20180702, 14:24  #1 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{6}×31 Posts 
July 2018

20180704, 17:14  #2 
Sep 2017
2·7^{2} Posts 
I can find a solution with 3 more years and 4 more, but 5 more seems impossible for ages <= 122 (this is the maximum I heard of any age: Jeanne Calment).
Did anyone managed to find a triplet that is obsucre and stays obscure for 5 more years? 
20180704, 18:08  #3 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{6}·31 Posts 
I thought there bonus was for 5 consecutive years.
5 more years can be interpreted as 6 or even 9 consecutive years. A solution for a total of 5 consecutive years is easy to find. I have not attempted to find anything further than that. 
20180705, 02:21  #4  
Jul 2015
3^{2} Posts 
Quote:
Is there any rules for the obscure triplets? 

20180705, 11:22  #5 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·2,909 Posts 
Am doing a search for a and b less than 1000 and c less than 2000, a < b < c
I have noticed that c is often 1.52x as big as b in the larger runs. My code is reaching its limits. I think I am going to have to rethink to go much further. I currently store all the tuples with common sums and products. For a b c and a' b' c' to have the same product and sum, I don't think max(c, c') can be prime if a < b < c and a' < b' < c'. Not sure this helps much. Last fiddled with by henryzz on 20180705 at 11:26 
20180705, 16:37  #6  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20180705 at 16:40 

20180706, 05:25  #7 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
1000111001000_{2} Posts 
I must be doing something wrong...
Taking every a,b,c from 1 to 100 I only get 98 of the 1,000,000 that are NOT obscure,
Every other triplet has at least one other triplet with the same sum and product. 
20180706, 06:01  #8  
"Rashid Naimi"
Oct 2015
Remote to Here/There
7C0_{16} Posts 
Quote:
That would mean that just about any triplet would be s solution to the bonus problem. I think that would be in conflict with what has been posted here so far. I did not run for loops for 1 to 100 but to a value smaller than 100 and found much less obscures than that. Then again, perhaps if I extend the loops to 100 steps, I would get the same results as you. But intuitively, I doubt that. FWIW. Last fiddled with by a1call on 20180706 at 06:02 

20180706, 08:46  #9 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·2,909 Posts 

20180706, 08:51  #10  
Jun 2003
2^{3}·607 Posts 
Quote:
With 1<=a<=b<=c<=100, I get 1 triplet the leads to 5 consecutive "obscures" (which is unfortunately not enough to earn a star), and 6 triplets with 4 consecutive "obscures"  although one of them is merely the second term of the 5er mentioned. With 1<=a<=b<=c<=500, I get 6 additional triplets with 5 terms (in addition to the one from above), but no 6 term ones yet. Incidentally, here are a few ones with 3 terms where c<=40. Code:
2:24:25 4:20:26 8:14:25 8:14:34 10:24:26 

20180706, 09:09  #11 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
1011010111010_{2} Posts 
c a decent amount larger was the trick that found it for me. Although my 1000 1000 2000 search found only one solution for 6 consecutive years it was overkill.
I think I have been using the wrong method storing all the obscure tuples. There are too many of them. The correct method is probably to go up through the products distributing factors. 
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