A353594
Triangle read by rows, T(n, k) = Sum_{j=0..n} binomial(j, k)*A352687(n, j).
Original entry on oeis.org
1, 1, 1, 2, 3, 1, 4, 8, 5, 1, 10, 25, 22, 8, 1, 28, 84, 95, 50, 12, 1, 84, 294, 406, 280, 100, 17, 1, 264, 1056, 1722, 1470, 700, 182, 23, 1, 858, 3861, 7260, 7392, 4410, 1554, 308, 30, 1, 2860, 14300, 30459, 36036, 25872, 11550, 3150, 492, 38, 1
Offset: 0
Triangle starts:
[0] 1;
[1] 1, 1;
[2] 2, 3, 1;
[3] 4, 8, 5, 1;
[4] 10, 25, 22, 8, 1;
[5] 28, 84, 95, 50, 12, 1;
[6] 84, 294, 406, 280, 100, 17, 1;
[7] 264, 1056, 1722, 1470, 700, 182, 23, 1;
[8] 858, 3861, 7260, 7392, 4410, 1554, 308, 30, 1;
[9] 2860, 14300, 30459, 36036, 25872, 11550, 3150, 492, 38, 1;
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S := (n, k) -> if n = k then 1 elif k = 0 then 0 else
binomial(n, k)^2*(k*(2*k^2 + (n + 1)*(n - 2*k)))/(n^2*(n - 1)*(n - k + 1)) fi:
T := (n, k) -> add(binomial(j, k)*S(n, j), j = 0..n):
A353279
Triangle read by rows, a Narayana related triangle whose rows are refinements of four times the Catalan numbers (for n >= 3).
Original entry on oeis.org
1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 5, 8, 5, 1, 0, 1, 8, 19, 19, 8, 1, 0, 1, 12, 41, 60, 41, 12, 1, 0, 1, 17, 81, 165, 165, 81, 17, 1, 0, 1, 23, 148, 406, 560, 406, 148, 23, 1, 0, 1, 30, 253, 910, 1666, 1666, 910, 253, 30, 1
Offset: 0
Triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1, 1;
[3] 0, 1, 2, 1
[4] 0, 1, 3, 3, 1
[5] 0, 1, 5, 8, 5, 1
[6] 0, 1, 8, 19, 19, 8, 1
[7] 0, 1, 12, 41, 60, 41, 12, 1
[8] 0, 1, 17, 81, 165, 165, 81, 17, 1
[9] 0, 1, 23, 148, 406, 560, 406, 148, 23, 1
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Q := proc(n, k) option remember; local A, B, j;
if n <= k then return [seq(binomial(n-1, j-1), j = 0..n)] fi; # A097805
A := [op(Q(n - 2, k)), 0, 0]; B := [op(Q(n - 1, k)), 1];
for j from n by -1 to 3 do
B[j] := ((B[j] + B[j-1])*(2*(n - k) + 1)
- (A[j] - 2*A[j-1] + A[j-2])*(n - k - 1)) / (n - k + 2);
od: B end:
Trow := n -> Q(n, 3): for n from 0 to 9 do print(Trow(n)) od:
-
Q[n_, k_] := Q[n, k] = Module[{A, B, j},
If[n <= k, Return[Table[Binomial[n-1, j-1], {j, 0, n}]]];
A = Join[Q[n-2, k], {0, 0}]; B = Join[Q[n-1, k], {1}];
For[j = n, j >= 3, j--,
B[[j]] = ((B[[j]] + B[[j-1]])*(2*(n-k)+1)-
(A[[j]]-2*A[[j-1]]+A[[j-2]])*(n-k-1))/(n-k+2)];
B];
Trow[n_] := Q[n, 3];
Table[Trow[n], {n, 0, 9}] // Flatten (* Jean-François Alcover, Jul 07 2022, translated from Maple code *)
-
from functools import cache
from math import comb
def comp(n, k): # compositions A097805
return comb(n-1, k-1) if k != 0 else k**n
@cache
def Trow(n, k):
if n <= k:
return [comp(n, j) for j in range(n + 1)]
A = Trow(n - 2, k) + [0, 0]
B = Trow(n - 1, k) + [1]
for j in range(n - 1, 1, -1):
B[j] = ((B[j] + B[j - 1]) * (2 * (n - k) + 1)
- (A[j] - 2 * A[j - 1] + A[j - 2]) * (n - k - 1)) // (n - k + 2)
return B
for n in range(10): print(Trow(n, 3)) # k=1 -> A090181, k=2 -> A352687
Showing 1-2 of 2 results.
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