The Art of Thinking

The lazy integer application  Let's apply formula log(w)  =  2 * Sum( ((w-1)/(w+1))^(2*n-1) / (2*n-1)  : n = 1 2 3 ...) for every real w > 0; see https://whmth.blogspot.com/2024/05/the-logarithmic-function.html EXAMPLES For  w := 2  we obtain         log(2)  =  2 * ( 1/3  +  1/(3*3^3)  +  1/(5*3^5)  +  1/(7*3^7)  + ... ) For  w:=3  we obtain         log(3)  =  2 * ( 1/2  +  1/(3*2^3)  +  1/(5*2^5)  +  1/(7*2^7)  +  ... ) For  w:=4  it's more efficient to apply log(4) = 2*log(2) The same goes for all composite integers: log(6) = log(2)+log(3)        log(8) = 3*log(2)        log(9) = 2*log(3) etc. Thus, let's concentrate on primes. We obtain log(5)  =  2 * ( 2/3  +  (2/3)^3 / 3  +  (2/3)^5 / 5  +  (2/3)^7 / 7  +  .....

The shifted logarithm, and the logarithmic function  Let  s t  be two arbitrary positive real numbers. The shifted logarithm from  s  to  t, symbolically LOG(s t),  is the (oriented) area of a curvilinear  trapezoid in plane xy that is bounded by the graph of function  y=1/x, and by the x-axis, and which stretches between parallel ("vertical") lines  x=s and x=t -- "oriented" means that this area is the customary geometric area when  s < t, and it is the negative of that geometric are when  s > t. Of course,  LOG(s t) = 0  when s=t. Thus, LOG(t s)   =   - LOG(s t) Now we define the logarithmic function log(t) := LOG(1 t) for arbitrary positive real number  t.  In particular: log(1) = 0 Obviously, LOG(q  s) + LOG(s  t)  =  LOG(q  t) for arbitrary positive real numbers  q s t. Logarithm as a translation from multiplication to addition Let lengths of the s...

Introduction A (2 3 5 7)-fraction is any fraction  (n+1)/n  such that  n  is a natural number (1 2 3 ...), and  n*(n+1) has no prime divisor > 7. Thus (2 3 5 7)-fractions form a subset of all near 1 fractions, see   https://whmth.blogspot.com/2024/05/near-1-fractions-and-computing-log2.html Earlier, in   https://whmth.blogspot.com/2024/05/team-2-3-5.html I listed all (2 3 5)-fractions, i.e. all fraction  (n+1)/n  such that  n  is a natural number (1 2 3 ...), and  n*(n+1) has no prime divisor > 5.  Obviously, the (2 3 5)-fractions form a subset of the set of all (2 3 5 7)-fractions. We saw that there are exactly 10 different (2 3 5)-fractions. A much more extensive table of near 1 fractions was included in a paper by D.H.Lehmer, see   https://whmth.blogspot.com/2024/05/dhlehmers-extensive-table-of-near-1.html The table of the (2 3 5 7)-fractions Either the numerator ot the denominator (or both) of a (2 3 5 ...

 To be exact, D.H.Lehmer ( in his paper received for publication on July 25, 1962 ) listed all integers  N := n*(n-1) such that integer  n>1, and no prime divisor of  N  is larger than 41. This of course is equivalent to studying near 1 fractions  n/(n-1) where n>1  is an integer. Actually, Lehmer presented also the prime decompositions of the related near 1 fractions  n/(n-1), for  n > 100000. The title of the mentioned D.H.Lehmer paper is: ON A PROBLEM OF STORMER. Furthermore, in addition to a theoretical discussion, Lehmer privided similar lists of integers  n*(n-a), where a = 2 3 4. D.H.Lehmer and Emma Lehmer were early pioneers of applying computers to Number Theory.

Introductions  The material below is original. I have created it in the year 1972. Originally, I applied my method right away to " team   2 3 5 7" while below I consider the simpler case, namely " team   2 3 5". Near 1 fractions In order to compute all three  log(2)  log(3)   log(5)   we will need three different near 1 fractions  (a+1)/a  such that  a  is a natural number, and  a*(a+1) is divisible by no prime different from  2 3 5. (Such three fractions should be in a sense independent, though they tend to be so). Thus, let's find all such near 1 fractions  (a+1)/a. Then we will select the three fractions with the largest denominators. It's clear that numerator  a+1  or  denominator  a  must be a power of  2  or. 3  or  5  while the "other floor" of the fraction can be divisible only by the other two (or one or none) of the primes  2  3  5, a...

We will pass from two fractions to three fraction. Let's start with: 5   =   (3/2)^4  *  (81/80)^(-1)  Fraction  3/2  is not impressive. A modest step toward greater efficiency (of computing log(5)) can be: 3/2  =  (5/4) * (6/5) Also, 5/4  =  (6/5) * (25/24) hence 3/2  =  (6/5)^2 * (25/24) and 5   =   (6/5)^8 * (25/24)^4 * (81/80)^(-1) Isn't this pretty good? -- maybe. The next note will present a better result, it will do so for all there integers 2 3 5 simultaneously.