[seqfan] Re: A new sequence in Dutch magazine "Pythagoras"
Neil Sloane
njasloane at gmail.com
Sat Dec 3 16:56:17 CET 2016
Arie, Richard:
Thank you very much for translating the article by Arnout Jaspers
in the April issue of Pythagoras. Especially Richard - you translated the
whole article, which was a lot of work: the article is quite long.
I have now created an entry for that sequence, see A278588.
Of course comments and improvements are welcomed.
In fact there is a conjecture about the numbers
on the main diagonal which appears to be an unsolved problem.
Of course if any English speaker wants to work on this they will need a
translation of the article,
so I did a bit of editing of Richard's translation, and added it to the
entry
as a plain .txt file (Richard, I hope that is OK).
Best regards
Neil
Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com
On Fri, Dec 2, 2016 at 8:59 AM, Richard J. Mathar <mathar at mpia-hd.mpg.de>
wrote:
> In response to http://list.seqfan.eu/pipermail/seqfan/2016-
> December/017115.html
>
> njas> By the way, what about the other new sequences in the same
> Pythagoras
> njas> article? See "Andere Rijen" - can someone explain what they are?
> (And then
> njas> they too should be added to the OEIS.)
> njas> (Here is the link again: http://www.pyth.eu/jaargangen/Pyth55-6.pdf
> See
> njas> page 21)
>
> Two of the series on page 21 are just a notice of the editor that some
> submitters forgot to include some members; after correction
> they are actually A016105 and A175746.
>
> To understand the remaining sequence on page 21 of PYTHAGORAS_JG55_No6.pdf
> one probably needs to read the article starting on page 26 of the earlier
> issue
> of the PYTHAGOROS_JG55_no5.pdf, which contains 21, 33, 57,.. as a table
> read down columns: Here is a rough translation:
>
> (by Arnout Jaspers)
> Creating mega-periodic Fractions
>
> If you write 1/7 as a decimal number, you get 0.bar(142857).
> The bar(..) above the 6 digits means that the group
> is repeated indefinitely. We say: the period of 1/7 has a period 6.
>
> In the article "The period of a decimal number" in the January issue
> we stated (without proof)
>
> Take two prime numbers a and b. If their inverses 1/a and 1/b, written
> as decimals, have periods p_a and p_b, the product 1/(a*b)
> has period lcm(p_a,p_b). Here lcm means the least common multiple.
>
> With the aid of 9-numbers of the previous article the proof is easy. We
> show it by example, that is a=7 and b=17. We observed that 1/7 has period
> 6.
> Also 1/17= 0.bar(0588235294117647) has period 16. Then the fraction
> 1/(7*17)
> =1/119 should have, written as decimals, have period lcm(6,16)=48.
>
> Earlier we showed that each periodic fraction may be written as a fraction
> with a 9-number in the denominator:
> 0.bar(142857) = 142'857/999'999
> and
> 0.bar(0.588235294117647)=588'235'294'117'647/9'999'999'999'999'999.
>
> You see: the number of 9's in the denominator equals the period of the
> decimal
> number.
> In addition we should have
> 1/7=142857/999'999
> and
> 1/17 = 588'235'294'117'647/9'999'999'999'999'999
> and this fits in, because 7*1542857 =999'999 and
> 17*588'235'294'117'647=9'999'999'999'999'999.
>
> In fact the period of a fraction (written as a decimal): 999'999 is
> the smallest 9-number with a factor of 7, and the smallest 9-number with
> a factor of 17 is 9'999'999'999'999'999.
>
> The writeup of a periodic fraction as a simple fraction with a 9-number
> in the denominator now demands:
> (*) Vor 1/(7*17), the product of 1/7 by 1/17, we need a 9-number in
> the denominator with a factor of 7 as well as a factor of 17, so we may
> write 1/119 = 1/(7*17)=(some integer number)/(9-number with factor 7 and
> factor 17).
>
> Now we use two properties of 9-numbers that we already proved:
> - a 9-number of P-1 digits has at least one factor P (if P is neither 2
> nor 5)
> - if the 9-number with N digits is the smallest 9-number with P as factor,
> then all the 9-numbers N, 2N, 3N.. have one or more factors P.
>
> Applied to our example this means we already know that 999'999 must contain
> a factor 7 and 9'999'999'999'999'999 a factor 17. How do we construct a
> 9-number
> that has a factor 7 as well as a factor 17? Easy: because of the second
> property
> all 9-numbers with 6, 12, 18... digits contain a factor 7, and all
> 9-numbers with
> 16, 32,.. digits contain a factor 17. The smallest 9-numbers that is in
> both
> sequences is the 9-number with lcm(6,16)=48 digits. By the second property
> these
> two sequences are also the *only* 9-number with the factors 7 and 17, and
> therefore the 9-number with 48 digits is indeed the smallest possible one.
>
> This means the faction 1/119 is written as (a integer with 48
> digits)/(9-number with 48 digits).
> In decimal writeup this is a fraction with period 48, i.e.
> 1/19=0.bar(0084033613....521)
>
> Quadratic base fractions.
>
> If a=b this rule cannot hold, as lcm(a,a)=a.
> We return to the requiremnt writen in (*), but in view of the case a=b,
> with a=7 as example:
>
> (*) Vor 1/(7*7) we need a 9-number in the denmoinator with two factors 7,
> so we can write 1/49=1/(7*7) = (an integer)/(9-number with two factors 7).
>
> We already know that 1/7 has period 6, so the 9-numbers with 6, 12, 18...
> digits all have a factor 7. But how does another factor 7 join? In general:
> if the smallest 9-number with a factor P with N digits is 10^N-1, what is
> the smallest 9-numbers with a factor P^2?
>
> As 10^N-1 has a factor P, it is a multiple V of P, so we can write
> 10^N-1=V*P
> whence
> 10^N=V*P+1
> and
> 10^(P*N)=(10^N)^P=(V*P+1)^P.
> Write this up with Newton's binomial expansion, (a+b)^M.
> (VP+1)^P=(VP)^P+P*(VP)^(P-1)+...+P(P-1)*(VP)^2+P(VP)+1.
> We now have
> 10^(PN)-1=(VP)^P+P*(VP)^(P-1)+...+P(P-1)*(VP)^2+P*(VP).
> All terms on the right hand side have a factor P^2, so 10^(PN)-1 is a
> 9-number
> with P*N digits that is divisible by P^2, precisely what we looked for.
>
> But: is 10^(PN)-1 the *smallest* 9-number with factor P^2? Assume that
> there
> is a smaller 9-number with factor P^2. This 9-number then has the form
> 10^(A*N)-1
> (because all these 9-numbers have one or more factors) with A<P. In the
> same
> manner as in the previous article happened for prime factor P one may
> proove
> that only the 9-numbers 10^(A*N)-1, 10^(2*A*N)-1, 10^(3*A*N)-1 have one or
> more
> factors P^2. In general: each 9-number 10^(Q*A*N)-1, with Q=1,2,3... has a
> factor P^2.
> But we have not shown that in each case 10^(P*N)-1 has a factor P^2,
> therefore
> that must show up in the sequence of 9-numbers. Is this possible?
> This may only occur as P*N=Q*A*N, therefore with P=Q*A. But here that
> states
> that P is the product of two integers, where P is a prime number. This
> cannot
> be, therefore there is no smaller 9-number with factor P^2 than 10^(P*N)-1.
>
> Back to the example 1/7. This fraction has period 6, so the denominator is
> the 9-number with 6 digit, 10^6-1. We like to add a factor 7, so P is 7,
> thus
> the first 9-number with two factors 7 is 10^(6*7)-1=10^42-1. The period of
> 1/(7*7)
> should be 42. Indeed
> 1/49 = 0.(020408...51).
>
> Base fractions with more than two prime factors.
> Vor small prime factors a the period fo 1/a is already the maximum, a-1.
> So 1/7 has period 6 and 1/17 period 16. That does not need to be always
> correct:
> 1/13 has period 6. Vor a larger prime G one cannot predict whether the
> inverse 1/G
> has the maximum period length G-1; this becomes more unlikely as G
> increases.
>
> This happens only as G appears in the (G-1)st 9-numbers as a prime factor.
> Look for example at G of approximately 1 million. Then a prime factor G
> has approximately a 1-millionth higher change to be earlier in the
> sequence of 9-numbers. The larger the 9-numbers get, the larger the
> probability that the number of prime factors (on the average) increases
> as the 9-number increases.
>
> So if you wish to construct a fraction with small numbers in the
> denominator, but a period as long as possible, you need to choose a small
> prime number P with longest possible period P-1. The fraction 1/P^k
> then has a period (P-1)*P^(k-1).
>
> All the prvious remarks are also valid for fractions 1/(a*b*c....) with
> more
> than two distinct prime factors. Then you need to find the smallest
> 9-number
> with factors a,b,c.., and this has lcm(p_a,p_b,p_c,...) digits.
>
> So periods may increase swiftly. For example 1/(17*29*47) = 1/23171 is a
> frctio
> with period lcm(16,28,46) =2576.
> You may check yourself the calculation of the fraction with online
> calculators
> for large numbers, for example https://defuls.ca/big-number-calculator.htm
> .
> This - and other calculators - don't show digits after the dot; so even
> a fraction 1/7 is shown as 0. But you may patch this simply: for the
> fraction
> 1/(17*29*47) key in:
> 10^10000/(17*29*47)
> and you get the answer
> 431 573 950 196 366 ...
> You may check that the first 2576 digits are indeed the period, by search
> with 'control-F' (Windows) or 'cmd-F' (Mac) for the first digits, then for
> '431 573 950' (dont forget the blanks). You find the block again after
> 2576,
> 5152 and 7728 digits, but nowhere alse.
>
> Product of arbitrary periodic fractions..
>
> Finally we look at an more general case: take two arbitrary periodic
> fractions, as 0.bar(523) and 0.bar(3054478), and multiply these. What is
> the period length of the product? We note
>
> 0.bar(523)*0.bar(3054478) = 523/999 * 3054478/9999999
> and search for a 9-number such that
> 523/9999 * 3054.378/9999999 = (an integer)/(a 9-number). (**)
> Note: the product of two 9-numbers is not itself a 9-number (Observe:
> (10^N-1)(10^M-1) = 10^(N+M)-10^N-10^M+1, which may not be written as 10^R-1
> for some R. Otherwise we could have simply calculated 999*9999999, put
> this 9-number in the denominator, 523*3054478, and that kills it.
>
> To find the 9-number in the denominator, we decompose both 9-numbers
> left of the equal sin in factors:
> 999=3^3*57 and 9999999=3^2*239*4649.
> To write the product as a fraction with a 9-number in the denominator
> we need a 9-number with the factors 3^5, 37, 239 and 4649. In the list
> of factorization of 9-numbers (see pages 25-26 in the January issue)
> we see that 239 and 4649 occur in the 9-numbers ith 7 and 14 digits,
> thus also in those with 21, 28...digits. 37 appears in the 9-numbers
> with 3, 6,9.. digits. Because lcm(3,7)=21, 239, 4649 and 37 occur in the
> 9-numbers with 21, 42, 63,...digits.
> With the factor 3 we have not to deal differently because it occurs
> in both denominators, comparable to what happens if you multiply
> two identical reduced fractions. We said that if a 9-number with
> N digits has a factor P, the 9-number with N*P ciffers has factor P^2.
> This concludes that a 9-number with N*P^2 ciffers has a factor P^3,
> and so forth.
>
> All 9-numbers are divisible by 9, so they have a factor 3^2. So the
> 9-numbers with 3, 6,9,... digits have a factor 3^3, the 9-numbes with
> 9,18,27.. digits have a factor 3^4, and the 9-numbers with 27, 54, 81,...
> digits
> have a fctor 3^5. The smallest 9-number with all the factors needed,
> 3^5, 37, 239 and 4649, therefore has lcm(3,7,27)=189 digits.
>
> Now look again at the expression (**). The denominator, the 9-number
> with 189 digits, has the factors 3^5, 37, 239 and 4649 and others, which
> we symbolically replace by [others]. The numerator is the
> number 523 *3054748 * [others]. Therefore
> 523/999 * 3054478/9999999 = 523*3054*478*others / (3^5*37*239*4649*others)
> [others] carries the minimum of factors that is needed to pump up the
> factors 3^5, 37, 239 and 4649 to a 9-number. Conclusion: the product
> of a decimal fraction with period 3 and a decimal fraction with period 7
> has at most period 189.
>
> Note: We write 'at most' because with an arbitrary fraction with 999
> in the denominator we cannot exclude that this may be reducable to
> a smaller fraction, and likewise for a fraction with 9999999 in the
> denominator. Example: each 9-number is divisible throuh 9 (therefore
> also by 3), so if you can divide the numerator through 9 (or 3), such
> a fraction can be reduced. All 9-numbers with an even number of digits
> are divisible through 11, so if you can divide the numerator through 11,
> the fraction can also be simplified. In that case the 9-number in the
> denominator of the product of the two periodic fractions must have a
> smaller
> number of factors, and may be replaced by a smaller 9-number. If you
> chose for both numerators prime numbers, this is exculded and you
> get a maximum period.
>
> This way you may step by step compute the maximum period of fractions
> with length 2,3,4.. This is summarized in the table on the bottom of the
> page.
> This table shows that the product of two periodic fractions with the
> same period delivers much longer periods than a product of two
> periodic fractions with different periods.
>
> This means for example, that if you chose two prime numbers of just 3
> digits, say 359 and 617, the product 0.bar(359)*0.bar(167) has period 2997.
> For two periodic fractions with period 5 you already arrive at a maximum
> period of 499995, close to half a million! Prime numbers with 5 digits
> are plentiful; lists of these are on the internet. The computer mentioned
> earlier may handle numbers to half a million digits, so you can create
> fractions with these enormous periods by yourself.
>
> At last: the table suggests that the maximum length for two fractions
> with period N equals N*(10^N-1). Whether this is true we leave as
> an open problem to the reader...
>
> 1 2 3 4 5 6 7
> ----------------------------------
> 1 9 18 27 36 45 54 63
> 2 198 54 396 90 594 126
> 3 2997 108 135 5994 189
> 4 39996 180 3564 252
> 5 499995 270 351
>
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