Nouvelle idée sur les nombres premiers

Capture du 2014-11-22 09:29:05

Quand dans la liste suivant on obtient un nombre à partir de n
La colonne B p=((n+1)*2-1)*2+1
Les nombres multiples de trois et cinq peuvent être facilement écartés
dans la colonne D
J’ai découvert pourquoi le nombre treize porte malheur
Pour chaque nombre premier trouvé on a son multiple de treize dans la liste de
La colonne B
Par exemple pour 7 on a 7*13=91
Et à partir de 91 tous les 7 cellules plus bas on a un multiple de 7.
Idem pour les autres nombres premiers une fois trouvé le nombre p*13, tous les p cellules on a un multiple de p

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3 réflexions au sujet de « Nouvelle idée sur les nombres premiers »

  1. A very little bit of algebra serves to prove a lot about your function, but the function may tell us nothing about prime numbers. Consider:
    Let f(n)=4n+3. (This is your column B function.)
    We can easily prove:

    If n is a multiple of 3, so is f(n) also.

    If n-3 is a multiple of 5, so is f(n) also.

    If n+4 is a multiple of 13, so is f(n) also.
    -Let n=13m-4. Then f(n)=13*(4m+1). This is easy to prove in two lines of algebra.
    –Let p=4m+1 and use the same n as above. Then f(n+kp)=p*(13+4k). For instance, in your table, starting at 39, every third cell you get 3*13=39, 3*17=51, 3*21=63, 3*25=75 et cetera. This is also easy to prove in two lines of algebra.

    n=m+k*f(m) for some positive integers k and m, then f(n) is a multiple of f(m). (This appears to be your main claim.) Again, two lines of algebra is enough to show that f(n)=f(m)*(4k+1). The above statements about 13 follow from this since clearly 13 is a number of the form 4k+1.
    –For your example, if m=5, then f(m)=23. If k=1, then m+k*f(m)=5+1*23=28. Then f(n)=f(28)=23*(4*1+1)=23*5=115.

    What this boils down to is that you’ve reconstructed the Sieve of Eratosthenes on numbers of the form 4n+3. Which is neat I guess. I bet you could do something similar for numbers of the form 4n+1 too.

  2. Actually, the last claim above (the most general one) can be generalized further: it holds for all linear functions with natural coefficients. Here’s the proof:

    Let f(x)=m*x+b be a function over the natural numbers where m and b are natural numbers.

    We want to show that f(n+k*f(n)) is also divisible by f(n) for all naturals k.

    f(n+k*f(n))=
    f(n+k*(m*n+b))=
    f(n+k*m*n+k*b)=
    m*(n+k*m*n+k*b)+b=
    m*n+k*m²*n+k*m*b+b=
    (k*m²*n+k*m*b)+(m*n+b)=
    k*m*(m*n+b)+(m*n+b)=
    (m*n+b)(k*m+1)=
    f(n)*(k*m+1)

    Since k and m are natural numbers, the result is a positive multiple of f(n), as was desired. In the particular case you described and I evaluated in the prior post, m=4, but you could have picked any arbitrary slope and it would have worked out. That’s the nice thing about linear functions.

    • Thanks a lot for your demonstration. It’s very clear for me now. So in my sort of sieve in the article there is other multiple 4k+1 as you say first.

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