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A050169 Triangle read by rows: T(n,k) = gcd(C(n,k), C(n,k-1)), n >= 1, 1 <= k <= n. 5

%I

%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

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Last modified February 18 00:37 EST 2020. Contains 332006 sequences. (Running on oeis4.)