I would prefer the logic used in Excel (or others?): date is an integer, normed.
Actually a float, to cover time over day

And all subsequent is formatting, depending on calendar conventions (which
have changed over centuries, if having such in mind).

Then a list for dates is just applying formatting as function (may be even to
regard working days only or some holiday calendar).

Including local conventions: 12/27/2000 is obvious, but 03/02/2000 is not
at all, it may be dd/mm/yyyy or mm/dd/yyyy and if using non-metric forms
it is always puzzling how to write 12:01 as p.m. or not while 00:01 ...

leap year if year = 0 mod 4

except if year = 0 mod 100

except if year = 0 mod 400

So there is 29 Feb in 2000, but not in 1900 or 2100

http://www.kzu.ch/fach/as/aktuell/2000/03_schalttag/2000_29_02.htm

Sorry, I did not recognise that you have a factorial of a factorial :-(

Sorry, I did not recognise that you have a factorial of a factorial :-(

I use the select command since the help on ifactor does not state
that the factors are sorted  (else one could take the first one and
simply checks, whether it holds a '2, making it more simple'):

 L:=NULL:
 for j from 2 to 2011 do
   tmp:=ifactors(j)[2]; # list of factors with positive multiplicity
   smallest:=op( select( 't -> 2=t[1]', tmp) );
   #is(smallest=NULL);
   if( NULL <> smallest and 2 = smallest[1]) then
     L := L,smallest[2];
   else
   end if;
 end do:
 L:=[L];

 theK:=convert(L, `+`); # log[10](2^%); evalf(%);

                             theK := 2002
 2^theK;

  4592522781097018096931332804710727936089270808354780801910570947\
        3030650455694812554266379452660250796737838585559498510937\
        8847584345222132570372542466661274156519956581249120002755\
        1165929601794857157079602539449791264618545855534557892681\
        0416186541588361998623264959523577851842249324659814465688\
        7881330725376675635937834022409517939231656235603739106001\
        7160108668265033220880325289451251670450715332688263959815\
        8567250808711943361616863941273300616355773560836768221983\
        1134358688156640327828866522331021521702334904062113395145\
        6777282180356611031355305007017421152032913710832718618150\
        48739404596117504

  length(%);
                                 603

I use the select command since the help on ifactor does not state
that the factors are sorted  (else one could take the first one and
simply checks, whether it holds a '2, making it more simple'):

 L:=NULL:
 for j from 2 to 2011 do
   tmp:=ifactors(j)[2]; # list of factors with positive multiplicity
   smallest:=op( select( 't -> 2=t[1]', tmp) );
   #is(smallest=NULL);
   if( NULL <> smallest and 2 = smallest[1]) then
     L := L,smallest[2];
   else
   end if;
 end do:
 L:=[L];

 theK:=convert(L, `+`); # log[10](2^%); evalf(%);

                             theK := 2002
 2^theK;

  4592522781097018096931332804710727936089270808354780801910570947\
        3030650455694812554266379452660250796737838585559498510937\
        8847584345222132570372542466661274156519956581249120002755\
        1165929601794857157079602539449791264618545855534557892681\
        0416186541588361998623264959523577851842249324659814465688\
        7881330725376675635937834022409517939231656235603739106001\
        7160108668265033220880325289451251670450715332688263959815\
        8567250808711943361616863941273300616355773560836768221983\
        1134358688156640327828866522331021521702334904062113395145\
        6777282180356611031355305007017421152032913710832718618150\
        48739404596117504

  length(%);
                                 603

Side remark (having the opposite problem): the deeper insight may come through
measuring hips and tighs as well.

Certainly a mistake.

Or ... may that be an ingenious idea of marketing? Pictures replacing substance?

sin(x^2) switches signs in x^2 = k*Pi and using the monoton transform T = sqrt(N*Pi)
it is convenient to write the task as

  Int(abs(sin(x^2)),x = 0 .. sqrt(N*Pi)) / sqrt(N*Pi)), N ---> oo

and split the integration range in those points, which leads to the following 

  's(n) = Int((sin(x^2)),x = sqrt((n-0)*Pi) .. sqrt((n+1)*Pi))'; 
  value(rhs(%)):
  simplify(%, size):
  s:=unapply(%,n); # symbolic solution as FresnelS

Checking sign in some points [s(0), s(1), s(100), s(101)]; evalf(%); thus 'shows'

  Int(abs(sin(x^2)),x = 0 .. T)/T = 'Sum( (-1)^k*(s(k)), k=0 .. N-1)/sqrt(N*Pi)'

Now evaluate in a large T:

  Digits:=14;
  N:=(10^13)^2;
  'Sum( (-1)^k*(s(k)), k=0 .. N-1)/sqrt(N*Pi)';
  ``=evalf(%);
  identify(%);

                           N - 1
                           -----
                            \        k
                             )   (-1)  s(k)
                            /
                           -----
                           k = 0
                           ----------------
                              sqrt(N Pi)


                           = 0.63661977236758


                                    2
                                 = ----
                                    Pi

sin(x^2) switches signs in x^2 = k*Pi and using the monoton transform T = sqrt(N*Pi)
it is convenient to write the task as

  Int(abs(sin(x^2)),x = 0 .. sqrt(N*Pi)) / sqrt(N*Pi)), N ---> oo

and split the integration range in those points, which leads to the following 

  's(n) = Int((sin(x^2)),x = sqrt((n-0)*Pi) .. sqrt((n+1)*Pi))'; 
  value(rhs(%)):
  simplify(%, size):
  s:=unapply(%,n); # symbolic solution as FresnelS

Checking sign in some points [s(0), s(1), s(100), s(101)]; evalf(%); thus 'shows'

  Int(abs(sin(x^2)),x = 0 .. T)/T = 'Sum( (-1)^k*(s(k)), k=0 .. N-1)/sqrt(N*Pi)'

Now evaluate in a large T:

  Digits:=14;
  N:=(10^13)^2;
  'Sum( (-1)^k*(s(k)), k=0 .. N-1)/sqrt(N*Pi)';
  ``=evalf(%);
  identify(%);

                           N - 1
                           -----
                            \        k
                             )   (-1)  s(k)
                            /
                           -----
                           k = 0
                           ----------------
                              sqrt(N Pi)


                           = 0.63661977236758


                                    2
                                 = ----
                                    Pi

  Int((sin(x^2)),x = sqrt((n-0)*Pi) .. sqrt((n+1)*Pi)); 
  value(%);
  simplify(%, size);
  s:=unapply(%,n); # = the integral as function of n, without abs-function

  MultiSeries:-asympt(s(n), n, 2); # 1st order 
  simplify(%) assuming 0 < n: 
  convert(%, polynom);
  simplify(%) assuming n::posint;

which gives (-1)^n / sqrt(n*Pi)

  Int((sin(x^2)),x = sqrt((n-0)*Pi) .. sqrt((n+1)*Pi)); 
  value(%);
  simplify(%, size);
  s:=unapply(%,n); # = the integral as function of n, without abs-function

  MultiSeries:-asympt(s(n), n, 2); # 1st order 
  simplify(%) assuming 0 < n: 
  convert(%, polynom);
  simplify(%) assuming n::posint;

which gives (-1)^n / sqrt(n*Pi)

"keep the feedback coming" ... one could not have expressed the message better

@Laurent: that explains only, why it is not possible 'without' effort. And not that it is not possible.
Its an objection, only - though a strong one.

The caveats of 'oldies' is not because of a different technology. But unsatisfactory functionality and
handling, not only for starters. Yet there are postings enough not to repeat argueing it again, there
lots of them. And in bad mood one might suspect, that the obstinate position towards that problems
might be related to the inventors of that interface.

For short: that brave 'new' appearence is not satisfactory.

Question: Doable?

Answer: "Yes"

But then Maple should provide a painless way, as option through installation.

And not leaving it as 'hack'