• Heegner numbers's related issue

    From apovolot@gmail.com@21:1/5 to All on Wed Jul 19 06:02:38 2017
    I have noticed that out of 9 Heegner numbers:
    the group of four leftmost of them and the group of four rightmost of
    them could be constructed together on the similar recurrence principle:

    a[n] = 5*(5*(EuilerPhi[((a[n-1]+a[n-5])/5 + 11)/5] + Prime[n-5]) - 11)
    - a[n-4]

    Simplify[RecurrenceTable[{a[n] ==5*(5*(EuilerPhi[((a[n-1]+a[n-5])/5 +
    11)/5] + Prime[n-5]) - 11) - a[n-4],a[1]==1,a[2]==2,a[3]==3,a[4]==7, a[5]==19},a[n],{n, 1, 8}]]


    Of course above recurrence is not much productive since it doesn't
    cover the fifth Heegner number and also requires five (out of total
    eight covered Heegner's numbers) to be given in the recurrence's
    initial conditions.

    Unfortunately RSolve (see below) is not capable to find the explicit
    direct solution for the above recurrence

    $Assumptions = n \[Element] Integers

    RSolve[{a[n]==-a[-4+n]+5(-11+5EulerPhi[1/5(11+1/5(a[-5+n]+a[-1+n]))]+Pri me[-5+n]),a[1]==1,a[2]==2,a[3]==3,a[4]==7,a[5]==19},a[n],n]

    However, I, seems to be, was able to figure out "by hand" some sort of
    explicit direct solution, which gives four leftmost (as a[1], a[2],
    a[3] and a[4]) and the four rightmost (as a[6], a[7], a[8] and a[9])
    Heegner numbers:

    a[n] =
    EulerPhi[Prime[Mod[4,n]!!]]+Floor[n/5]+((1+Floor[n/5])^2)!*((1+Sqrt[3])^ (n-(1+2*(Floor[n/5]))!)-(1-Sqrt[3])^(n-(1+2*(Floor[n/5]))!))/(2*Sqrt[3])

    Simplify[Table[EulerPhi[Prime[Mod[4,n]!!]]+Floor[n/5]+((1+Floor[n/5])^2)! *((1+Sqrt[3])^(n-(1+2*(Floor[n/5]))!)-(1-Sqrt[3])^(n-(1+2*(Floor[n/5]))!) )/(2*Sqrt[3]), {n,1,9}]]


    (Note that unfortunately a[5]=31 in above is "off" - the actual Heegner
    number for index 5 should be not 31 but 11 ...)

    and as follows:

    a[n-1] = EulerPhi[Prime[Mod[4,(n-1)]!!]]+Floor[(n-1)/5]+((1+Floor[(n-1)/5])^2)!*( (1+Sqrt[3])^((n-1)-(1+2*(Floor[(n-1)/5]))!)-(1-Sqrt[3])^((n-1)-(1+2*(Flo or[(n-1)/5]))!))/(2*Sqrt[3])

    a[n-4] = EulerPhi[Prime[Mod[4,(n-4)]!!]]+Floor[(n-4)/5]+((1+Floor[(n-4)/5])^2)!*( (1+Sqrt[3])^((n-4)-(1+2*(Floor[(n-4)/5]))!)-(1-Sqrt[3])^((n-4)-(1+2*(Flo or[(n-4)/5]))!))/(2*Sqrt[3])

    a[n-5]= EulerPhi[Prime[Mod[4,(n-5)]!!]]+Floor[(n-5)/5]+((1+Floor[(n-5)/5])^2)!*( (1+Sqrt[3])^((n-5)-(1+2*(Floor[(n-5)/5]))!)-(1-Sqrt[3])^((n-5)-(1+2*(Flo or[(n-5)/5]))!))/(2*Sqrt[3])

    Would it be possible to combine (utilize) above explicit and recurrent
    formulas (please keep in mind that indexes for the last four Heegner
    numbers are off by one if to compare explicit and recurrent formulas)
    towards deriving (either explicit or recurrent) formula which would
    cover all nine Heegner numbers?

    Alexander R. Povolotsky

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)