Numbers of the form 1!+2!+3!+...

ricky · 2018-09-18, 16:06

For no special reasons I started getting interested in factoring numbers of the form $<br /> \sum_{n=0}^k n!<br />$ or $<br /> \sum_{n=1}^k n!<br />$ . Do you know if someone has already looked into these numbers? Of course a lot of them are done on factodb, and some are quite easy with ECM, but sometimes the factorization is not so easy.

CRGreathouse · 2018-09-18, 16:35

For one thing, Zivkovic proved that there are only finitely many primes of the latter form. You may find more information on their OEIS entries:
https://oeis.org/A007489
https://oeis.org/A003422

a1call · 2018-09-18, 16:58

That is
1+2+6+24+120....

So except for the first term it is always divisible by 3 and except for the 2nd term it is never prime.
After the 6th term it will always be divisible by 3 only once and the same type of progression will apply to infinity.

a1call · 2018-09-18, 22:53

As usual spoke before checking first.
Apparently after and including the 5th term all results are divisible by 3 exactly 2 times which is 9.

a1call · 2018-09-18, 23:07

Quote:

Originally Posted by CRGreathouse

For one thing, Zivkovic proved that there are only finitely many primes of the latter form. You may find more information on their OEIS entries:
https://oeis.org/A007489
https://oeis.org/A003422

The sequence A0074789 seems to me to be wrong.
0! Equals 1 not 0.

ETA For k 1 to n how did the 1st term becomes 0?

Dr Sardonicus · 2018-09-18, 23:56

The sums starting with 0! are even starting with k = 1, and greater than 2 for k > 1.

The sums starting with 1! are (as already observed) divisible by 3^2 for k > 4, and also by 11 for all k > 9.

The sum 0! + ... + 29! is 2*prime, and 1! + ... + 30! is 3^2 * 11 * prime.

a1call · 2018-09-19, 04:55

The mechanics of it is:
valuation (factorial sum, prime) locks in value as soon as the valuation of the addends exceed the valuation of the running sum.
So iff the running sum ever factors to a valuation higher than one (such as is the case with 3), just before the addend's valuation exceeds the valuation of the running sum, the valuation can lock in a value greater than one.

General rules:
https://www.mersenneforum.org/showthread.php?t=22434

Would be interesting to see what other prime factors lock in valuations of greater than one(if at all possible).

ricky · 2018-09-19, 08:14

Letting $a_k$ and $b_k$ the two sequences, it is clear that if $d \leq k+1$ divides $a_k$ or $b_k$ then $d$ divides all the following terms of the sequences. As far as I've looked, this happens only for $d=2$ that divides $a_1 = 2$ , for $d=3$ that divides $b_2 = 3$ , for $d=9$ that divides $b_8 = 46233$ and for $d=11$ that divides $b_{10} = 4037913$ . It would be interesting to know whether this happen again, but it seems quite unlike.

I do not see any other easy properties of these number, I will think about it.

CRGreathouse · 2018-09-19, 12:06

Quote:

Originally Posted by a1call

The sequence A0074789 seems to me to be wrong.
0! Equals 1 not 0.

ETA For k 1 to n how did the 1st term becomes 0?

The first term is 1! = 1. The sequence starting with 0! = 1 is A003422.

a1call · 2018-09-19, 12:28

Quote:

Originally Posted by CRGreathouse

The first term is 1! = 1.

This is what I see in data section:

Quote:

0, 1, 3, 9, 33, 153, 873, 5913, 46233, 409113, 4037913, 43954713, 522956313, 6749977113, 93928268313, 1401602636313, 22324392524313, 378011820620313, 6780385526348313, 128425485935180313, 2561327494111820313, 53652269665821260313

https://oeis.org/A007489

With the 1st term being 0 not 1.

axn · 2018-09-19, 13:04

Quote:

Originally Posted by a1call

This is what I see in data section:

https://oeis.org/A007489

With the 1st term being 0 not 1.

If a(n) = Sum_{k=1..n} k!, what is a(0)?

2018-09-18, 16:06	#1
ricky May 2018 2B₁₆ Posts	Numbers of the form 1!+2!+3!+... For no special reasons I started getting interested in factoring numbers of the form $<br /> \sum_{n=0}^k n!<br />$ or $<br /> \sum_{n=1}^k n!<br />$ . Do you know if someone has already looked into these numbers? Of course a lot of them are done on factodb, and some are quite easy with ECM, but sometimes the factorization is not so easy. Last fiddled with by ricky on 2018-09-18 at 16:07

2018-09-18, 23:56	#6
Dr Sardonicus Feb 2017 Nowhere 47×89 Posts	The sums starting with 0! are even starting with k = 1, and greater than 2 for k > 1. The sums starting with 1! are (as already observed) divisible by 3^2 for k > 4, and also by 11 for all k > 9. The sum 0! + ... + 29! is 2prime, and 1! + ... + 30! is 3^2 11 * prime. Last fiddled with by Dr Sardonicus on 2018-09-18 at 23:56

2018-09-19, 04:55	#7
a1call "Rashid Naimi" Oct 2015 Remote to Here/There 1982₁₀ Posts	The mechanics of it is: valuation (factorial sum, prime) locks in value as soon as the valuation of the addends exceed the valuation of the running sum. So iff the running sum ever factors to a valuation higher than one (such as is the case with 3), just before the addend's valuation exceeds the valuation of the running sum, the valuation can lock in a value greater than one. General rules: https://www.mersenneforum.org/showthread.php?t=22434 Would be interesting to see what other prime factors lock in valuations of greater than one(if at all possible). Last fiddled with by a1call on 2018-09-19 at 05:27

2018-09-19, 08:14	#8
ricky May 2018 43 Posts	Letting $a_k$ and $b_k$ the two sequences, it is clear that if $d \leq k+1$ divides $a_k$ or $b_k$ then $d$ divides all the following terms of the sequences. As far as I've looked, this happens only for $d=2$ that divides $a_1 = 2$ , for $d=3$ that divides $b_2 = 3$ , for $d=9$ that divides $b_8 = 46233$ and for $d=11$ that divides $b_{10} = 4037913$ . It would be interesting to know whether this happen again, but it seems quite unlike. I do not see any other easy properties of these number, I will think about it. Last fiddled with by ricky on 2018-09-19 at 08:16

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2018-09-18, 16:35	#2
CRGreathouse Aug 2006 3·1,987 Posts	For one thing, Zivkovic proved that there are only finitely many primes of the latter form. You may find more information on their OEIS entries: https://oeis.org/A007489 https://oeis.org/A003422

2018-09-18, 16:58	#3
a1call "Rashid Naimi" Oct 2015 Remote to Here/There 2×991 Posts	That is 1+2+6+24+120.... So except for the first term it is always divisible by 3 and except for the 2nd term it is never prime. After the 6th term it will always be divisible by 3 only once and the same type of progression will apply to infinity.

2018-09-18, 22:53	#4
a1call "Rashid Naimi" Oct 2015 Remote to Here/There 2·991 Posts	As usual spoke before checking first. Apparently after and including the 5th term all results are divisible by 3 exactly 2 times which is 9.