Heegner number

Euler's prime-generating polynomial

$$n^2\; +\; n\; +\; 41,$$

which gives (distinct) primes for n = 0, ..., 39, is related to the Heegner number 163 = 4 · 41 − 1.

RabinowitschRabinovitch, Georg [https://babel.hathitrust.org/cgi/pt?id=miun.aag4063.0001.001;view=1up;seq=420 "Eindeutigkeit der Zerlegung in Primzahlfaktoren in quadratischen Zahlkörpern."] Proc. Fifth Internat. Congress Math. ( Cambridge) 1, 418–421, 1913. proved that

$$n^2\; +\; n\; +\; p$$

gives primes for $n=0,\backslash dots,p-2$ if and only if this quadratic's discriminant $1-4p$ is the negative of a Heegner number.

(Note that $p-1$ yields $p^2$, so $p-2$ is maximal.)

1, 2, and 3 are not of the required form, so the Heegner numbers that work are 7, 11, 19, 43, 67, 163, yielding prime generating functions of Euler's form for 2, 3, 5, 11, 17, 41; these latter numbers are called lucky numbers of Euler by F. Le Lionnais.Le Lionnais, F. Les nombres remarquables. Paris: Hermann, pp. 88 and 144, 1983.

Ramanujan's constant is the transcendental number{{MathWorld|title=Transcendental Number|urlname=TranscendentalNumber}} gives $e^\{\backslash pi\backslash sqrt\{d\}\},\; d\; \backslash in\; Z^*$, based on

Nesterenko, Yu. V. "On Algebraic Independence of the Components of Solutions of a System of Linear Differential Equations." Izv. Akad. Nauk SSSR, Ser. Mat. 38, 495–512, 1974. English translation in Math. USSR 8, 501–518, 1974.

$e^\{\backslash pi\; \backslash sqrt\{163\}\}$, which is an almost integer:[http://mathworld.wolfram.com/RamanujanConstant.html Ramanujan Constant – from Wolfram MathWorld]

$$e^\{\backslash pi\; \backslash sqrt\{163\}\}\; =\; 262\backslash ,537\backslash ,412\backslash ,640\backslash ,768\backslash ,743.999\backslash ,999\backslash ,999\backslash ,999\backslash ,25\backslash ldots\backslash approx\; 640\backslash ,320^3+744.$$

This number was discovered in 1859 by the mathematician Charles Hermite.{{cite book

| last = Barrow

| first = John D

| title = The Constants of Nature

| publisher = Jonathan Cape

| page = 72

| year = 2002

| location = London

| isbn = 0-224-06135-6 }}

In a 1975 April Fool article in Scientific American magazine,{{cite journal

| last = Gardner

| first = Martin

| title = Mathematical Games

| journal = Scientific American

| volume = 232

| issue = 4

| page = 127

| date = April 1975

| publisher = Scientific American, Inc | doi = 10.1038/scientificamerican0475-126

| bibcode = 1975SciAm.232d.126G

}}

"Mathematical Games" columnist Martin Gardner made the hoax claim that the number was in fact an integer, and that the Indian mathematical genius Srinivasa Ramanujan had predicted it – hence its name. In this wise it has as a spurious provenance as the Feynman point.

This coincidence is explained by complex multiplication and the q-expansion of the j-invariant.

In what follows, j(z) denotes the j-invariant of the complex number z. Briefly, $\backslash textstyle\; j\backslash left(\backslash frac\{1+\backslash sqrt\{-d\}\}\{2\}\backslash right)$ is an integer for d a Heegner number, and

$$e^\{\backslash pi\; \backslash sqrt\{d\}\}\; \backslash approx\; -j\backslash left(\backslash frac\{1+\backslash sqrt\{-d\}\}\{2\}\backslash right)\; +\; 744$$

via the q-expansion.

If $\backslash tau$ is a quadratic irrational, then its j-invariant $j(\backslash tau)$ is an algebraic integer of degree $\backslash left|\backslash mathrm\{Cl\}\backslash bigl(\backslash mathbf\{Q\}(\backslash tau)\backslash bigr)\backslash right|$, the class number of $\backslash mathbf\{Q\}(\backslash tau)$ and the minimal (monic integral) polynomial it satisfies is called the 'Hilbert class polynomial'. Thus if the imaginary quadratic extension $\backslash mathbf\{Q\}(\backslash tau)$ has class number 1 (so d is a Heegner number), the j-invariant is an integer.

The q-expansion of j, with its Fourier series expansion written as a Laurent series in terms of $q=e^\{2\; \backslash pi\; i\; \backslash tau\}$, begins as:

$$j(\backslash tau)\; =\; \backslash frac\{1\}\{q\}\; +\; 744\; +\; 196\backslash ,884\; q\; +\; \backslash cdots.$$

The coefficients $c\_n$ asymptotically grow as

$$\backslash ln(c\_n)\; \backslash sim\; 4\backslash pi\; \backslash sqrt\{n\}\; +\; O\backslash bigl(\backslash ln(n)\backslash bigr),$$

and the low order coefficients grow more slowly than $200\backslash ,000^n$, so for $\backslash textstyle\; q\; \backslash ll\; \backslash frac\{1\}\{200\backslash ,000\}$, j is very well approximated by its first two terms. Setting $\backslash textstyle\backslash tau\; =\; \backslash frac\{1+\backslash sqrt\{-163\}\}\{2\}$ yields

$$q=-e^\{-\backslash pi\; \backslash sqrt\{163\}\}\; \backslash quad\backslash therefore\backslash quad\; \backslash frac\{1\}\{q\}=-e^\{\backslash pi\; \backslash sqrt\{163\}\}.$$

Now

$$j\backslash left(\backslash frac\{1+\backslash sqrt\{-163\}\}\{2\}\backslash right)=\backslash left(-640\backslash ,320\backslash right)^3,$$

so,

$$\backslash left(-640\backslash ,320\backslash right)^3=-e^\{\backslash pi\; \backslash sqrt\{163\}\}+744+O\backslash left(e^\{-\backslash pi\; \backslash sqrt\{163\}\}\backslash right).$$

Or,

$$e^\{\backslash pi\; \backslash sqrt\{163\}\}=640\backslash ,320^3+744+O\backslash left(e^\{-\backslash pi\; \backslash sqrt\{163\}\}\backslash right)$$

where the linear term of the error is,

$$\backslash frac\{-196\backslash ,884\}\{e^\{\backslash pi\; \backslash sqrt\{163\}\}\}\; \backslash approx\; \backslash frac\{-196\backslash ,884\}\{640\backslash ,320^3+744\}$$

\approx -0.000\,000\,000\,000\,75

explaining why $e^\{\backslash pi\; \backslash sqrt\{163\}\}$ is within approximately the above of being an integer.

The Chudnovsky brothers found in 1987 that

$$\backslash frac\{1\}\{\backslash pi\}\; =\; \backslash frac\{12\}\{640\backslash ,320^\backslash frac32\}\; \backslash sum\_\{k=0\}^\backslash infty\; \backslash frac\{(6k)!\; (163\; \backslash cdot\; 3\backslash ,344\backslash ,418k\; +\; 13\backslash ,591\backslash ,409)\}\{(3k)!(k!)^3\; (-640\backslash ,320)^\{3k\}\},$$

a proof of which uses the fact that

$$j\backslash left(\backslash frac\{1+\backslash sqrt\{-163\}\}\{2\}\backslash right)\; =\; -640\backslash ,320^3.$$

For similar formulas, see the Ramanujan–Sato series.

For the four largest Heegner numbers, the approximations one obtainsThese can be checked by computing

$$\backslash sqrt[3]\{e^\{\backslash pi\backslash sqrt\{d\}\}-744\}$$

on a calculator, and

$$\backslash frac\{196\backslash ,884\}\{e^\{\backslash pi\backslash sqrt\{d\}\}\}$$

for the linear term of the error. are as follows.

$$\backslash begin\{align\}$$

e^{\pi \sqrt{19}} &\approx \phantom{000\,0}96^3+744-0.22\\

e^{\pi \sqrt{43}} &\approx \phantom{000\,}960^3+744-0.000\,22\\

e^{\pi \sqrt{67}} &\approx \phantom{00}5\,280^3+744-0.000\,0013\\

e^{\pi \sqrt{163}} &\approx 640\,320^3+744-0.000\,000\,000\,000\,75

\end{align}

Alternatively,{{Cite web|url=http://groups.google.com.ph/group/sci.math.research/browse_thread/thread/3d24137c9a860893?hl=en|title=More on e^(pi*SQRT(163))|access-date=2008-04-19|archive-date=2009-08-11|archive-url=https://web.archive.org/web/20090811214935/http://groups.google.com.ph/group/sci.math.research/browse_thread/thread/3d24137c9a860893?hl=en|url-status=dead}}

$$\backslash begin\{align\}$$

e^{\pi \sqrt{19}} &\approx 12^3\left(3^2-1\right)^3\phantom{00}+744-0.22\\

e^{\pi \sqrt{43}} &\approx 12^3\left(9^2-1\right)^3\phantom{00}+744-0.000\,22\\

e^{\pi \sqrt{67}} &\approx 12^3\left(21^2-1\right)^3\phantom{0}+744-0.000\,0013\\

e^{\pi \sqrt{163}} &\approx 12^3\left(231^2-1\right)^3+744-0.000\,000\,000\,000\,75

\end{align}

where the reason for the squares is due to certain Eisenstein series. For Heegner numbers , one does not obtain an almost integer; even $d\; =\; 19$ is not noteworthy.The absolute deviation of a random real number (picked uniformly from unit interval, say) is a uniformly distributed variable on {{closed-closed|0, 0.5|size=120%}}, so it has absolute average deviation and median absolute deviation of 0.25, and a deviation of 0.22 is not exceptional. The integer j-invariants are highly factorisable, which follows from the form

:$$12^3\backslash left(n^2-1\backslash right)^3=\backslash left(2^2\backslash cdot\; 3\; \backslash cdot\; (n-1)\; \backslash cdot\; (n+1)\backslash right)^3,$$

and factor as,

$$\backslash begin\{align\}$$

j\left(\frac{1+\sqrt{-19}}{2}\right) &= \phantom{000\,0}-96^3 = -\left(2^5 \cdot 3\right)^3\\

j\left(\frac{1+\sqrt{-43}}{2}\right) &= \phantom{000\,}-960^3 = -\left(2^6 \cdot 3 \cdot 5\right)^3\\

j\left(\frac{1+\sqrt{-67}}{2}\right) &= \phantom{00}-5\,280^3 = -\left(2^5 \cdot 3 \cdot 5 \cdot 11\right)^3\\

j\left(\frac{1+\sqrt{-163}}{2}\right)&= -640\,320^3 = -\left(2^6 \cdot 3 \cdot 5 \cdot 23 \cdot 29\right)^3.

\end{align}

These transcendental numbers, in addition to being closely approximated by integers (which are simply algebraic numbers of degree 1), can be closely approximated by algebraic numbers of degree 3,{{cite web|url=http://sites.google.com/site/tpiezas/001|title=Pi Formulas}}

$$\backslash begin\{align\}$$

e^{\pi \sqrt{19}} &\approx x^{24}-24.000\,31 ; & x^3-2x-2&=0\\

e^{\pi \sqrt{43}} &\approx x^{24}-24.000\,000\,31 ; & x^3-2x^2-2&=0\\

e^{\pi \sqrt{67}} &\approx x^{24}-24.000\,000\,0019 ; & x^3-2x^2-2x-2&=0\\

e^{\pi \sqrt{163}} &\approx x^{24}-24.000\,000\,000\,000\,0011 ; &\quad x^3-6x^2+4x-2&=0

\end{align}

The roots of the cubics can be exactly given by quotients of the Dedekind eta function η(τ), a modular function involving a 24th root, and which explains the 24 in the approximation. They can also be closely approximated by algebraic numbers of degree 4,{{cite web|url=http://sites.google.com/site/tpiezas/ramanujan|title=Extending Ramanujan's Dedekind Eta Quotients}}

$$\backslash begin\{align\}$$

e^{\pi \sqrt{19}} &\approx 3^5 \left(3-\sqrt{2\left(1- \tfrac{96}{24}+1\sqrt{3\cdot19}\right)} \right)^{-2}-12.000\,06\dots\\

e^{\pi \sqrt{43}} &\approx 3^5 \left(9-\sqrt{2\left(1- \tfrac{960}{24}+7\sqrt{3\cdot43}\right)} \right)^{-2}-12.000\,000\,061\dots\\

e^{\pi \sqrt{67}} &\approx 3^5 \left(21-\sqrt{2\left(1- \tfrac{5\,280}{24} +31\sqrt{3\cdot67}\right)} \right)^{-2}-12.000\,000\,000\,36\dots\\

e^{\pi \sqrt{163}} &\approx 3^5 \left(231-\sqrt{2\left(1- \tfrac{640\,320}{24}+2\,413\sqrt{3\cdot163}\right)} \right)^{-2}-12.000\,000\,000\,000\,000\,21\dots

\end{align}

If $x$ denotes the expression within the parenthesis (e.g. $x=3-\backslash sqrt\{2\backslash left(1-\; \backslash tfrac\{96\}\{24\}+1\backslash sqrt\{3\backslash cdot19\}\backslash right)\}$), it satisfies respectively the quartic equations

$$\backslash begin\{align\}$$

x^4 -\phantom{00} 4\cdot 3 x^3 + \phantom{000\,0}\tfrac23( 96 +3) x^2 - \phantom{000\,000}\tfrac23\cdot3(96-6)x - 3&=0\\

x^4 -\phantom{00} 4\cdot 9x^3 + \phantom{000\,}\tfrac23( 960 +3) x^2 - \phantom{000\,00}\tfrac23\cdot9(960-6)x - 3&=0\\

x^4 -\phantom{0} 4\cdot 21x^3 + \phantom{00}\tfrac23( 5\,280 +3) x^2 - \phantom{000}\tfrac23\cdot21(5\,280-6)x - 3&=0\\

x^4 - 4\cdot 231x^3 + \tfrac23( 640\,320 +3) x^2 - \tfrac23\cdot231(640\,320-6)x - 3&=0\\

\end{align}

Note the reappearance of the integers $n\; =\; 3,\; 9,\; 21,\; 231$ as well as the fact that

$$\backslash begin\{align\}$$

2^6 \cdot 3\left(-\left(1- \tfrac{96}{24}\right)^2+ 1^2 \cdot3\cdot 19 \right) &= 96^2\\

2^6 \cdot 3\left(-\left(1- \tfrac{960}{24}\right)^2+ 7^2\cdot3 \cdot 43 \right) &= 960^2\\

2^6 \cdot 3\left(-\left(1- \tfrac{5\,280}{24}\right)^2+ 31^2 \cdot 3\cdot67 \right) &= 5\,280^2\\

2^6 \cdot 3\left(-\left(1- \tfrac{640\,320}{24}\right)^2+ 2413^2\cdot 3 \cdot163 \right) &= 640\,320^2

\end{align}

which, with the appropriate fractional power, are precisely the j-invariants.

Similarly for algebraic numbers of degree 6,

$$\backslash begin\{align\}$$

e^{\pi \sqrt{19}} &\approx \left(5x\right)^3-6.000\,010\dots\\

e^{\pi \sqrt{43}} &\approx \left(5x\right)^3-6.000\,000\,010\dots\\

e^{\pi \sqrt{67}} &\approx \left(5x\right)^3-6.000\,000\,000\,061\dots\\

e^{\pi \sqrt{163}} &\approx \left(5x\right)^3-6.000\,000\,000\,000\,000\,034\dots

\end{align}

where the xs are given respectively by the appropriate root of the sextic equations,

$$\backslash begin\{align\}$$

5x^6-\phantom{000\,0}96x^5-10x^3+1&=0\\

5x^6-\phantom{000\,}960x^5-10x^3+1&=0\\

5x^6-\phantom{00}5\,280x^5-10x^3+1&=0\\

5x^6-640\,320x^5-10x^3+1&=0

\end{align}

with the j-invariants appearing again. These sextics are not only algebraic, they are also solvable in radicals as they factor into two cubics over the extension $\backslash Q\backslash sqrt\{5\}$ (with the first factoring further into two quadratics). These algebraic approximations can be exactly expressed in terms of Dedekind eta quotients. As an example, let $\backslash textstyle\; \backslash tau\; =\; \backslash frac\{1+\backslash sqrt\{-163\}\}\{2\}$, then,

$$\backslash begin\{align\}$$

e^{\pi \sqrt{163}} &= \left( \frac{e^\frac{\pi i}{24} \eta(\tau)}{\eta(2\tau)} \right)^{24}-24.000\,000\,000\,000\,001\,05\dots\\

e^{\pi \sqrt{163}} &= \left( \frac{e^\frac{\pi i}{12} \eta(\tau)}{\eta(3\tau)} \right)^{12}-12.000\,000\,000\,000\,000\,21\dots\\

e^{\pi \sqrt{163}} &= \left( \frac{e^\frac{\pi i}{6} \eta(\tau)}{\eta(5\tau)} \right)^{6}-6.000\,000\,000\,000\,000\,034\dots

\end{align}

where the eta quotients are the algebraic numbers given above.

The three numbers 88, 148, 232, for which the imaginary quadratic field $\backslash Q\backslash left[\backslash sqrt\{-d\}\backslash right]$ has class number 2, are not Heegner numbers but have certain similar properties in terms of almost integers. For instance,{{Cite web|author=Titus Piezas|url=https://www.oocities.org/titus_piezas/Ramanujan_a.pdf|title=Ramanujan’s Constant e^(pv163) And Its Cousins}}

$$\backslash begin\{align\}$$

e^{\pi \sqrt{88}} +8\,744 &\approx \phantom{00\,00}2\,508\,952^2-0.077\dots\\

e^{\pi \sqrt{148}} +8\,744 &\approx \phantom{00\,}199\,148\,648^2-0.000\,97\dots\\

e^{\pi \sqrt{232}} +8\,744 &\approx 24\,591\,257\,752^2-0.000\,0078\dots\\

\end{align}

and

$$\backslash begin\{align\}$$

e^{\pi \sqrt{22}} -24 &\approx \phantom{00}\left(6+4\sqrt{2}\right)^{6} +0.000\,11\dots\\

e^{\pi \sqrt{37}} +24 &\approx \left(12+ 2 \sqrt{37}\right)^6 -0.000\,0014\dots\\

e^{\pi \sqrt{58}} -24 &\approx \left(27 + 5 \sqrt{29}\right)^6 -0.000\,000\,0011\dots\\

\end{align}

Given an odd prime p, if one computes $k^2\; \backslash mod\; p$ for $\backslash textstyle\; k=0,1,\backslash dots,\backslash frac\{p-1\}\{2\}$ (this is sufficient because $\backslash left(p-k\backslash right)^2\backslash equiv\; k^2\backslash mod\; p$), one gets consecutive composites, followed by consecutive primes, if and only if p is a Heegner number.{{Cite web|url=http://www.mathpages.com/home/kmath263.htm|title=Simple Complex Quadratic Fields}}

For details, see "Quadratic Polynomials Producing Consecutive Distinct Primes and Class Groups of Complex Quadratic Fields" by Richard Mollin.{{cite journal|author=Mollin, R. A.|title=Quadratic polynomials producing consecutive, distinct primes and class groups of complex quadratic fields|journal=Acta Arithmetica|volume=74|year=1996|pages=17–30|doi=10.4064/aa-74-1-17-30|url=http://matwbn.icm.edu.pl/ksiazki/aa/aa74/aa7412.pdf}}

{{reflist}}

• {{MathWorld|title=Heegner Number|urlname=HeegnerNumber}}
• {{OEIS el|sequencenumber=A003173|name=Heegner numbers: imaginary quadratic fields with unique factorization|formalname=Heegner numbers: imaginary quadratic fields with unique factorization (or class number 1)}}
• [https://www.ams.org/bull/1985-13-01/S0273-0979-1985-15352-2/S0273-0979-1985-15352-2.pdf Gauss' Class Number Problem for Imaginary Quadratic Fields, by Dorian Goldfeld]: Detailed history of problem.
• {{cite web|last=Clark|first=Alex|title=163 and Ramanujan Constant|url=http://www.numberphile.com/videos/163.html|work=Numberphile|publisher=Brady Haran|access-date=2013-04-02|archive-url=https://web.archive.org/web/20130516045906/http://www.numberphile.com/videos/163.html|archive-date=2013-05-16|url-status=dead}}

Category:Algebraic number theory