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A290448 Triangle read by rows: T(n,k) = (Eulerian(n+1,k)-binomial(n,k))/2, for 0 <= k <= n. 0
0, 0, 0, 0, 1, 0, 0, 4, 4, 0, 0, 11, 30, 11, 0, 0, 26, 146, 146, 26, 0, 0, 57, 588, 1198, 588, 57, 0, 0, 120, 2136, 7792, 7792, 2136, 120, 0, 0, 247, 7290, 44089, 78060, 44089, 7290, 247, 0, 0, 502, 23902, 227554, 655114, 655114, 227554, 23902, 502, 0 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,8
COMMENTS
The entries in the triangle of Eulerian numbers (A173018) and Pascal's triangle of binomial coefficients (A007318) have the same parity (see A047999) provided the final diagonal of zeros in the Eulerian triangle is removed and all its rows are moved up a notch. Compare
1
1 0
1 1 0
1 4 1 0 (A173018)
1 11 11 1 0
1 26 66 26 1 0
...
and
1
1 1
1 2 1 (A007318)
1 3 3 1
1 4 6 4 1
...
The present triangle is the difference, halved.
LINKS
EXAMPLE
The triangle begins:
[0]
[0, 0]
[0, 1, 0]
[0, 4, 4, 0]
[0, 11, 30, 11, 0]
[0, 26, 146, 146, 26, 0]
[0, 57, 588, 1198, 588, 57, 0]
[0, 120, 2136, 7792, 7792, 2136, 120, 0]
[0, 247, 7290, 44089, 78060, 44089, 7290, 247, 0]
...
MAPLE
E:= proc(n, k) option remember;
if k=0 and n>=0 then 1
elif k<0 or k>n then 0
else (n-k) * E(n-1, k-1) + (k+1) * E(n-1, k)
fi
end:
T:=(n, k)->(E(n+1, k)-binomial(n, k))/2;
for n from 0 to 12 do lprint([seq(T(n, k), k=0..n)]); od:
MATHEMATICA
Eu[n_, k_] := Eu[n, k] = Which[k == 0 && n >= 0, 1, k < 0 || k > n, 0, True, (n-k)*Eu[n-1, k-1] + (k+1)*Eu[n-1, k]];
T[n_, k_] := (Eu[n+1, k] - Binomial[n, k])/2;
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 19 2023, after Maple code *)
CROSSREFS
Sequence in context: A165727 A284609 A363174 * A282593 A276339 A260318
KEYWORD
nonn,tabl
AUTHOR
N. J. A. Sloane, Aug 05 2017
STATUS
approved

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Last modified March 29 08:45 EDT 2024. Contains 371267 sequences. (Running on oeis4.)