[seqfan] Re: A new sequence in Dutch magazine "Pythagoras"
njasloane at gmail.com
Sat Dec 3 16:56:17 CET 2016
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
as a plain .txt file (Richard, I hope that is OK).
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>
> In response to http://list.seqfan.eu/pipermail/seqfan/2016-
> njas> By the way, what about the other new sequences in the same
> 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
> 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
> 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
> Also 1/17= 0.bar(0588235294117647) has period 16. Then the fraction
> =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
> You see: the number of 9's in the denominator equals the period of the
> In addition we should have
> 1/17 = 588'235'294'117'647/9'999'999'999'999'999
> and this fits in, because 7*1542857 =999'999 and
> 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
> that has a factor 7 as well as a factor 17? Easy: because of the second
> 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
> sequences is the 9-number with lcm(6,16)=48 digits. By the second property
> 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.
> 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
> Write this up with Newton's binomial expansion, (a+b)^M.
> We now have
> All terms on the right hand side have a factor P^2, so 10^(PN)-1 is a
> 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
> is a smaller 9-number with factor P^2. This 9-number then has the form
> (because all these 9-numbers have one or more factors) with A<P. In the
> manner as in the previous article happened for prime factor P one may
> that only the 9-numbers 10^(A*N)-1, 10^(2*A*N)-1, 10^(3*A*N)-1 have one or
> 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,
> 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
> that P is the product of two integers, where P is a prime number. This
> 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,
> the first 9-number with two factors 7 is 10^(6*7)-1=10^42-1. The period of
> 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
> 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
> 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
> than two distinct prime factors. Then you need to find the smallest
> 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
> with period lcm(16,28,46) =2576.
> You may check yourself the calculation of the fraction with online
> 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
> 1/(17*29*47) key in:
> 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
> 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,...
> 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
> 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
> 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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