[seqfan] AN coprime to 210 and x+a sequence

[zbi74583.boat at orange.zero.jp]

    Hi,Seqfans


    [Amicable Number coprime to 210 and x+a sequence]
    In early time Mathrmaticians thought that no AN coprime to 6 exist.
    "Bratley and Mckay's  Conjecture"
    http://mathworld.wolfram.com/AmicablePair.html

    But, Te Riele, Battiato and Borho, Kohmoto found the counter example.
    After that I found AN coprime to 30.

    Still, no AN cooprime to 210 is known.

    I explain the method to solve this problem.

    [ How to compute AN coprime 210 ]
    Assume that x,y is AN coprime 210.

    Sigma(x)=Sigma(y)=x+y .... E0
    GCD(x,210)=1, GCD(y,210)=1

    Let x=c*u, y=c*v      .... E1
    GCD(c,x)=1, GCD(c,y)=1

    Generally for any number n, it is represented as follows.
    Sigma(n)=2^e*3^f*5^g*7^h*k
    where e,f,g,h are high.

    I think the probability  that n is of this form is high.

    So,
    Sigma(x)=2^e*3^f*5^g*7^h*k
    where e,f,g,h are high.

    From E0  nd E1
    c*(u+v) =2^e*3^f*5^g*7^h*k .... E2
    where e,f,g,h are high.

    x,y are coprime to 210.
    Hence c is coprome to 210.
    From E2
    u+v must be of the form as follows
    u+v = 2^e*3^f*5^g*7^h*k .... E3
    where e,f,g,h are high.

    u,v is a seed, so
    Sigma(u)=Sigma(v) .... E4

    we assume the easest form.
    Let u=p*q, v=r
    where p,q,r are prime

    From E4
    (p+1)*(q+1)=r+1
    r=p*q+p+q
    u+v=2*p*q+p+q

    From E3
    2*p*q+p+q=2^e*3^f*5^g*7^h*k .... E5
    where e,f,g,h are high.

    We consider about x+a sequense

    Take any prime p.

    p+(2*p+1)-p = 0 Mod 2*p+1
    2^e_1*x_1-p = 0 Mod 2*p+1
    Where 2^e_1 is the highest power of 2 dividing p+(2*p+1)

    2^e_1*(x_1+2*p+1)-p = 0 Mod 2*p+1
    2^e_1*2^e_2*x_2-p = 0   Mod 2*p+1
    Where 2^e_2 is the highest power of 2 dividing x_1+(2*p+1)
    ....
    Repeat
    ....
    If Sum_i e_i = e then End

    2^e*x_j-p = 0 Mod 2*p+1
    2^e*x_j-p = (2*p+1)*q
    q=(2^e*x_j-p)/(2*p+1)
    If q is prime then compute r
    r=(p+1)*q+p
    If r is prime then
    2^e*x_j = (2*p+1)*q+p

    It saisfies the part of prime 2 of E5

    If we do the same thing for 3,5,7 then we obtain the following
    2^e*3^f*5^g*7^h*x_m = (2*p+1)*q+p

    The explanation is rather complecated but program is easy.

    X=p, A=2*p+1, E=0
    X=X+A
    While X=0 Mod 2, X=X/2,E=E+1
    If E=e_0 Then Q=(2^E*X-p)/(2*p+1)
    If PrimeQ(Q)=1 Then R=(p+1)*Q+p
    If PrimeQ(R)=1 Then End

    p,Q,R is the solution of the following

    Sigma(p*q)=Sigma(r)
    p*q+r=2^E*X

    Ex
    p=11
    X=11, A=23
    X+A=34
    X=34/2=17, E=0+1
    X+A=40
    X=40/2^3=5,   E=1+3
    X+A=28
    X=28/2^2=7,   E=4*2

    Q=(2^6*7-11)/23=19
    19 is Prime
    R=12*19+11=239
    239 is Prime
    So
    Sigma(11*19)=Sigma(239)
    11*19+23 =2^6*7
    209,239 is a seed. Compute sprout.
    c=3^2*5*13
    x,y=3^2*5*13*{11*19,239}

    e in E5 is high hence c is coprime 2
    So, we are able to say
    e,f,g.h in E5 are high hence c is coprime to 210

    We consider about the maximal number of X.
    Assume that each X is devided by 2 only once.
    This is the maximal case.

    (x+a)/2
    ((x+a)/2+a)/2=(x+a)/4+a/2
    ((x+a)/4+a/2+a)/2=(x+a)/8+a/4+a/2
    ....
    ....
    Limit of the sequence X = a
    So, it must be periodic

    [x+a Sequens]

    s(n)=(s(n-1)+a)/p^m
    Where p^m is the highest power of p dividing (s(n-1)+a)

    To Neil :
    Tell me how to submit it to OEIS.
    It is dificult because the sequence has some ambiguous  of a and p.

    For example
    s(0)=11, a=23, p=2
    11,17,5,7,15,19,21,11,....
    3,13,9,1,3,....

    OEIS doesn't have both.
    But if it has the two sequences then it must have the sequences for all s(0)
and a and p.

    The number of them is aleph(0).



    Yasutoshi