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A046899
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Triangle in which n-th row is {binomial(n+k,k), k=0..n}, n >= 0.
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23
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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
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OFFSET
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0,3
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COMMENTS
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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
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
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REFERENCES
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H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83.
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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)
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MAPLE
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for n from 0 to 10 do seq( binomial(n+m, n), m = 0 .. n) od; # Zerinvary Lajos, Dec 09 2007
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MATHEMATICA
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t[n_, k_] := Binomial[n + k, n]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 12 2013 *)
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PROG
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(PARI) /* same as in A092566 but use */
steps=[[1, 0], [1, 0] ];
(Haskell)
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
(Magma) /* As triangle */ [[Binomial(n+k, n): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 18 2015
(SageMath)
for n in (0..9):
print([multinomial(n, k) for k in (0..n)]) # Peter Luschny, Dec 24 2020
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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