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Jul 19, 2017, 8:02:42 AM7/19/17

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I have noticed that out of 9 Heegner numbers:

{1,2,3,7,11,19,43,67,163}

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}]]

{1,2,3,7,19,43,67,163}

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}]]

{1,2,3,7,31,19,43,67,163}

(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?

Thanks,

Alexander R. Povolotsky

{1,2,3,7,11,19,43,67,163}

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}]]

{1,2,3,7,19,43,67,163}

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}]]

{1,2,3,7,31,19,43,67,163}

(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?

Thanks,

Alexander R. Povolotsky

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