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
Examples
Triangle starts: 1; 1, 1; 1, 3, 1; 1, 2, 2, 1; 1, 5, 10, 5, 1; 1, 3, 5, 5, 3, 1; ...
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]
Programs
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GAP
Flat(List([1..12],n->List([1..n],k->Gcd(Binomial(n,k),Binomial(n,k-1))))); # Muniru A Asiru, Oct 24 2018
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Magma
/* As triangle */ [[Gcd(Binomial(n,k), Binomial(n,k-1)): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 25 2018
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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
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Mathematica
Table[GCD@@{Binomial[n,k],Binomial[n,k-1]},{n,20},{k,n}]//Flatten (* Harvey P. Dale, Aug 06 2017 *)
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PARI
T(n,k)=if(n<1 || k<1,0,gcd(n,k)*(n+k-1)!/n!/k!)
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PARI
T(n,k)=if(k<1 || k>n,0,gcd(n+1,k)*binomial(n,k-1)/k) /* Michael Somos, Mar 03 2004 */
Formula
a(2n, n) = n-th Catalan number; see A000108.
Also T(n, k) = gcd(C(n, k), C(n+1, k)).
Extensions
Offset set to 1 by R. J. Mathar, Dec 21 2010
Comments