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a(n) = number of squarefree numbers among C(n,k), k=0..n.
6

%I #25 Jan 18 2018 08:23:13

%S 1,2,3,4,3,6,6,8,3,2,6,12,4,10,12,14,2,6,2,8,8,10,12,24,4,4,8,2,4,12,

%T 6,12,2,4,8,8,2,8,14,12,4,12,14,26,16,8,20,42,2,2,2,4,6,18,4,6,2,6,10,

%U 22,8,26,40,8,2,4,6,8,8,16,12,18,2,8,18,4,6,14,18,34,2,2,4,6,4,10,12,16,4

%N a(n) = number of squarefree numbers among C(n,k), k=0..n.

%C The only odd numbers are at n = 0, 2, 4, and 8. So this sequence is essentially twice A238337. - _T. D. Noe_, Mar 07 2014

%H T. D. Noe, <a href="/A048276/b048276.txt">Table of n, a(n) for n = 0..5000</a>

%F a(n) = n+1-A048277(n). - _R. J. Mathar_, Jan 18 2018

%e If n=20, then C(20, k) is squarefree for k = 0,2,4,8,12,16,18,20, that is, for 8 cases of k, so a(20)=8.

%p A048276 := proc(n)

%p local a,k ;

%p a := 0 ;

%p for k from 0 to n do

%p if issqrfree(binomial(n,k)) then

%p a := a+1 ;

%p end if;

%p end do:

%p a ;

%p end proc:

%p seq(A048276(n),n=0..40) ; # _R. J. Mathar_, Jan 18 2018

%t Table[Length[Select[Binomial[n, Range[0, n]], SquareFreeQ[#] &]], {n, 0, 100}]

%o (PARI) a(n) = sum(k=0, n, issquarefree(binomial(n, k))); \\ _Michel Marcus_, Dec 19 2013

%Y Cf. A005117, A046098, A048277, A238337.

%K nonn

%O 0,2

%A _Labos Elemer_