mersenneforum.org Numbers of the form 1!+2!+3!+...
 Register FAQ Search Today's Posts Mark Forums Read

 2018-09-18, 16:06 #1 ricky   May 2018 538 Posts Numbers of the form 1!+2!+3!+... For no special reasons I started getting interested in factoring numbers of the form $ \sum_{n=0}^k n!$ or $ \sum_{n=1}^k n!$. 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, 16:35 #2 CRGreathouse     Aug 2006 3·52·79 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 2×13×73 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 2·13·73 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.
2018-09-18, 23:07   #5
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

2×13×73 Posts

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?

Last fiddled with by a1call on 2018-09-18 at 23:12

 2018-09-18, 23:56 #6 Dr Sardonicus     Feb 2017 Nowhere 22×53×7 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 2*prime, 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 2·13·73 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
2018-09-19, 12:06   #9
CRGreathouse

Aug 2006

10111001001012 Posts

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.

2018-09-19, 12:28   #10
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

2·13·73 Posts

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.

Last fiddled with by a1call on 2018-09-19 at 12:30

2018-09-19, 13:04   #11
axn

Jun 2003

4,703 Posts

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)?

 Similar Threads Thread Thread Starter Forum Replies Last Post a1call Information & Answers 17 2017-02-26 22:01 carpetpool carpetpool 1 2017-01-30 13:36 carpetpool carpetpool 3 2017-01-26 01:29 Dougy Math 8 2009-09-03 02:44 juergen Math 2 2004-04-17 12:19

All times are UTC. The time now is 17:16.

Thu Oct 1 17:16:19 UTC 2020 up 21 days, 14:27, 0 users, load averages: 1.63, 1.77, 1.68