[seqfan] Re: 10 integers dividing the sum of the 10 integers

[Éric Angelini]

Lars:

> This illustrates a problem with lexicofirst: One can obtain a smaller value for some term, at the expense of larger values for subsequent terms, and a larger sum.

... personnaly I would stick to the
true lexicographically first.
But the second idea (smallest sum)
is nice too.
What about submitting both?

à+
É.
Catapulté de mon aPhone


> Le 15 mai 2019 à 16:44, Lars Blomberg <larsl.blomberg at comhem.se> a écrit :
> 
> Hello SeqFans,
> 
> For n=6 I get 1 2 3 4 6 8 (same as reported by Peter Munn).
> For n=13 I get 
> 1 2 3 4 5 6 7 8 9 15 360 840 1260   sum 2520 not
> 1 2 3 4 5 6 7 8 10 24 70 280 420 sum 840.                
> This illustrates a problem with lexicofirst: One can obtain a smaller value for some term, at the expense of larger values for subsequent terms, and a larger sum.
> So can one put a limit on the sums to be tested?
> 
> /Lars B
> 
> -----Ursprungligt meddelande-----
> Från: SeqFan <seqfan-bounces at list.seqfan.eu> För Éric Angelini
> Skickat: den 13 maj 2019 11:43
> Till: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Ämne: [seqfan] 10 integers dividing the sum of the 10 integers
> 
> Hello SeqFans,
> this was suggested to me by an "Enigma" posted here :
> https://bit.ly/2Vyu9Mv
> 
> "Find 10 different integers {a, b, c, ... i, j} such that they all divide the sum (a+b+c+d+e+f+g+h+i+j)"
> 
> The given solution was not mine -- as I wanted to find the lexicographically earliest set of this kind.
> 
> Then I started to search for such lexico-sets of size "n" (with n > 2). Carole Dubois and I found:
> 
> n=3, integers = 1 2 3          
> n=4, integers = 1 2 3 6         
> n=5, integers = 1 2 3 6 12        
> n=6, integers = 1 2 3 4 20 30       
> n=7, integers = 1 2 3 4 5 15 30      
> n=8, integers = 1 2 3 4 5 10 15 20     
> n=9, integers = 1 2 3 4 5 6 9 60 90    
> n=10, integers = 1 2 3 4 5 6 7 42 140 210   
> n=11, integers = 1 2 3 4 5 6 7 12 30 140 210 n=12, integers = 1 2 3 4 5 6 7 8 20 84 280 420 n=13, integers = 1 2 3 4 5 6 7 8 10 24 70 280 420
> 
> We stopped there.
> 
> Questions:
> Could someone (if this is of interest and not old hat) extend this array to, say, n = 100?
> And how could this array enter the OEIS?
> Carine said to me that the successive row-sums might be a possibility. We would then have this 11-term start:
> 
> S = 6, 12, 24, 60, 60, 60, 180, 420, 420, 840, 840,...
> (this succession is not in the OEIS)
> 
> Best,
> É.
> 
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> 
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