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Problem 333
Posted: Sun Apr 17, 2011 10:44 pm
by elr
i think that i miss something
the problem statement saying that :
"The sum of the primes q <100 such that P(q)=1 equals 233."
however my calculations saying that the sum is 630 :
Expand
5:3+2
7:4+3
13:9+4
17:9+8
31:27+4
41:32+9
43:27+16
47:27+12+8
59:32+27
61:27+18+16
71:36+27+8
73:64+9
79:36+27+16
83:81+2
i am wondering how a sum of 233 could be made ?
Re: Problem 333
Posted: Sun Apr 17, 2011 11:32 pm
by jaap
You have some false positives, for example:
31 = 27 + 4 = 9 + 6 + 16
41 = 9 + 32 = 27 + 6 + 8
and some missing ones, for example:
23 = 9 + 6 + 8
It looks like you are only using powers of 2 and powers of 3, and missing out terms like 6 which are neither.
Re: Problem 333
Posted: Tue May 03, 2011 4:50 am
by gsidebottom
I'm having problems with this one as well. In particular, my program is getting the sum of the primes q <100 such that P(q)=1 equals 490. Below are the primes less than 100 with unique partitions as far as I understand. Anybody see what my problem is?
Expand
2=2
3=3
5=3+2
7=4+3
13=9+4
17=9+8
23=9+8+6
43=27+16
59=32+27
61=27+18+16
71=32+27+12
89=81+8
97=81+16
Re: Problem 333
Posted: Tue May 03, 2011 5:36 am
by jaap
gsidebottom wrote:71=32+27+12
89=81+8
97=81+16
71 = 27 + 36 + 8
89 = 27 + 18 + 12 + 32
97 = 27 + 6 + 64
Re: Problem 333
Posted: Tue Jan 19, 2016 3:56 pm
by PhilLeTaxi
Hi,
for q < 10000, is 1404597 correct ?
Re: Problem 333
Posted: Sun May 21, 2017 5:05 pm
by vamsikal3
<deleted post>
Re: Problem 333
Posted: Sun May 21, 2017 7:59 pm
by hk
Confirmed.
Re: Problem 333
Posted: Mon May 22, 2017 4:28 am
by vamsikal3
<deleted post>
Re: Problem 333
Posted: Wed Mar 28, 2018 7:24 pm
by Jochen_P
Resuming the problems after a long break.
My terms for q <100 are
[2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96]
my result for P(q)=1 for q<100 is 228 (instead of 233)
Has anybody any clue where I'm going wrong?
My primes with P(q) = 1 are
5
7
13
17
23
43
59
61
Re: Problem 333
Posted: Wed Mar 28, 2018 7:46 pm
by traxex
Jochen_P wrote: Wed Mar 28, 2018 7:24 pm
my result for P(q)=1 for q<100 is 228 (instead of 233)
Perhaps you are missing two small primes that sum to 5?

Re: Problem 333
Posted: Wed Mar 28, 2018 8:09 pm
by Jochen_P
Aye, thanx a lot!
I kind of left them out by purpose.. thought once again incorrectly that a partition must consist of n>1 elements. m(
Now to make this bugger efficient. Runs way too long to get P(q)=1 for q<10**6
Re: Problem 333
Posted: Tue Apr 03, 2018 8:48 pm
by Jochen_P
... way too long to get even sum P(q)=1 for q<10**3.
4600 in a bit more than an hour m(
Can't think of any other good way to optimize that anymore, there are way too many partitions to check.
There has to be a simple rule/formula for excluding terms in this, but damn, do I suck at maths

Re: Problem 333
Posted: Thu Jan 27, 2022 7:29 pm
by Col_Kernel
Is it possible that "Confimed." above refers to something in the deleted post? First I believed it refers to the preceding question for q < 10000 and started to debug my algorithm. But I get 24853 for q < 10000 (and my answer for 10^6 got accepted

).