| Displaying 1-8 of 8 results found.
page
1
Maximum number of rational points on a smooth absolutely irreducible projective curve of genus 1 over the field F_7^n.
+10
15
13, 64, 381, 2500, 17066, 118336, 825358, 5769604, 40366312, 282508864, 1977415678, 13841522500, 96889632947, 678224719936, 4747565867723, 33232942099204, 232630544491667, 1628413678617664, 11398895398904361, 79792266862562500, 558545865578002528, 3909821052537641536
FORMULA
a(n) = 7^n + 1 + floor(2*7^(n/2)) if 7 does not divide floor(2*7^(n/2)), n is even, or n = 1. Otherwise a(n) = 7^n + floor(2*7^(n/2)) [Deuring-Waterhouse]. - Robin Visser, Aug 17 2023
PROG
(Sage)
def a(n):
if (n==1) or (n%2 == 0) or (floor(2*7^(n/2))%7 != 0):
return 7^n + 1 + floor(2*7^(n/2))
else:
Maximal number of rational points on a curve of genus 2 over GF(q), where q = A246655(n) is the n-th prime power > 1.
(Formerly M4067)
+10
5
6, 8, 10, 12, 16, 18, 20, 24, 26, 33, 32, 36, 42, 46, 48, 50, 52, 53, 60, 66, 68, 74, 78, 82, 90, 92, 97, 100, 104, 106, 114, 118, 120, 126, 136, 140, 144, 148, 150, 156, 166, 170, 172, 172, 176, 184, 186, 198, 200, 206, 214, 218, 222, 226, 232, 234, 246, 248, 252, 256, 268, 282
COMMENTS
The successive values of q are 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, ... (see A246655).
REFERENCES
J. W. P. Hirschfeld, Linear codes and algebraic curves, pp. 35-53 of F. C. Holroyd and R. J. Wilson, editors, Geometrical Combinatorics. Pitman, Boston, 1984. See N_q(2) on page 51.
J.-P. Serre, Oeuvres, vol. 3, pp. 658-663 and 664-669.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
a(n) <= q + 1 + 4*sqrt(q) where q = A246655(n) [Hasse-Weil theorem]. - Robin Visser, Aug 03 2023 a(n) >= q - 1 + 2*floor(2*sqrt(q)) for all n except for 3 and 7, where q = A246655(n) [Serre]. - Robin Visser, Aug 03 2023
EXAMPLE
a(2) = 8 because 8 is the maximal number of rational points on a genus 2 curve over GF(3). One example of such a maximal curve is the genus 2 curve y^2 = x^6 + 2*x^2 + 1 which consists of the rational points (x,y) = (0, 1), (0, 2), (1, 1), (1, 2), (1, 1), (1, 2), and two points at infinity. - Robin Visser, Aug 03 2023
PROG
(Sage)
for q in range(1, 1000):
if Integer(q).is_prime_power():
p = Integer(q).prime_factors()[0]
if q==4: print(10)
elif q==9: print(20)
elif (Integer(q).is_square()): print(q + 1 + 4*sqrt(q))
elif ((floor(2*sqrt(q))%p == 0) or (q-1).is_square()
or (4*q-3).is_square() or (4*q-7).is_square()):
if (frac(2*sqrt(q)) > ((sqrt(5)-1)/2)): print(q + 2*floor(2*sqrt(q)))
else: print(q + 2*floor(2*sqrt(q)) - 1)
else: print(q + 1 + 2*floor(2*sqrt(q))) # Robin Visser, Aug 03 2023
Values k arising from a construction of Hirschfeld of k-arcs on elliptic curves over GF(q), where q = A246655(n) is the n-th prime power > 1.
(Formerly M0475)
+10
2
2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 16, 18, 19, 20, 21, 22, 25, 27, 28, 30, 32, 34, 37, 38, 40, 42, 44, 45, 48, 50, 51, 54, 58, 61, 62, 64, 65, 67, 72, 74, 75, 75, 77, 80, 81, 87, 88, 91, 94, 96, 98, 100, 103, 104, 109, 110, 113, 114, 120, 126, 129, 130, 132, 135, 136, 137, 141
COMMENTS
Let E be an elliptic curve over GF(q). A k-arc on E is a set of k points in E(GF(q)) such that no three are collinear (in the projective plane over GF(q)). Hirschfeld showed that if the number #E(GF(q)) of GF(q)-rational points on E is even, then there exists a k-arc on E for k = #E(GF(q))/2. Here, a(n) denotes the largest possible k arising from this construction, hence a(n) = floor( A005523(n)/2). Note that a(n) is not necessarily the maximal k such that there exists a k-arc on an elliptic curve over GF(q); e.g. the elliptic curve y^2 = x^3 + x + 1 over GF(5) contains a 6-arc consisting of the points {(0,1), (3,1), (4,2), (4,3), (0,4), (3,4)}. - Robin Visser, Aug 26 2023
REFERENCES
J. W. P. Hirschfeld, Linear codes and algebraic curves, pp. 35-53 of F. C. Holroyd and R. J. Wilson, editors, Geometrical Combinatorics. Pitman, Boston, 1984. See M_q(1) on page 51.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Mathematica Information Center, Item 5175, for full code.
EXAMPLE
For n = 4, the elliptic curve E : y^2 = x^3 + 3*x over GF(5) has 10 rational points. As this is the maximal number of rational points an elliptic curve over GF(5) can have, this implies a(4) = 10/2 = 5. - Robin Visser, Aug 26 2023
PROG
(Sage)
for q in range(1, 1000):
if Integer(q).is_prime_power():
p = Integer(q).prime_factors()[0]
if (floor(2*sqrt(q))%p != 0) or (Integer(q).is_square()) or (q==p):
print(floor((q + 1 + floor(2*sqrt(q)))/2))
else:
print(floor((q + floor(2*sqrt(q)))/2)) # Robin Visser, Aug 26 2023
a(n) is the number of isogeny classes of elliptic curves over GF(q), where q = A246655(n) is the n-th prime power > 1.
+10
2
5, 7, 9, 9, 11, 9, 13, 13, 15, 13, 17, 17, 19, 20, 17, 21, 23, 15, 25, 25, 27, 27, 27, 29, 31, 31, 21, 33, 33, 35, 35, 29, 37, 37, 39, 41, 41, 41, 41, 43, 45, 37, 45, 25, 45, 47, 47, 49, 49, 51, 51, 51, 50, 53, 53, 53, 55, 55, 57, 57, 59, 59, 61, 61, 61, 61, 63, 45, 63, 37, 65, 65
COMMENTS
Two elliptic curves over a finite field F_q are isogenous if and only if they have the same trace of Frobenius, or equivalently, have the same number of points over F_q.
Thus a(n) is the number of integers k such that there exists an elliptic curve over GF(q) with trace k, where q = A246655(n).
FORMULA
a(n) = 2*floor(2*sqrt(q)) + 1 if q is prime, where q = A246655(n).
EXAMPLE
For n = 1, the a(1) = 5 isogeny classes of elliptic curves over GF(2) are parametrized by the 5 possible values for the trace of Frobenius: -2, -1, 0, 1, 2.
For n = 2, the a(2) = 7 isogeny classes of elliptic curves over GF(3) are parametrized by the 7 possible values for the trace of Frobenius: -3, -2, -1, 0, 1, 2, 3.
PROG
(Sage)
for q in range(1, 1000):
if Integer(q).is_prime_power():
p, ans = Integer(q).prime_factors()[0], 0
for a in range(-floor(2*sqrt(q)), floor(2*sqrt(q))+1):
if (a%p != 0) or (Integer(q).is_square() and ((abs(a) == 2*sqrt(q))
or ((p%3 != 1) and (abs(a) == sqrt(q))) or ((p%4 != 1) and
(a==0)))) or ((not Integer(q).is_square()) and
(((p in [2, 3]) and (abs(a) == sqrt(p*q))) or (a==0))):
ans += 1
print(ans)
Maximum number of rational points on a smooth absolutely irreducible projective curve of genus 1 over the field F_2^n.
+10
1
5, 9, 14, 25, 44, 81, 150, 289, 558, 1089, 2138, 4225, 8374, 16641, 33130, 66049, 131796, 263169, 525736, 1050625, 2100048, 4198401, 8394400, 16785409, 33566018, 67125249, 134240898, 268468225, 536917252, 1073807361, 2147576330, 4295098369, 8590119956, 17180131329, 34360109096
FORMULA
a(n) = 2^n + 1 + floor(2^(n/2 + 1)) if floor(2^(n/2 + 1)) is odd, n is even, or n = 1. Otherwise a(n) = 2^n + floor(2^(n/2 + 1)) [Deuring-Waterhouse]. - Robin Visser, Aug 17 2023
PROG
(Sage)
def a(n):
if (n==1) or (n%2 == 0) or (floor(2^(n/2+1))%2 != 0):
return 2^n + 1 + floor(2^(n/2+1))
else:
Maximum number of rational points on a smooth absolutely irreducible projective curve of genus 1 over the field F_3^n.
+10
1
7, 16, 38, 100, 275, 784, 2280, 6724, 19964, 59536, 177989, 532900, 1596849, 4787344, 14356482, 43059844, 129162891, 387459856, 1162329651, 3486902500, 10460557755, 31381413904, 94143792483, 282430599364, 847290450408, 2541869016976, 7625603007884, 22876802020900, 68630393933574
FORMULA
a(n) = 3^n + 1 + floor(2*3^(n/2)) if 3 does not divide floor(2*3^(n/2)), n is even, or n = 1. Otherwise a(n) = 3^n + floor(2*3^(n/2)) [Deuring-Waterhouse]. - Robin Visser, Aug 17 2023
PROG
(Sage)
def a(n):
if (n==1) or (n%2 == 0) or (floor(2*3^(n/2))%3 != 0):
return 3^n + 1 + floor(2*3^(n/2))
else:
Maximum number of rational points on a smooth absolutely irreducible projective curve of genus 1 over the field F_5^n.
+10
1
10, 36, 148, 676, 3237, 15876, 78685, 391876, 1955920, 9771876, 48842100, 244171876, 1220773003, 6103671876, 30517927510, 152588671876, 762941200054, 3814701171876, 19073495062765, 95367451171876, 476837201876328, 2384185888671876, 11920929173444139, 59604645263671876
FORMULA
a(n) = 5^n + 1 + floor(2*5^(n/2)) if 5 does not divide floor(2*5^(n/2)), n is even, or n = 1. Otherwise a(n) = 5^n + floor(2*5^(n/2)) [Deuring-Waterhouse]. - Robin Visser, Aug 17 2023
PROG
(Sage)
def a(n):
if (n==1) or (n%2 == 0) or (floor(2*5^(n/2))%5 != 0):
return 5^n + 1 + floor(2*5^(n/2))
else:
Prime powers q such that there does not exist an elliptic curve E over GF(q) with cardinality q + 1 + floor(2*sqrt(q)).
+10
1
128, 2048, 2187, 16807, 32768, 131072, 524288, 1953125, 2097152, 8388608, 14348907, 48828125, 134217728, 536870912, 30517578125, 549755813888, 847288609443, 2199023255552, 19073486328125, 140737488355328, 562949953421312, 36028797018963968, 144115188075855872, 450283905890997363
COMMENTS
By Hasse's theorem, every elliptic curve E over GF(q) has cardinality at most q + 1 + floor(2*sqrt(q)). Moreover, for every prime power q, there exists an elliptic curve E over GF(q) with cardinality at least q + floor(2*sqrt(q)). Thus these are the prime powers q for which A005523(n) = q + floor(2*sqrt(q)), where q = A246655(n). By a theorem of Deuring and Waterhouse, these are exactly the prime powers q = p^k such that q is not prime, q is not a square, and p divides floor(2*sqrt(q)).
EXAMPLE
The first few values of the sequence (factorized) are 2^7, 2^11, 3^7, 7^5, 2^15, 2^17, 2^19, 5^9, 2^21, 2^23, 3^15, 5^11, 2^27, 2^29, ...
PROG
(Sage)
for q in range(1, 100000):
if Integer(q).is_prime_power():
p = Integer(q).prime_factors()[0]
if (floor(2*sqrt(q))%p == 0) and (not Integer(q).is_square()) and (q != p):
print(q)
Search completed in 0.010 seconds
| 