  1, 22, 253, 2024, 12650,  65780,  296010,  1184040;
  1, 29, 435, 4495, 35960, 237336, 1344904,  6724520,  30260340;
  1, 37, 703, 9139, 91390, 749398, 5245786, 32224114, 177232627, 886163135;

G. C. Greubel, Rows n = 0..50 of the triangle, flatten

Cf. A107868 (rows sums), A158498.

[Binomial(Binomial(n, 2) + k, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 09 2021

T[n_, k_]= Binomial[Binomial[n, 2] + k, k];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten

row(n) = vector(n+1, k, k--; binomial(binomial(n,2) + k, k)); \\ Michel Marcus, Jul 10 2021

flatten([[binomial(binomial(n,2) +k, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jul 09 2021

T(n, k) = binomial(binomial(n, 2) + k, k).

Sum_{k=0..n} T(n, k) = A107868(n).

cp's OEIS Frontend

A059481 Triangle read by rows. T(n, k) = binomial(n+k-1, k) for 0 <= k <= n.

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A158497 Triangle T(n,k) formed by the coordination sequences and the number of leaves for trees.

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A174952 Triangle t(n,m)= binomial(n+m-1,n-1) + binomial(2*n-m-1,n-1) -binomial(2*n-1,n-1) read by rows, 0<=m<=n.

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A176566 Triangle T(n, k) = binomial(n*(n+1)/2 + k, k), read by rows.

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A174952 Triangle t(n,m)= binomial(n+m-1,n-1) + binomial(2n-m-1,n-1) -binomial(2n-1,n-1) read by rows, 0<=m<=n.