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Old 2018-09-19, 13:25   #12
Dr Sardonicus
 
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Quote:
Originally Posted by ricky View Post
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.
Obviously the prime p divides n! for all n greater than p-1, so one can check whether p divides ap-1 or bp-1.

I was unable to find a "nice" way to do this. Brute-force checking turned up the examples already found, and showed no primes p, 3 < p < 20000 had this property for either sequence.
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Old 2018-09-19, 13:34   #13
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Quote:
Originally Posted by axn View Post
If a(n) = Sum_{k=1..n} k!, what is a(0)?
Not the 1st term, as nether is a(-19).

Further more for
a(n) = Sum_{k=0..n} k!
a(0) = 1

Last fiddled with by a1call on 2018-09-19 at 13:40
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Old 2018-09-19, 13:41   #14
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Quote:
Originally Posted by a1call View Post
Not the 1st term, as nether is a(-19).
I think OFFSET 0,3 is relevant here

Last fiddled with by axn on 2018-09-19 at 13:43
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Old 2018-09-19, 14:25   #15
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Quote:
Originally Posted by axn View Post
If a(n) = Sum_{k=1..n} k!, what is a(0)?
The empty sum
\sum_{k=1}^0 k! = \sum_{x\in\{\,\}} x = 0.

Last fiddled with by CRGreathouse on 2018-09-19 at 14:26
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Old 2018-09-19, 21:19   #16
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The sum starting from 0 is sometimes denoted !n, the sum starting from 1, K(n). See early results at http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/. I believe all those have been moved to factordb.com though.
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Old 2018-09-19, 22:51   #17
a1call
 
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Quote:
Originally Posted by axn View Post
I think OFFSET 0,3 is relevant here
Yes you are correct. It is relevant. It specifies the first subscript is 0 for a function that is defined to be valid for subscripts greater than or equal to 1.
Not really fixing the problem in my opinion.
What is the point of including a(0) in a sequence that does not define it?
How is that more significant than a(-19)?
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Old 2018-09-20, 02:19   #18
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Quote:
Originally Posted by a1call View Post
Yes you are correct. It is relevant. It specifies the first subscript is 0 for a function that is defined to be valid for subscripts greater than or equal to 1.
No, that function is valid for 0 as well, as I explained above. It gives the empty sum.

Last fiddled with by CRGreathouse on 2018-09-20 at 02:20
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Old 2018-09-20, 03:55   #19
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I think this is a better reference. It much more directly explains that the concept is a convention. It is interesting that the entry does not have the usual history segment. This makes me suspect it is as new of a concept as the notion of series implying summation. That would make it younger than I am.
As of this moment Wikipedia has equivalent entries in only 15 languages.

Last fiddled with by a1call on 2018-09-20 at 04:41
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Old 2018-09-20, 06:44   #20
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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.


By the way, it seems that quite a lot of effort has already been made in factoring these numbers, but there is nothing particularly interesting, it seems.
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Old 2018-09-20, 12:24   #21
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Quote:
Originally Posted by a1call View Post
I think this is a better reference. It much more directly explains that the concept is a convention. It is interesting that the entry does not have the usual history segment. This makes me suspect it is as new of a concept as the notion of series implying summation. That would make it younger than I am.
I don't agree with either of those: your conclusion that it is young or that article's take that it is (merely) a convention.
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Old 2018-09-20, 12:25   #22
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Quote:
Originally Posted by ricky View Post
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.
Precisely.
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