Yes, Pascal found the same two pentagonal numbers I found:
*pn=.3 :'y*(1-~3*y)%2'*
* p=:pn>:i.5000x*
Paschal's solution: P(2166)-P(1019)
*2166 1019{p*
*7042750 1560090*
My solution:
*]n=.m#p2*
*1560090 7042750*
The question is, do these two pentagonal numbers have the smallest D?
Skip Cave
Cave Consulting LLC
On Sat, Jul 13, 2019 at 7:02 PM Louis de Forcrand <[email protected]> wrote:
> Quick correction: the answer given by Pascal is not 1019 but
> P(2166)-P(1019), and my email should use that instead of 1019 wherever it
> is mentioned.
>
> Sorry for the noise,
> Louis
>
> > On 14 Jul 2019, at 01:58, Louis de Forcrand <[email protected]> wrote:
> >
> > I’m with Skip here: how do y’all guarantee that your brute-force answers
> (which search through a list of the first few pentagonal numbers) actually
> return the pair with the smallest _difference_? How do you know there isn’t
> a pair outside your search range, where each number in the pair is much
> larger than any you searched but whose difference is smaller?
> >
> > One possibility is to find a lower bound B as a function of k for the
> difference P(n)-P(m), where n>m>k. If B(k) is larger than 1019 (the answer
> Pascal Jasmin gave) for k outside your search range, then 1019 is indeed
> the answer.
> > A possibility for B is B(k) = P(k+1)-P(k).
> >
> > Another idea I had is that since the sum and the difference must both be
> pentagonal, we can instead search in the same way for them and make sure
> their half-sum and half-difference (which equal the original numbers) are
> indeed pentagonal. The first pair we find will be the one we want since by
> then we’ll have checked all pairs with a smaller difference.
> >
> > Do you have a better justification?
> >
> > Thanks!
> > Louis
> >
> >> On 13 Jul 2019, at 01:31, Skip Cave <[email protected]> wrote:
> >>
> >> Another approach to Euler 44:
> >> NB. make a pentagonal number generator verb:
> >> *pn=.3 :'y*(1-~3*y)%2'*
> >>
> >> NB. Generate the first 5000 pentagonal numbers and store them in p. Then
> >> find all possible pair combinations of the first 5000 pentagonal numbers
> >> and store the pairs in p2. Then store the two integers in each pair in
> the
> >> vectors a & b respectively:
> >> * 'a b'=.|:p2=.(2 comb 5000){p=:pn>:i.5000x*
> >>
> >> NB. Find and mark all pairs where a+b and a-b are both pentagonal
> numbers.
> >> Store that mark vector in m. Also sum the marks in m to see how many
> >> solution pairs we have:
> >> * +/m=.(p e. ~a+b) *. p e. ~|a-b*
> >>
> >> *1*
> >>
> >> NB. So there is only one solution pair in the first 5000 pentagonal
> >> numbers. Now use the mark vector to extract that one pair of pentagonal
> >> numbers we found that meets all the criteria, store the pair in n, and
> >> display them.
> >> * ]n=.m#p2*
> >>
> >> *1560090 7042750*
> >>
> >>
> >> NB. List a, b, a+b, & a-b of the discovered pair in n=a,b:
> >>
> >> * (,n),(+/"1 n),(|-/"1 n) *
> >>
> >> *1560090 7042750 8602840 5482660*
> >>
> >>
> >> NB. Are the four integers a, b, a+b, & a-b all pentagonal numbers?
> >>
> >> *p e.~(,n),(+/"1 n),(|-/"1 n) *
> >>
> >> *1 1 1 1*
> >>
> >>
> >> NB. Yes. So what is "D" (the difference between the pair) which is
> required
> >> to get the final answer in the euler question?
> >>
> >>
> >> *]D=.|-/"1 n*
> >>
> >> *5482660*
> >>
> >>
> >> However the Euler 44 question wants to find a pentagonal pair where D is
> >> minimised, so we will need to expand our search so see if there are
> >> pentagonal pairs with a smaller D. Unfortunately, I run into memory
> limits
> >> when I try to examine a larger range of pentagonal numbers. So I'll
> need to
> >> break the one-shot approach into a loop and segment the array of
> pentagonal
> >> numbers (sigh). I'll leave that exercise for the reader...
> >>
> >>
> >> Skip Cave
> >> ----------------------------------------------------------------------
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> >
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