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%I #23 Sep 08 2022 08:46:16
%S 1,1,1,1,2,1,1,3,3,1,1,1,6,1,1,1,5,10,10,5,1,1,6,15,5,15,6,1,1,7,21,
%T 35,35,21,7,1,1,2,7,14,70,14,7,2,1,1,1,1,21,14,14,21,1,1,1,1,10,5,30,
%U 210,7,210,30,5,10,1,1,11,55,165,330,462,462,330,165,55,11,1,1,3,66,55,55,22,231,22,55,55,66,3,1,1,13,78,286,715,143,429,429,143,715,286,78,13,1
%N Triangle T(n,k) read by rows: T(n,k) is the squarefree part of C(n,k).
%e Triangle starts:
%e [1]
%e [1, 1]
%e [1, 2, 1]
%e [1, 3, 3, 1]
%e [1, 1, 6, 1, 1]
%e [1, 5, 10, 10, 5, 1]
%e [1, 6, 15, 5, 15, 6, 1]
%e [1, 7, 21, 35, 35, 21, 7, 1]
%e [1, 2, 7, 14, 70, 14, 7, 2, 1]
%e [1, 1, 1, 21, 14, 14, 21, 1, 1, 1]
%e [1, 10, 5, 30, 210, 7, 210, 30, 5, 10, 1]
%e [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1]
%e [1, 3, 66, 55, 55, 22, 231, 22, 55, 55, 66, 3, 1]
%e [1, 13, 78, 286, 715, 143, 429, 429, 143, 715, 286, 78, 13, 1]
%e [1, 14, 91, 91, 1001, 2002, 3003, 858, 3003, 2002, 1001, 91, 91, 14, 1]
%e [1, 15, 105, 455, 1365, 3003, 5005, 715, 715, 5005, 3003, 1365, 455, 105, 15, 1]
%e ...
%p T:= (n, k)-> mul(i[1]^irem(i[2], 2),
%p i=ifactors(binomial(n, k))[2]):
%p seq(seq(T(n, k), k=0..n), n=0..15); # _Alois P. Heinz_, Apr 01 2016
%t Table[Times @@ Apply[Power, {First@ #, Mod[Last@ #, 2]} & /@ FactorInteger@ Binomial[n, k], {1}], {n, 0, 13}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Apr 01 2016, after _Zak Seidov_ at A007913 *)
%o (PARI) for(n=0,20,for(k=0,n,print1(core(binomial(n,k)),", ")));
%o (Magma) /* As triangle: */ [[Squarefree(Binomial(n,k)): k in [0..n]]: n in [0..20]]; // _Wesley Ivan Hurt_, Apr 01 2016
%o (Sage) [squarefree_part(binomial(n,k)) for n in (0..15) for k in (0..n) ] # _Bruno Berselli_, Apr 01 2016
%Y Cf. A007318 (binomial coefficients), A007913 (squarefree parts).
%Y T(2n,n) gives A069113.
%K nonn,tabl
%O 0,5
%A _Joerg Arndt_, Apr 01 2016
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