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A038719
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Triangle T(n,k) (0 <= k <= n) giving number of chains of length k in partially ordered set formed from subsets of n-set by inclusion.
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14
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1, 2, 1, 4, 5, 2, 8, 19, 18, 6, 16, 65, 110, 84, 24, 32, 211, 570, 750, 480, 120, 64, 665, 2702, 5460, 5880, 3240, 720, 128, 2059, 12138, 35406, 57120, 52080, 25200, 5040, 256, 6305, 52670, 213444, 484344, 650160, 514080, 221760, 40320, 512, 19171
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OFFSET
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0,2
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COMMENTS
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The relation of this triangle to A143494 given in the Formula section leads to the following combinatorial interpretation: T(n,k) gives the number of partitions of the set {1,2,...,n+2} into k + 2 blocks where 1 and 2 belong to two distinct blocks and the remaining k blocks are labeled from a fixed set of k labels. - Peter Bala, Jul 10 2014
Also, the number of distinct k-level fuzzy subsets of a set consisting of n elements ordered by set inclusion. - Rajesh Kumar Mohapatra, Mar 16 2020
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LINKS
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FORMULA
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T(n, k) = Sum_{j=0..k} (-1)^j*C(k, j)*(k+2-j)^n.
T(n+1, k) = k*T(n, k-1) + (k+2)*T(n, k), T(0,0) = 1, T(0,k) = 0 for k>0.
Binomial transform of n-th row = 2^n + 3^n + 4^n + ...; e.g., binomial transform of [8, 19, 18, 6] = 2^3 + 3^3 + 4^3 + 5^3 + ... = 8, 27, 64, 125, ... - Gary W. Adamson, May 15 2005
T(n,k) = k!*( Stirling2(n+2,k+2) - Stirling2(n+1,k+2) ).
n-th row polynomial = 1/(1 + x)*( sum {k >= 0} (k + 2)^n*(x/(1 + x))^k ). Cf. A028246. (End)
The row polynomials have the form (2 + x) o (2 + x) o ... o (2 + x), where o denotes the black diamond multiplication operator of Dukes and White. See example E12 in the Bala link. - Peter Bala, Jan 18 2018
Z(P,m) = Sum_{k=0..n} T(n,k)Binomial(m-2,k) = m^n, the zeta polynomial of the poset B_n. Each length m multichain from 0 to 1 in B_n corresponds to a function from [n] into [m]. - Geoffrey Critzer, Dec 25 2020
The entries in row n are the first terms in a table of the successive differences of the sequence [2^n, 3^n, 4^n, ...]. Examples are given below. - Peter Bala, Feb 02 2022
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EXAMPLE
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Triangle begins
1;
2, 1;
4, 5, 2;
8, 19, 18, 6;
16, 65, 110, 84, 24;
...
Table of successive differences of k^2 starting at k = 2
4 9 16
5 7
2
gives [4, 5, 2] as row 2 of this triangle.
Table of successive differences of k^3 starting at k = 2
8 27 64 125
19 37 61
18 24
6
gives [8, 19, 8, 6] as row 3 of this triangle. (End)
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MAPLE
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T:= proc(n, k) option remember;
`if` (n=0, `if`(k=0, 1, 0), k*T(n-1, k-1) +(k+2)*T(n-1, k))
end:
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MATHEMATICA
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t[n_, k_] := Sum[ (-1)^(k-i)*Binomial[k, i]*(2+i)^n, {i, 0, k}]; Flatten[ Table[ t[n, k], {n, 0, 9}, {k, 0, n}]] (* Jean-François Alcover, after Pari *)
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PROG
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(PARI) T(n, k)=sum(i=0, k, (-1)^(k-i)*binomial(k, i)*(2+i)^n)
(Haskell)
a038719 n k = a038719_tabl !! n !! k
a038719_row n = a038719_tabl !! n
a038719_tabl = iterate f [1] where
f row = zipWith (+) (zipWith (*) [0..] $ [0] ++ row)
(zipWith (*) [2..] $ row ++ [0])
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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More terms from Larry Reeves (larryr(AT)acm.org), May 09 2000
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STATUS
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approved
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