A178300 -id:A178300 - cp's OEIS frontend

A046899 Triangle in which n-th row is {binomial(n+k,k), k=0..n}, n >= 0.

1, 1, 2, 1, 3, 6, 1, 4, 10, 20, 1, 5, 15, 35, 70, 1, 6, 21, 56, 126, 252, 1, 7, 28, 84, 210, 462, 924, 1, 8, 36, 120, 330, 792, 1716, 3432, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 1, 11, 66, 286, 1001

Offset: 0

N. J. A. Sloane

C(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Row sums are A001700.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) of waist k (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
If T(r,c), r=0,1,2,..., c=1,2,...,(r+1), are the triangle elements, then for r > 0, T(r,c) = binomial(r+c-1,c-1) = M(r,c) is the number of monotonic mappings from an ordered set of r elements into an ordered set of c elements. For example, there are 15 monotonic mappings from an ordered set of 4 elements into an ordered set of 3 elements. For c > r+1, use the identity M(r,c) = M(c-1,r+1) = T(c-1,r+1). For example, there are 210 monotonic mappings from an ordered set of 4 elements into an ordered set of 7 elements, because M(4,7) = T(6,5) = 210. Number of monotonic endomorphisms in a set of r elements, M(r,r), therefore appear on the second diagonal of the triangle which coincides with A001700. - Stanislav Sykora, May 26 2012
Start at the origin. Flip a fair coin to determine steps of (1,0) or (0,1). Stop when you are a (perpendicular) distance of n steps from the x axis or the y axis. For k = 0,1,...,n-1, C(n-1,k)/2^(n+k) is the probability that you will stop on the point (n,k). This is equal to the probability that you will stop on the point (k,n). Hence, Sum_{k=0..n} C(n,k)/2^(n+k) = 1. - Geoffrey Critzer, May 13 2017

			The triangle is the lower triangular part of the square array:
  1|  1,  1,   1,   1,    1,    1,     1,     1,     1, ...
  1,  2|  3,   4,   5,    6,    7,     8,     9,    10, ...
  1,  3,  6|  10,  15,   21,   28,    36,    45,    55, ...
  1,  4, 10,  20|  35,   56,   84,   120,   165,   220, ...
  1,  5, 15,  35,  70|  126,  210,   330,   495,   715, ...
  1,  6, 21,  56, 126,  252|  462,   792,  1287,  2002, ...
  1,  7, 28,  84, 210,  462,  924|  1716,  3003,  5005, ...
  1,  8, 36, 120, 330,  792, 1716,  3432|  6435, 11440, ...
  1,  9, 45, 165, 495, 1287, 3003,  6435, 12870| 24310, ...
  1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620| ...
The array read by antidiagonals gives the binomial triangle.
From _Reinhard Zumkeller_, Jul 27 2012: (Start)
Take the first n elements of the n-th diagonal (NW to SE) of left half of Pascal's triangle and write it as n-th row on the triangle on the right side, see above
  0:                 1                    1
  1:               1   _                  1  2
  2:             1   2  __                1  3  6
  3:           1   3  __  __              1  4 10 20
  4:         1   4   6  __  __            1  5 15 35 70
  5:       1   5  10  __  __  __          1  6 21 56 .. ..
  6:     1   6  15  20  __  __  __        1  7 28 .. .. .. ..
  7:   1   7  21  35  __  __  __  __      1  8 .. .. .. .. .. ..
  8: 1   8  28  56  70  __  __  __  __    1 .. .. .. .. .. .. .. .. (End)

H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83.

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Karl Dilcher and Maciej Ulas, Arithmetic properties of polynomial solutions of the Diophantine equation P(x)x^(n+1)+Q(x)(x+1)^(n+1)=1, arXiv:1909.11222 [math.NT], 2019. See Qn(x) Table 1 p. 2.
H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83. (Annotated scanned copy)
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving full transformations, Semigroup Forum 72 (2006), 51-62.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A000108, A046900, A001700, A007318, A034868, A092392, A239103, A178300.

import Data.List (transpose)
a046899 n k = a046899_tabl !! n !! k
a046899_row n = a046899_tabl !! n
a046899_tabl = zipWith take [1..] $ transpose a007318_tabl
-- Reinhard Zumkeller, Jul 27 2012

/* As triangle */ [[Binomial(n+k, n): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 18 2015

for n from 0 to 10 do seq( binomial(n+m,n), m = 0 .. n) od; # Zerinvary Lajos, Dec 09 2007

t[n_, k_] := Binomial[n + k, n]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 12 2013 *)

/* same as in A092566 but use */
steps=[[1, 0], [1, 0] ];
/* Joerg Arndt, Jul 01 2011 */

for n in (0..9):
    print([multinomial(n, k) for k in (0..n)]) # Peter Luschny, Dec 24 2020

T(n,k) = A092392(n,n-k), k = 0..n. - Reinhard Zumkeller, Jul 27 2012
T(n,k) = A178300(n,k), n>0, k = 1..n. - L. Edson Jeffery, Jul 23 2014
T(n,k) = (n + 1)*hypergeom([-n, 1 - k], [2], 1). - Peter Luschny, Jan 09 2022
T(n,k) = hypergeom([-n, -k], [1], 1). - Peter Luschny, Mar 21 2024
G.f.: 1/((1-2x*y*C(x*y))*(1-x*C(x*y))), where C(x) is the g.f. for A000108, the Catalan numbers. - Michael D. Weiner, Jul 31 2024

More terms from James Sellers

A227075 A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.

Original entry on oeis.org

1, 3, 3, 9, 6, 9, 27, 15, 15, 27, 81, 42, 30, 42, 81, 243, 123, 72, 72, 123, 243, 729, 366, 195, 144, 195, 366, 729, 2187, 1095, 561, 339, 339, 561, 1095, 2187, 6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561, 19683, 9843, 4938, 2556, 1578, 1578, 2556, 4938

Offset: 0

T. D. Noe, Aug 01 2013

All rows except the zeroth are divisible by 3. Is there a closed-form formula for these numbers, like for binomial coefficients?

Let b=3 and T(n,k) = A(n-k,k) be the associated reading of the symmetric array A by antidiagonals, then A(n,k) = sum_{r=1..n} b^r*A178300(n-r,k) + sum_{c=1..k} b^c*A178300(k-c,n). Similarly with b=4 and b=5 for A227074 and A227076. - R. J. Mathar, Aug 10 2013

			Triangle:
1,
3, 3,
9, 6, 9,
27, 15, 15, 27,
81, 42, 30, 42, 81,
243, 123, 72, 72, 123, 243,
729, 366, 195, 144, 195, 366, 729,
2187, 1095, 561, 339, 339, 561, 1095, 2187,
6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A166060 (row sums: 4*3^n - 3*2^n), A227074 (4^n edges), A227076 (5^n edges).

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 3^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

A178301 Triangle T(n,k) = binomial(n,k)*binomial(n+k+1,n+1) read by rows, 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 8, 10, 1, 15, 45, 35, 1, 24, 126, 224, 126, 1, 35, 280, 840, 1050, 462, 1, 48, 540, 2400, 4950, 4752, 1716, 1, 63, 945, 5775, 17325, 27027, 21021, 6435, 1, 80, 1540, 12320, 50050, 112112, 140140, 91520, 24310, 1, 99, 2376, 24024, 126126, 378378, 672672, 700128, 393822, 92378

Offset: 0

Alford Arnold, May 30 2010

Antidiagonal sums are given by A113682. - Johannes W. Meijer, Mar 24 2013
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial binomial(x+n,n)*binomial(x+n,n-1) in the basis made of the binomial(x+i,i). - F. Chapoton, Nov 01 2022
Chapoton's observation above is correct: the precise expansion is  binomial(x+n,n)*binomial(x+n,n-1) = Sum_{k = 0..n-1} (-1)^k*T(n-1,n-1-k)*binomial(x+2*n-1-k,2*n-1-k), as can be verified using the WZ algorithm. For example, n = 4 gives binomial(x+4,4)*binomial(x+4,3)  = 35*binomial(x+7,7) - 45*binomial(x+6,6) + 15*binomial(x+5,5) - binomial(x+4,4). - Peter Bala, Jun 24 2023

			n=0: 1;
n=1: 1,  3;
n=2: 1,  8,  10;
n=3: 1, 15,  45,   35;
n=4: 1, 24, 126,  224,   126;
n=5: 1, 35, 280,  840,  1050,   462;
n=6: 1, 48, 540, 2400,  4950,  4752,  1716;
n=7: 1, 63, 945, 5775, 17325, 27027, 21021, 6435;

David A. Corneth, Table of n, a(n) for n = 0..10010
Author?, Norm of a continuous function, dxdy.ru (in Russian).

Cf. A007318, A047781 (row sums), A178300, A182626, A123160, A132813.

A178301 := proc(n,k)
        binomial(n,k)*binomial(n+k+1,n+1) ;
end proc: # R. J. Mathar, Mar 24 2013
R := proc(n) add((-1)^(n+k)*(2*k+1)*orthopoly:-P(k,2*x+1)/(n+1), k=0..n) end:
for n from 0 to 6 do seq(coeff(R(n), x, k), k=0..n) od; # Peter Luschny, Aug 25 2021

Flatten[Table[Binomial[n,k]Binomial[n+k+1,n+1],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Aug 23 2014 *)

create_list(binomial(n,k)*binomial(n+k+1,n+1),n,0,12,k,0,n); /* Emanuele Munarini, Dec 16 2016 */

R(n,x) = sum(k=0,n, (-1)^(n+k) * (2*k+1) * pollegendre(k,2*x+1)) / (n+1); \\ Max Alekseyev, Aug 25 2021

T(n,k) = A007318(n,k) * A178300(n+1,k+1).
From Peter Bala, Jun 18 2015: (Start)
n-th row polynomial R(n,x) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k+1,n+1)*x^k = Sum_{k = 0..n} (-1)^(n+k)*binomial(n+1,k+1)*binomial(n+k+1,n+1)*(1 + x)^k.
Recurrence: (2*n - 1)*(n + 1)*R(n,x) = 2*(4*n^2*x + 2*n^2 - x - 1)*R(n-1,x) - (2*n + 1)(n - 1)*R(n-2,x) with R(0,x) = 1, R(1,x) = 1 + 3*x.
A182626(n) = -R(n-1,-2) for n >= 1. (End)
From Peter Bala, Jul 20 2015: (Start)
n-th row polynomial R(n,x) = Jacobi_P(n,0,1,2*x + 1).
(1 + x)*R(n,x) gives the row polynomials of A123160. (End)
G.f.: (1+x-sqrt(1-2*x+x^2-4*x*y))/(2*(1+y)*x*sqrt(1-2*x+x^2-4*x*y)). - Emanuele Munarini, Dec 16 2016
R(n,x) = Sum_{k=0..n} (-1)^(n+k)*(2*k+1)*P(k,2*x+1)/(n+1), where P(k,x) is the k-th Legendre polynomial (cf. A100258) and P(k,2*x+1) is the k-th shifted Legendre polynomial (cf. A063007). - Max Alekseyev, Jun 28 2018; corrected by Peter Bala, Aug 08 2021
Polynomial g(n,x) = R(n,-x)/(n+1) delivers the maximum of f(1)^2/(Integral_{x=0..1} f(x)^2 dx) over all polynomials f(x) with real coefficients and deg(f(x)) <= n. This maximum equals (n+1)^2. See dxdy.ru link. - Max Alekseyev, Jun 28 2018

A176992 Triangle T(n,m) = binomial(2n-k+1, n+1) read by rows, 0 <= k <= n.

Original entry on oeis.org

1, 3, 1, 10, 4, 1, 35, 15, 5, 1, 126, 56, 21, 6, 1, 462, 210, 84, 28, 7, 1, 1716, 792, 330, 120, 36, 8, 1, 6435, 3003, 1287, 495, 165, 45, 9, 1, 24310, 11440, 5005, 2002, 715, 220, 55, 10, 1, 92378, 43758, 19448, 8008, 3003, 1001, 286, 66, 11, 1, 352716, 167960, 75582, 31824, 12376, 4368, 1365, 364, 78, 12, 1

Offset: 0

Roger L. Bagula, Dec 08 2010

Row sums are A001791.
Obtained from A059481 by removal of the last two terms in each row, followed by row reversal.
Riordan array (c(x)/sqrt(1 - 4*x), x*c(x)) where c(x) is the g.f. of A000108. - Philippe Deléham, Jul 12 2015

			Triangle begins:
       1;
       3,      1;
      10,      4,     1;
      35,     15,     5,     1;
     126,     56,    21,     6,    1;
     462,    210,    84,    28,    7,     1;
    1716,    792,   330,   120,   36,     8,    1;
    6435,   3003,  1287,   495,   165,   45,    9,   1;
   24310,  11440,  5005,  2002,   715,  220,   55,  10,  1;
   92378,  43758, 19448,  8008,  3003, 1001,  286,  66, 11,  1;
  352716, 167960, 75582, 31824, 12376, 4368, 1365, 364, 78, 12, 1;

Cf. A092392, A001791, A078812.
Cf. Similar triangle: A033184, A054445.
Cf. A178300 (reversal).

/* As triangle */ [[Binomial(2*n-k+1,n+1): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 12 2015

A176992 := proc(n,k) binomial(1+2*n-k,n+1) ; end proc: # R. J. Mathar, Dec 09 2010

p[t_, j_] = ((-1)^(j + 1)/2)*Sum[Binomial[k - j - 1, j + 1]*t^k, {k, 0, Infinity}];
Flatten[Table[CoefficientList[ExpandAll[p[t, j]], t], {j, 0, 10}]]

n-th row of the triangle = top row of M^n, where M is the following infinite square production matrix:
  3, 1, 0, 0, 0, ...
  1, 1, 1, 0, 0, ...
  1, 1, 1, 1, 0, ...
  1, 1, 1, 1, 1, ...
  ... - Philippe Deléham, Jul 12 2015

A130746 Triangle read by rows: T(n,m) = binomial(n+m,1+n), 1<=m<=n.

Original entry on oeis.org

1, 1, 4, 1, 5, 15, 1, 6, 21, 56, 1, 7, 28, 84, 210, 1, 8, 36, 120, 330, 792, 1, 9, 45, 165, 495, 1287, 3003, 1, 10, 55, 220, 715, 2002, 5005, 11440, 1, 11, 66, 286, 1001, 3003, 8008, 19448, 43758, 1, 12, 78, 364, 1365, 4368, 12376, 31824, 75582, 167960

Offset: 1

Roger L. Bagula, Jul 12 2007

Row sums are A002054.

			1;
1, 4;
1, 5, 15;
1, 6, 21, 56;
1, 7, 28, 84, 210;
1, 8, 36, 120, 330, 792;
1, 9, 45, 165, 495, 1287, 3003;
1, 10, 55, 220, 715, 2002, 5005, 11440;

Cf. A178300, A046899, A059481, A002054, A000292.

Table[Table[Binomial[n + i, i + 1], {n, 1, i}], {i, 1, 10}] Flatten[%]

A176991 Triangle t(n,m) = binomial(n+m,m) - binomial(n-m,m), 1<=m<=n, read by rows.

Original entry on oeis.org

2, 2, 6, 2, 10, 20, 2, 14, 35, 70, 2, 18, 56, 126, 252, 2, 22, 83, 210, 462, 924, 2, 26, 116, 330, 792, 1716, 3432, 2, 30, 155, 494, 1287, 3003, 6435, 12870, 2, 34, 200, 710, 2002, 5005, 11440, 24310, 48620, 2, 38, 251, 986, 3002, 8008, 19448, 43758, 92378, 184756

Offset: 1

Roger L. Bagula, Dec 08 2010

Row sums are binomial(2n+1,n+1)-1-A000071(n+1) = A001700(n)-A000045(n+1) = 2, 8, 32, 121, 454, 1703, 6414, 24276, 92323, 352627,....

			2;
2, 6;
2, 10, 20;
2, 14, 35, 70;
2, 18, 56, 126, 252;
2, 22, 83, 210, 462, 924;
2, 26, 116, 330, 792, 1716, 3432;
2, 30, 155, 494, 1287, 3003, 6435, 12870;
2, 34, 200, 710, 2002, 5005, 11440, 24310, 48620;
2, 38, 251, 986, 3002, 8008, 19448, 43758, 92378, 184756;

Cf. A046899, A178300, A001791

t[n_, m_] = Binomial[n + (m - 1), (m - 1)] - Binomial[n - (m - 1), (m - 1)];
Table[Table[t[n, m], {m, 2, n + 1}], {n, 1, 10}];
Flatten[%]

t(n,m) = A046899(n,m) - A011973(n,m), 0<=m<=n/2.

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A178302 Multiply the irregular Array A125108 by A178300;compute a(n)the vertical sums.

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A046899 Triangle in which n-th row is {binomial(n+k,k), k=0..n}, n >= 0.

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A227075 A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.

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A178301 Triangle T(n,k) = binomial(n,k)*binomial(n+k+1,n+1) read by rows, 0 <= k <= n.

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A176992 Triangle T(n,m) = binomial(2n-k+1, n+1) read by rows, 0 <= k <= n.

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A130746 Triangle read by rows: T(n,m) = binomial(n+m,1+n), 1<=m<=n.

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A176991 Triangle t(n,m) = binomial(n+m,m) - binomial(n-m,m), 1<=m<=n, read by rows.

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