- Heegner number
In
number theory , a Heegner number is a (square-free)positive integer "d" such that theimaginary quadratic field mathbf{Q}(sqrt{-d}) has class number 1. Equivalently, itsring of integers has aunique factorization .The determination of such numbers is a special case of the
class number problem , and they underlie several striking results in number theory.According to the
Stark–Heegner theorem there are precisely nine Heegner numbers::num|1, num|2, num|3, num|7, num|11, num|19, num|43, num|67, num|163.This result was conjectured by Gauss and proven byKurt Heegner in1952 .Euler's prime-generating polynomial
Euler's prime-generating polynomial:n^2 + n + 41,which gives (distinct) primes for n=0,dots,39, is related to the Heegner number 163 = 4 cdot 41 - 1.
Rabinowitz [Rabinowitz, G. "Eindeutigkeit der Zerlegung in Primzahlfaktoren in quadratischen Zahlkörpern." Proc. Fifth Internat. Congress Math. (Cambridge) 1, 418-421, 1913.] proved that:n^2 + n + pgives primes for n=0,dots,p-2 if and only if its discriminant 1-4p is 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.]Almost integers and Ramanujan's constant
Ramanujan's constant is thetranscendental number Fact|date=June 2008 e^{pi sqrt{163, which is analmost integer , in that it is very close to aninteger ::e^{pi sqrt{163 = 262,537,412,640,768,743.999 999 999 999 25ldots [ [http://mathworld.wolfram.com/RamanujanConstant.html Ramanujan Constant - from Wolfram MathWorld ] ] approx 640,320^3+744
This coincidence is explained by
complex multiplication and the "q"-expansion of thej-invariant .Detail
Briefly, j((1+sqrt{-d})/2) is an integer for "d" a Heegner number, and e^{pi sqrt{d approx -j((1+sqrt{-d})/2) + 744 via the "q"-expansion.
If au is a quadratic irrational, then the "j"-invariant is an
algebraic integer of degree mbox{Cl}(mathbf{Q}( au))|, the class number of mathbf{Q}( au) and the minimal (monic integral) polynomial it satisfies is called the Hilbert class polynomial.Thus if the imaginary quadratic extension mathbf{Q}( au) 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 aLaurent series in terms of q=exp(2 pi i au), begins as::j(q) = frac{1}{q} + 744 + 196,884 q + cdotsThe coefficients c_n asymptotically grow as ln(c_n) sim 4pi sqrt{n} + O(ln(n)), and the low order coefficients grow more slowly than 200,000^n, so for q gg 200,000, "j" is very well approximated by its first two terms.Setting au = (1+sqrt{-163})/2 yields q=-exp(-pi sqrt{163}) or equivalently, frac{1}{q}=-exp(pi sqrt{163}).Now j((1+sqrt{-163})/2)=(-640,320)^3, so:640,320)^3=e^{pi sqrt{163+744+O(e^{-pi sqrt{163)Or,:e^{pi sqrt{163=640,320^3+744+O(e^{-pi sqrt{163)where the linear term of the error is:196,884/e^{pi sqrt{163 approx 196,884/(640,320^3+744)approx -.00000000000075explaining why e^{pi sqrt{163 is within approximately the above of being an integer.
Other Heegner numbers
For the four largest Heegner numbers, the approximations one obtains [These can be checked by computing sqrt [3] {e^{pisqrt{d-744} on a calculator, and196,884/e^{pisqrt{d for the linear term of the error.] are as follows.:egin{align}e^{pi sqrt{19 &approx 96^3+744-.22\e^{pi sqrt{43 &approx 960^3+744-.00022\e^{pi sqrt{67 &approx 5,280^3+744-.0000013\e^{pi sqrt{163 &approx 640,320^3+744-.00000000000075end{align}Alternatively [http://groups.google.com.ph/group/sci.math.research/browse_thread/thread/3d24137c9a860893?hl=en#] ,:egin{align}e^{pi sqrt{19 &approx 12^3(3^2-1)^3+744-.22\e^{pi sqrt{43 &approx 12^3(9^2-1)^3+744-.00022\e^{pi sqrt{67 &approx 12^3(21^2-1)^3+744-.0000013\e^{pi sqrt{163 &approx 12^3(231^2-1)^3+744-.00000000000075end{align}where the reason for the squares is due to certain Eisenstein series. For Heegner numbers d<19, 196,884/e^{pisqrt{d > 1, so one does not obtain an almost integer; even d=19 is not noteworthy.
The "j"-invariants associated to the cubes above are highly factorable and factor as:egin{align}j((1+sqrt{-19})/2) &= 96^3 =(2^5 cdot 3)^3\j((1+sqrt{-43})/2) &= 960^3=(2^6 cdot 3 cdot 5)^3\j((1+sqrt{-67})/2) & =5,280^3=(2^5 cdot 3 cdot 5 cdot 11)^3\j((1+sqrt{-163})/2) &=640,320^3=(2^6 cdot 3 cdot 5 cdot 23 cdot 29)^3end{align}
For d=3, the fit is reasonably good, but does not follow from the above::e^{pi sqrt{3 approx (-2^3)^3+744-1.2
Consecutive primes
Given an odd prime p, if one computes k^2 pmod{p} for k=0,1,dots,(p-1)/2 (this is sufficient because p-k)^2equiv k^2 pmod{p}), one gets consecutive composites, followed by consecutive primes, if and only if p is a Heegner number. [http://www.mathpages.com/home/kmath263.htm]
For details, see "Quadratic Polynomials Producing Consecutive Distinct Primes and Class Groups of Complex Quadratic Fields" by Richard Mollin.
References
External links
*
* [http://www.research.att.com/~njas/sequences/?Anum=A003173 A003173] from theOn-Line Encyclopedia of Integer Sequences
* [http://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.
Wikimedia Foundation. 2010.