A000522 vs A131178 ?

[Maximilian Hasler]

Indeed, from
sum( 1/k!, k=0..n+1 ) < e < sum( 1/k!, k=0..n )+1/n!
one gets: (subtract 1/(n+1)! and multiply by n!)
n! sum( 1/k!, k=0..n ) < e n! - 1/(n+1) < n! sum( 1/k!, k=0..n ) + n/(n+1)

The l.h.s. is an integer, and the r.h.s. is less than 1 more.
Only for n=0, the second inequality is not valid,
but the expression in the middle equals 1.71828..., thus floor(...)=1=lhs
i.e. the formula is also valid for n=0 :

n! sum( 1/k!, k=0..n ) = floor( e n! - 1/(n+1) ) for all n>=0.

B.t.w.,
1/ The relation between A000522 and A131178  is already given as the
3rd entry in "FORMULA" for A000522, but the reference to A131178   is
missing there and should be added (also, n>=0 should be corrected to
n>=1)

2/ using Taylor's formula with integral rest for exp(1) yields yet
another formula for A000522:

A000522[n] = e*n! - int( (1-t)^n*e^t, t=0..1)

Maple code:

> n!*exp(1)-int((1-t)^n*exp(t),t=0..1)$n=0..15;

  1, 2, 5, 16, 65, 326, 1957, 13700, 109601, 986410, 9864101,
108505112, 1302061345, 16926797486, 236975164805, 3554627472076

M.H.

On 10/30/07, Alexander Povolotsky <apovolot at gmail.com> wrote:
> It appears that David Cantrell's suggested formula (if math works in
> it - for all range of n's  - I havn't checked), being converted though
> from the starting index 1 to the starting index 0 (to comply with the
> current EIS's definition of the A000522 ),
>
> floor(e (n)! - 1/(n +1 )), valid for all n >= 0
>
> might be essentially equivalent
> to the formula derived by Maximilian Hasler:
>
> sum( n! / k!, k=0..n ) = floor( e n! ) - delta[n,0]   for all n >= 0.
>
> where delta[n,0] (=1 for n=0, =0 else):
>
>
> On 10/29/07, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> > From the Taylor series of exp(1), one has the well-known formula
> > sum( 1/k!, k=0..n+1 ) < e < sum( 1/k!, k=0..n )+1/n!
> > for all n >= 1.
> > Multiplying by n! and applying floor() preserves the strictness of the
> > second inequality
> > since the rightmost expression is an integer but not the middle
> > expression (irrational).
> > Thus we have the equality : floor( e n! ) = n! sum( 1/k!, k=0..n )
> > for all n >= 1, and it is seen that for n=0, we must add delta[n,0]
> > (=1 for n=0, =0 else):
> >
> >  sum( n! / k!, k=0..n ) = floor( e n! ) - delta[n,0]   for all n >= 0.
> >
> > M.H.
> >
> > On 10/29/07, David W. Cantrell <DWCantrell at sigmaxi.net> wrote:
> > > My quick guess is that a formula in closed form for A000522 is
> > >
> > > floor(e (n - 1)! - 1/n), valid for n >= 1.
> > >
> > > David W. Cantrell
> > >
> > >
> > > ----- Original Message -----
> > > From: "Alexander Povolotsky" <apovolot at gmail.com>
> > > To: <njas at research.att.com>
> > > Cc: "seqfan" <seqfan at ext.jussieu.fr>
> > > Sent: Monday, October 29, 2007 22:58
> > > Subject: A000522 vs A131178 ?
> > >
> > >
> > > > A000522  Total number of arrangements of a set with n elements:
> > > > a(n) = Sum_{k=0..n} n!/k!.
> > > >
> > > > 1, 2, 5, 16, 65, 326, 1957, 13700, 109601, 986410, 9864101,
> > > > 108505112, 1302061345, 16926797486, 236975164805, 3554627472076,
> > > > 56874039553217, 966858672404690, 17403456103284421,
> > > > 330665665962404000, 6613313319248080001
> > > >
> > > > AUTHOR   njas
> > > >
> > > > A131178  Floor( e n! ).
> > > > 2, 2, 5, 16, 65, 326, 1957, 13700, 109601, 986410, 9864101,
> > > > 108505112, 1302061345, 16926797486, 236975164805, 3554627472076,
> > > > 56874039553217, 966858672404690, 17403456103284421,
> > > > 330665665962404000, 6613313319248080001, 138879579704209680022
> > > >
> > > > AUTHOR   njas
> > >
> >
>




Shouldn't perhaps the old name be made part of the
Comment line, so that if (say) someone happens to do a

Yes, of course. That goes without saying.
Neil



I merged A000522 and A131178 yesterday

Neil