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

Dr Sardonicus · 2018-09-20, 15:57

I checked the sums 1! + ... + (p-1)! (mod p) and 0! + 1! + ... + (p-1)! (mod p) for p < 50,000 and found no new examples of either sum being 0 (mod p).

I did notice that for p > 2,1! + ... + (p-1)! (mod p) may be re-expressed as -(1 - 1/1! + 1/2! - ... + 1/(p-1)!) (mod p) (for p = 2, the sum is -(1 - 1/1!) = 0 (mod 2). Alas, this seemed to be computationally useless.

The only somewhat-similar-looking sum with a simple formula I could recall, was an exercise in mathematical induction from a class I took when dinosaurs ruled the earth,

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

a1call · 2018-09-22, 00:08

Quote:

Originally Posted by ricky

The empty sum is equal to $0$ because in this way formulas like $\sum_{i \in I} a_i + \sum_{j \in J} a_j = \sum_{k \in I \cup J} a_k$ are correct, if $I \cap J = \emptyset$ . It is the only consistent way to define it.

I think I understand your notation, but can't figure out how an empty sum plays into making it consistent.

If I understand it correctly this:
$\sum ^{19}_{i=1}i+\sum ^{38}_{j=20}j=\sum ^{38}_{k=1}k$
Be an example of such a unity.
If not, can you please give a numeric example of when an empty-sum is significant for the consistency of the sum of two mutually exclusive sequence-sums?
Thank you in advance.

ricky · 2018-09-22, 09:20

It just means that if $I$ is empty than $J = J\cup I$ , so the sums $\sum_{k \in J} a_k$ and $\sum_{k \in J \cup I} a_k$ are equal. If you also want that the second sum is equal to $\sum_{k \in J} a_k + \sum_{k \in I} a_k$ , then you must have $\sum_{k \in I} a_k = 0$ .

There is nothing deep of difficult, usually in a proof the case when the index set is empty is trivial, so you can treat it easily directly, but with this definition you can treat it uniformly with the other cases.

For the same reasons the empty product is equal to $1$ , or the union of zero sets is the empty set. (But pay attention to the intersection of zero sets, that is trickier.)

ricky · 2018-09-22, 11:10

If someone is interested, it is true that $\sum_{n=0}^{k-1} n! \, \not \equiv \,0 \; \bmod{k}$ if $k \neq 2$ . See

http://www.numdam.org/article/JTNB_2004__16_1_1_0.pdf

Not so fast! The paper has been withdrawn...

a1call · 2018-09-29, 16:03

Quote:

Originally Posted by CRGreathouse

No, that function is valid for 0 as well, as I explained above. It gives the empty sum.

Is that function also valid for (-19)?
If not, why not?
If it is, then why is it not included in the sequence?
What is the cutoff criteria here?
Thank you in advance.

axn · 2018-09-29, 16:09

Quote:

Originally Posted by a1call

Is that function also valid for (-19)?
If not, why not?
If it is, then why is it not included in the sequence?
What is the cutoff criteria here?
Thank you in advance.

Zero terms of a series is meaningful. -19 terms is not.

a1call · 2018-09-29, 16:36

Quote:

Originally Posted by axn

Zero terms of a series is meaningful. -19 terms is not.

Why do you think the following statement is meaningless?
What's wrong with it?

$<br /> \sum ^{20}_{i=-19}i=20<br /> <br />$

VBCurtis · 2018-09-29, 20:46

How many terms does your example series have?

Now, how do you propose to define a series with -19 terms?

a1call · 2018-09-29, 20:53

Quote:

Originally Posted by VBCurtis

How many terms does your example series have?

Now, how do you propose to define a series with -19 terms?

-19 is the lower-bound of the variable i,not the number of terms in the sequence.
My example is equivalent to:

-19 + -18 + ... + 20 = 20
Number of the terms in the sequence is

19*2+2= 40

I find your question very revealing.

science_man_88 · 2018-09-29, 20:55

Quote:

Originally Posted by a1call

-19 is the lower-bound of the variable i,not the number of terms in the sequence.
My example is equivalent to:

-19 + -18 + ... + 20 = 20
Number of the terms in the sequence is

19*2+2= 40

unless you define an analytic continuation, factorial isn't defined below 0, is one point to be made.

a1call · 2018-09-29, 21:04

Quote:

Originally Posted by science_man_88

unless you define an analytic continuation, factorial isn't defined below 0, is one point to be made.

The text that you quoted does not deal with Factorials.
As for the function in discussion a(0) does not involve any Factorials either so it does not make a difference since factorial of 0 is not equal to 0. As is neither any Factorials what so ever.

2018-09-20, 15:57	#23
Dr Sardonicus Feb 2017 Nowhere 1001001000010₂ Posts	primes dividing all but finitely many of these sums I checked the sums 1! + ... + (p-1)! (mod p) and 0! + 1! + ... + (p-1)! (mod p) for p < 50,000 and found no new examples of either sum being 0 (mod p). I did notice that for p > 2,1! + ... + (p-1)! (mod p) may be re-expressed as -(1 - 1/1! + 1/2! - ... + 1/(p-1)!) (mod p) (for p = 2, the sum is -(1 - 1/1!) = 0 (mod 2). Alas, this seemed to be computationally useless. The only somewhat-similar-looking sum with a simple formula I could recall, was an exercise in mathematical induction from a class I took when dinosaurs ruled the earth, 11! + 22! + ... + nn! = (n+1)! - 1. Last fiddled with by Dr Sardonicus on 2018-09-20 at 16:01*

2018-09-22, 11:10	#26
ricky May 2018 43₁₀ Posts	If someone is interested, it is true that $\sum_{n=0}^{k-1} n! \, \not \equiv \,0 \; \bmod{k}$ if $k \neq 2$ . See http://www.numdam.org/article/JTNB_2004__16_1_1_0.pdf Not so fast! The paper has been withdrawn... Last fiddled with by ricky on 2018-09-22 at 11:31

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2018-09-22, 09:20	#25
ricky May 2018 2B₁₆ Posts	It just means that if $I$ is empty than $J = J\cup I$ , so the sums $\sum_{k \in J} a_k$ and $\sum_{k \in J \cup I} a_k$ are equal. If you also want that the second sum is equal to $\sum_{k \in J} a_k + \sum_{k \in I} a_k$ , then you must have $\sum_{k \in I} a_k = 0$ . There is nothing deep of difficult, usually in a proof the case when the index set is empty is trivial, so you can treat it easily directly, but with this definition you can treat it uniformly with the other cases. For the same reasons the empty product is equal to $1$ , or the union of zero sets is the empty set. (But pay attention to the intersection of zero sets, that is trickier.)

2018-09-29, 20:46	#30
VBCurtis "Curtis" Feb 2005 Riverside, CA 2·2,437 Posts	How many terms does your example series have? Now, how do you propose to define a series with -19 terms?