%I #43 Sep 08 2022 08:44:58
%S 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,
%T 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,
%U 165,66,462,66,165,55,11,1,1,6,22,55,99,132,132,99,55,22,6,1
%N Triangle read by rows: T(n,k) = gcd(C(n,k), C(n,k-1)), n >= 1, 1 <= k <= n.
%C Equivalently, table T(n,k) = gcd(n,k)*(n+k-1)!/(n!*k!) read by antidiagonals. - _Michael Somos_, Jul 19 2002
%C Apparently, T(n,k)*gcd(C(n+1,k),n+1) = C(n+1,k). - _Thomas Anton_, Oct 24 2018
%D H. Gupta, On a problem in parity, Indian J. Math., 11 (1969), 157-163. MR0260659
%H Muniru A Asiru, <a href="/A050169/b050169.txt">Table of n, a(n) for n = 1..1275</a>(Rows n=1..50,flattened)
%H H. Gupta, <a href="/A002783/a002783_1.pdf">On a problem in parity</a>, Indian J. Math., 11 (1969), 157-163. [Annotated scanned copy]
%F a(2n, n) = n-th Catalan number; see A000108.
%F Also T(n, k) = gcd(C(n, k), C(n+1, k)).
%e Triangle starts:
%e 1;
%e 1, 1;
%e 1, 3, 1;
%e 1, 2, 2, 1;
%e 1, 5, 10, 5, 1;
%e 1, 3, 5, 5, 3, 1;
%e ...
%p 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
%t Table[GCD@@{Binomial[n,k],Binomial[n,k-1]},{n,20},{k,n}]//Flatten (* _Harvey P. Dale_, Aug 06 2017 *)
%o (PARI) T(n,k)=if(n<1 || k<1,0,gcd(n,k)*(n+k-1)!/n!/k!)
%o (PARI) T(n,k)=if(k<1 || k>n,0,gcd(n+1,k)*binomial(n,k-1)/k) /* _Michael Somos_, Mar 03 2004 */
%o (GAP) Flat(List([1..12],n->List([1..n],k->Gcd(Binomial(n,k),Binomial(n,k-1))))); # _Muniru A Asiru_, Oct 24 2018
%o (Magma) /* As triangle */ [[Gcd(Binomial(n,k), Binomial(n,k-1)): k in [1..n]]: n in [1.. 15]]; // _Vincenzo Librandi_, Oct 25 2018
%Y Cf. A002784, A178252.
%K nonn,tabl
%O 1,5
%A _Clark Kimberling_
%E Offset set to 1 by _R. J. Mathar_, Dec 21 2010