In the second, you have the reference operator followed by a subroutine name. This returns a reference to the named subroutine.

In the first, you have the reference operator followed by a subroutine call. Subroutine calls evaluate to the value returned by a call to the subroutine. You're making a reference to that.

The solution is to wrap the call in a subroutine that takes no argument, and take a reference to that subroutine instead.

my $subref = sub { testing(1,2,3,4,6) };
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