A050169 Triangle read by rows: T(n,k) = gcd(C(n,k), C(n,k-1)), n >= 1, 1 <= k <= n.

1, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 5, 10, 5, 1, 1, 3, 5, 5, 3, 1, 1, 7, 7, 35, 7, 7, 1, 1, 4, 28, 14, 14, 28, 4, 1, 1, 9, 12, 42, 126, 42, 12, 9, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 11, 55, 165, 66, 462, 66, 165, 55, 11, 1, 1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1

Offset

1,5

Comments

Equivalently, table T(n,k) = gcd(n,k)*(n+k-1)!/(n!*k!) read by antidiagonals. - Michael Somos, Jul 19 2002

Apparently, T(n,k)*gcd(C(n+1,k),n+1) = C(n+1,k). - Thomas Anton, Oct 24 2018

References

H. Gupta, On a problem in parity, Indian J. Math., 11 (1969), 157-163. MR0260659

Links

Muniru A Asiru, Table of n, a(n) for n = 1..1275(Rows n=1..50,flattened)

H. Gupta, On a problem in parity, Indian J. Math., 11 (1969), 157-163. [Annotated scanned copy]

Formula

a(2n, n) = n-th Catalan number; see A000108.

Also T(n, k) = gcd(C(n, k), C(n+1, k)).

Example

Triangle starts:
  1;
  1,  1;
  1,  3,  1;
  1,  2,  2,  1;
  1,  5, 10,  5,  1;
  1,  3,  5,  5,  3,  1;
  ...

Maple

a:=(n, k)->gcd(binomial(n, k), binomial(n, k-1)): seq(seq(a(n, k), k=1..n), n=1..12); # Muniru A Asiru, Oct 24 2018

Mathematica

Table[GCD@@{Binomial[n, k], Binomial[n, k-1]}, {n, 20}, {k, n}]//Flatten (* Harvey P. Dale, Aug 06 2017 *)

Prog

(PARI) T(n, k)=if(n<1 || k<1, 0, gcd(n, k)*(n+k-1)!/n!/k!)
(PARI) T(n, k)=if(k<1 || k>n, 0, gcd(n+1, k)*binomial(n, k-1)/k) /* Michael Somos, Mar 03 2004 */
(GAP) Flat(List([1..12], n->List([1..n], k->Gcd(Binomial(n, k), Binomial(n, k-1))))); # Muniru A Asiru, Oct 24 2018
(Magma) /* As triangle */ [[Gcd(Binomial(n, k), Binomial(n, k-1)): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 25 2018

Crossrefs

Cf. A002784, A178252.

Sequence in context: A176346 A338878 A073166 * A143214 A300380 A300682

Adjacent sequences: A050166 A050167 A050168 * A050170 A050171 A050172

Keyword

nonn,tabl

Author

Clark Kimberling

Extensions

Offset set to 1 by R. J. Mathar, Dec 21 2010

Status

approved