A096006 - OEIS

%I #15 Dec 18 2024 21:02:12

%S 2,3,3,2,3,2,5,5,5,5,3,5,5,5,3,7,7,7,7,7,7,2,7,7,7,7,7,2,3,3,7,7,7,7,

%T 3,3,5,5,5,7,7,7,5,5,5,11,11,11,11,11,11,11,11,11,11,3,11,11,11,11,11,

%U 11,11,11,11,3,13,13,13,13,13,13,13,13,13,13,13,13,7,13,13,13,13,13,13,13

%N Triangle read by rows: T(n,k) = largest prime factor of binomial(n,k), 1 <= k <= n-1.

%H Andrew Howroyd, <a href="/A096006/b096006.txt">Table of n, a(n) for n = 2..1276</a> (first 50 rows)

%F T(n,k) = A006530(A007318(n,k)).

%F T(n,k) = A098802(n,k).

%e Triangle begins:

%e   2;

%e   3, 3;

%e   2, 3, 2;

%e   5, 5, 5, 5;

%e   3, 5, 5, 5, 3;

%e   7, 7, 7, 7, 7, 7;

%e   2, 7, 7, 7, 7, 7, 2;

%e   3, 3, 7, 7, 7, 7, 3, 3;

%e   5, 5, 5, 7, 7, 7, 5, 5, 5;

%e   ...

%e n Pascal's Triangle

%e 1 1

%e 2 1 2 1

%e 3 1 3 3 1

%e 4 1 4 6 4 1

%e so 2,3,2 = largest prime factors of row 4 = entries position 4,5,6 in the sequence.

%e 4' 2 3 2

%o (PARI) T(n,k) = { my(f=factor(binomial(n,k))[,1]); if(!#f, 1, f[#f]) }

%o { for(n=2, 10, for(k=1, n-1, print1(T(n,k), ", ")); print) }

%Y Cf. A006530, A007318, A096007.

%Y Essentially A098802 without first column and last diagonal.

%K nonn,tabl,easy

%O 2,1

%A _Cino Hilliard_, Jul 25 2004

%E Offset corrected by _Andrew Howroyd_, Dec 18 2024