A213999 Denominators of the triangle of fractions read by rows: pf(n,0) = 1, pf(n,n) = 1/(n+1) and pf(n+1,k) = pf(n,k) + pf(n,k-1) with 0 < k < n; denominators: A213998.

1, 1, 2, 1, 2, 3, 1, 2, 6, 4, 1, 2, 3, 12, 5, 1, 2, 6, 12, 60, 6, 1, 2, 3, 4, 10, 20, 7, 1, 2, 6, 12, 20, 20, 140, 8, 1, 2, 3, 12, 15, 10, 35, 280, 9, 1, 2, 6, 4, 20, 30, 70, 280, 2520, 10, 1, 2, 3, 12, 10, 12, 21, 56, 252, 2520, 11, 1, 2, 6, 12, 60, 60, 84, 168, 504, 2520, 27720, 12

Offset

0,3

Comments

T(n,0) = 1;

T(n,1) = A007395(n) for n > 0;

T(n,2) = A010704(n) for n > 1;

T(n,n-3) = A124838(n-2) for n > 2;

T(n,n-2) = A027611(n-1) for n > 1;

T(n,n-1) = A002805(n) for n > 0;

T(n,n) = n + 1;

A003418(n+1) = least common multiple of n-th row;

A214075(n,k) = floor(A213998(n,k) / T(n,k)).

Links

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened

Index entries for triangles and arrays related to Pascal's triangle

Example

See A213998.

Mathematica

T[_, 0] = 1; T[n_, n_] := 1/(n + 1);
T[n_, k_] := T[n, k] = T[n - 1, k] + T[n - 1, k - 1];
Table[T[n, k] // Denominator, {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 10 2021 *)

Prog

(Haskell)
import Data.Ratio ((%), denominator, Ratio)
a213999 n k = a213999_tabl !! n !! k
a213999_row n = a213999_tabl !! n
a213999_tabl = map (map denominator) $ iterate pf [1] where
   pf row = zipWith (+) ([0] ++ row) (row ++ [-1 % (x * (x + 1))])
            where x = denominator $ last row

Crossrefs

Sequence in context: A087730 A263736 A126247 * A249026 A263597 A263905

Adjacent sequences: A213996 A213997 A213998 * A214000 A214001 A214002

Keyword

nonn,frac,tabl,look

Author

Reinhard Zumkeller, Jul 03 2012

Status

approved