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A158464 Number of distinct squares in row n of Pascal's triangle.

%I A158464 #25 Nov 08 2017 22:30:34
%S A158464 1,1,1,1,2,1,1,1,1,3,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,
%T A158464 1,1,2,1,1,1,1,1,1,1,1,1,1,1,1,2,3,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,
%U A158464 1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1
%N A158464 Number of distinct squares in row n of Pascal's triangle.
%C A158464 It seems that some subset of terms in A055997 (A115599) gives the positions of 3's. E.g., we have a(9) = a(50) = a(289) = a(9801) = 3, but on the other hand, a(1682) = a(57122) = 2. - _Antti Karttunen_, Nov 03 2017
%H A158464 Antti Karttunen, <a href="/A158464/b158464.txt">Table of n, a(n) for n = 0..11111</a>
%H A158464 <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F A158464 a(n) = Sum_{k=0..floor(n/2)} A010052(A007318(n,k));
%F A158464 a(A000290(n)) > 1 for n > 1.
%e A158464 a(8) = #{1} = 1;
%e A158464 a(9) = #{1,9,36} = 3.
%p A158464 A158464 := proc(n)
%p A158464     local sqset,k ;
%p A158464     sqset := {} ;
%p A158464     for k from 0 to n do
%p A158464         P := binomial(n,k) ;
%p A158464         if issqr(P) then
%p A158464             sqset := sqset union {P} ;
%p A158464         end if;
%p A158464     end do:
%p A158464     nops(sqset) ;
%p A158464 end proc:
%p A158464 seq(A158464(n),n=0..120) ; # _R. J. Mathar_, Jul 09 2016
%t A158464 CountDistinct /@ Table[Sqrt@ Binomial[n, k] /. k_ /; ! IntegerQ@ k -> Nothing, {n, 0, 104}, {k, 0, n}] (* _Michael De Vlieger_, Nov 03 2017 *)
%o A158464 (PARI) A158464(n) = sum(k=0,n\2,issquare(binomial(n,k))); \\ _Antti Karttunen_, Nov 03 2017
%Y A158464 Cf. A000290, A055997 (A115599).
%K A158464 nonn
%O A158464 0,5
%A A158464 _Reinhard Zumkeller_, Mar 19 2009
%E A158464 More terms from _Antti Karttunen_, Nov 03 2017