Is there any significance in such Heegner numbers (or class number 1) representation symmetry?

166 Views Asked by At

$$\mathrm{A003173}(n) = 1+\frac{(1 + \sqrt{3})^{n-1} - (1 - \sqrt{3})^{n-1}}{2\sqrt{3}}$$

for $n = 1,2,3,4$ .

$$\mathrm{A003173}(n) = 19+24\frac{(1 + \sqrt{3})^{n-6} - (1 - \sqrt{3})^{n-6}}{2\sqrt{3}}$$

for $n = 6,7,8,9$ .

In general the recursive form of generalization:

$$\mathrm{A003173}(n) = \mathrm{A003173}(k) + (\mathrm{A003173}(k+1)-\mathrm{A003173}(k))\frac{(1 + \sqrt{3})^{n-k} - (1 - \sqrt{3})^{n-k)}}{2\sqrt{3}}$$

Where $k=1$ for $n = 1,2,3,4$ and $k=6$ for $n = 6,7,8,9$ .

Below is the explicit form of generalization:

$$\mathrm{A003173}(n) = \phi(prime((4\bmod n)!!)) + \left\lfloor n/5 \right\rfloor + ((1 + \left\lfloor n/5 \right\rfloor)^2)!\frac{(1+\sqrt{3})^{n-(1+2\left\lfloor n/5 \right\rfloor)!}-(1-\sqrt{3})^{n-(1+2 \left\lfloor n/5 \right\rfloor)!}}{2\sqrt{3}}$$

For all $n$ except $n=5$ .

In raw Mathematica (WolframAlpha) format the above formula is:

A003173(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))

Note that all of above covers first four and last four Heegner numbers (eight out of a total of nine) but doesn't cover the middle term: $\mathrm{A003173}(5)= 11$.

Here are other Mathematica expressions for first and last 4 Heegner Numbers

Simplify[Table[1 + ((1 + Sqrt[3])^(n - 1) - (1 - Sqrt[3])^(n - 1))/(2 Sqrt[3]), {n, 1, 4}]]

{1,2,3,7}

Simplify[Table[19 + 24((1 + Sqrt[3])^(n - 6) - (1 - Sqrt[3])^(n - 6))/(2 Sqrt[3]), {n, 6,9 }]]

{19,43,67,163}

Simplify[Table[6*j^2-5 + ((j^2)!)*((1 + Sqrt[3])^(n-1) - (1 - Sqrt[3])^(n-1))/(2*Sqrt[3]), {j,1,2}, {n,1,4}]]

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

Is there any significance in such Heegner numbers (or class number $1$) representation symmetry?