n!-(n-1)!-(n-2)!...-1!

Mr. Odd · 2012-11-20, 19:52

Playing around with number sequences, I started factoring factorial subtractions - n!-(n-1)!-(n-2)!-...-1!. It's clear there's a pattern in the numbers but my algebra is too rusty to figure it out. For n>=11 it's 3^2*11*m. Anyone have a simplified formula?

R. Gerbicz · 2012-11-20, 20:20

Quote:

Originally Posted by Mr. Odd

Anyone have a simplified formula?

A132371(n)

Dubslow · 2012-11-20, 20:38

a(n) = n! - (n-1)! - ...

a(n+1) = (n+1)! - n! - (n-1)! - ... = a(n) + (n+1)! - 2*n! = a(n) + n!*[(n+1) - 2] = a(n) + (n-1)*n!

a(n) = a(n-1) + (n-2)*(n-1)! = (n-2)*(n-1)! + (n-3)*(n-2)! + a(n-2) = ...

a(1) = 1! = 1

a(2) = 2! - 1! = 1
a(2) = a(1) + 0*1! = 1 + 0 = 1, good

a(3) = 3! - 2! - 1! = 6 - 2 - 1 = 3
a(3) = 1*2! + 0*1! + a(1) = 2 + 0 + 1 = 3, good

=> a(n) = 1 + $\sum_{i=2}^n{(i-2)*(i-1)!}$ = [j=i-1] 1 + $\sum_{j=1}^{n-1}{(j-1)*j!}$ , but the first term is 0, so
a(n) = 1 + $\sum_{j=2}^{n-1}{(j-1)*j!}$

...

n! + (n-1)! + ... = $\sum_{i=1}^n{\left(\prod_{j=1}^ij\right)}$

science_man_88 · 2012-11-20, 22:03

Quote:

Originally Posted by Dubslow

a(n) = n! - (n-1)! - ...

a(n+1) = (n+1)! - n! - (n-1)! - ... = a(n) + (n+1)! - 2*n! = a(n) + n!*[(n+1) - 2] = a(n) + (n-1)*n!

a(n) = a(n-1) + (n-2)*(n-1)! = (n-2)*(n-1)! + (n-3)*(n-2)! + a(n-2) = ...

a(1) = 1! = 1

a(2) = 2! - 1! = 1
a(2) = a(1) + 0*1! = 1 + 0 = 1, good

a(3) = 3! - 2! - 1! = 6 - 2 - 1 = 3
a(3) = 1*2! + 0*1! + a(1) = 2 + 0 + 1 = 3, good

=> a(n) = 1 + $\sum_{i=2}^n{(i-2)*(i-1)!}$ = [j=i-1] 1 + $\sum_{j=1}^{n-1}{(j-1)*j!}$ , but the first term is 0, so
a(n) = 1 + $\sum_{j=2}^{n-1}{(j-1)*j!}$

...

n! + (n-1)! + ... = $\sum_{i=1}^n{\left(\prod_{j=1}^ij\right)}$

simplest I can get is: a(n)=a(n-1)+(n!-2*(n-1)!) I believe, reasoning:

a(3) = 3!-2!-1! = 3
a(4) = 4!-3!-2!-1! = 15

the lime is the same, the sign on 3! has opposite indicating a change from 1 to -1 (-1)-1=-2 the coefficient I give the other change is the addition of 4! , n=4 so n!-2*(n-1)! is the difference:

Code:

a=[1];for(x=1,100,a=concat(a,a[#a]+(#a+1)!-2*eval(#a)!));a

Dubslow · 2012-11-20, 22:06

Quote:

Originally Posted by science_man_88

so n!-2*(n-1)! is the difference

Of course it is, and that's what I did. n! - 2*(n-1)! = (n-1)!*(n - 2). (If you don't believe me, distribute the latter multiplication.)

Of course, the sum I reduced it to is basically no simpler than the sum of factorials at the bottom of the post. If you can sum that, then you'll have a simple formula.

PS @OP, what do you mean by m? n! ?

science_man_88 · 2012-11-20, 22:13

Quote:

Originally Posted by Dubslow

Of course it is, and that's what I did. n! - 2*(n-1)! = (n-1)!*(n - 2). (If you don't believe me, distribute the latter multiplication.)

Of course, the sum I reduced it to is basically no simpler than the sum of factorials at the bottom of the post. If you can sum that, then you'll have a simple formula.

PS @OP, what do you mean by m? n! ?

maybe I got confused since now that i look at it your last equation suggest a sum of factorials is what's wanted by the OP, but I read it as subtraction in the OP.

Dubslow · 2012-11-20, 22:22

Quote:

Originally Posted by science_man_88

maybe I got confused since now that i look at it your last equation suggest a sum of factorials is what's wanted by the OP, but I read it as subtraction in the OP.

No, it is the subtraction, but I was initially trying to see if there was a way to do the simple factorial sum, call if F(n) = sum_i=1^n of n!.

Then a(n) = n! - F(n-1), so finding a formula for F would produce an easy formula for a. Of course, as I showed by reducing a(n) to a simpler series, that manner is no more promising than finding a formula for F itself.

In shorter words, finding a simple formula for both F and a is, AFAICT, essentially equivalent. That's what I was trying to say in my previous post.

science_man_88 · 2012-11-20, 22:38

Quote:

Originally Posted by Dubslow

No, it is the subtraction, but I was initially trying to see if there was a way to do the simple factorial sum, call if F(n) = sum_i=1^n of n!.

Then a(n) = n! - F(n-1), so finding a formula for F would produce an easy formula for a. Of course, as I showed by reducing a(n) to a simpler series, that manner is no more promising than finding a formula for F itself.

In shorter words, finding a simple formula for both F and a is, AFAICT, essentially equivalent. That's what I was trying to say in my previous post.

but if you look at the OEIS sequence for it, it doesn't start with 0 as you stated.

Dubslow · 2012-11-20, 22:52

Quote:

Originally Posted by science_man_88

but if you look at the OEIS sequence for it, it doesn't start with 0 as you stated.

0? Who said 0? a(1) = 1 = a(2), a(3) = 3.

PS: Sums of factorials. (Hardly a simple form.)

$\sum_{k=1}^n{k!} = \frac{1}{e}\left(-e + \int_\infty^{-1}{\frac{e^{-t}}{t}dt}+\pi i + (n+1)! \int_1^\infty{\frac{e^t}{t^{n+2}}dt} \right)$

science_man_88 · 2012-11-20, 22:57

Quote:

Originally Posted by Dubslow

but the first term is 0, so

thought this was part of it, apparently I'm mistaken.

LaurV · 2012-11-21, 03:49

Quote:

Originally Posted by Dubslow

PS @OP, what do you mean by m? n! ?

He means the decomposition of a_n into factors, last factor (prime or composite) being m. He claims after n>11, all have 3^2*11 as a common factor, which is totally true, because all factorials bigger then 10 contain 9 and 11, and the sum S=1!+2!+....+10! is divisible by 9 and 11, so their sum/difference with S is too.

(edit limit)

1. because 1!+2!=3 is divisible by 3, all a(n), n>=3, will be divisible by 3

2. because 1!+2!+3!+4!+5!=152=9*17 is divisible by 3^2, all a(n), n>=6, will be divisible by 3^2 (all factorials bigger or equal with 6 contain 3^2)

3. because 1!+2!+...+10! is divisible by 3^2*11, all a(n), n>=11, will be divisible by 3^2*11 (all factorials bigger or equal with 11 contain 3^2*11)

This probably happens an infinite number of times, but it is difficult to spot due to the rapid growth of a(n) series. The necessary condition is that the factorial sum s(n) to have factors lower then n, and the probability of this is higher, as n is higher. The next value where it is happening would look something like:

4. because 1!+2!+...+xxxxxx! is divisible by 3^2*11*yyy, with yyy<xxxxxx+1, then all a(n) for n>xxxxxx, will be divisible by 3^2*11*yyy (all factorials bigger then xxxxxx contain 3^2*11*yyy)

(tip: you don't need to factor the sums, only to TF them with primes under n, which is very fast, and you don't need to do it sequentially, just pick a very big xxxxxx, like the binary search, and if lucky, come down by halving n. I tested s(20000) and it has not other smaller factors except 9 and 11)

This only in case I don't miss some theoretic trick which forbids the factorial sums s(n) to have factors smaller than n.

2012-11-20, 19:52	#1
Mr. Odd Mar 2010 110111₂ Posts	n!-(n-1)!-(n-2)!...-1! Playing around with number sequences, I started factoring factorial subtractions - n!-(n-1)!-(n-2)!-...-1!. It's clear there's a pattern in the numbers but my algebra is too rusty to figure it out. For n>=11 it's 3^211m. Anyone have a simplified formula?

2012-11-20, 20:38	#3
Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40<A<43 -89<O<-88 3×29×83 Posts	a(n) = n! - (n-1)! - ... a(n+1) = (n+1)! - n! - (n-1)! - ... = a(n) + (n+1)! - 2n! = a(n) + n![(n+1) - 2] = a(n) + (n-1)n! a(n) = a(n-1) + (n-2)(n-1)! = (n-2)(n-1)! + (n-3)(n-2)! + a(n-2) = ... a(1) = 1! = 1 a(2) = 2! - 1! = 1 a(2) = a(1) + 01! = 1 + 0 = 1, good a(3) = 3! - 2! - 1! = 6 - 2 - 1 = 3 a(3) = 12! + 01! + a(1) = 2 + 0 + 1 = 3, good => a(n) = 1 + $\sum_{i=2}^n{(i-2)(i-1)!}$ = [j=i-1] 1 + $\sum_{j=1}^{n-1}{(j-1)j!}$ , but the first term is 0, so a(n) = 1 + $\sum_{j=2}^{n-1}{(j-1)j!}$ ... n! + (n-1)! + ... = $\sum_{i=1}^n{\left(\prod_{j=1}^ij\right)}$ Last fiddled with by Dubslow on 2012-11-20 at 21:36