

A127311


a(n) = E(GF(p))/H where E(GF(p)) is the group of rational points on the elliptic curve E: y^2 + y = x^3  x^2 mod p, the prime p is p(n) or p(n+1) according as n < 5 or n >= 5 and H = {oo, (0,0), (0,1), (1,0), (1,1)}.


2



1, 1, 1, 2, 2, 4, 4, 5, 6, 5, 7, 10, 10, 8, 12, 11, 10, 15, 15
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OFFSET

1,4


COMMENTS

H is a subgroup of E(GF(p)) of order 5 so a(n) = E(GF(p))/5 where p is p(n) or p(n+1) according as n < 5 or n >= 5.
E is singular over GF(p(5)) = GF(11) so we take p != 11.
Hasse proved that 5*a(n)  (p+1) <= 2*sqrt(p) where p is p(n) or p(n+1) according as n < 5 or n >= 5.
Elkies proved that 5*a(n) = p(n+1) + 1 for infinitely many n.


REFERENCES

B. Mazur, The Structure of Error Terms in Number Theory and an Introduction to the SatoTate Conjecture, Current Events Bulletin, Amer. Math. Soc., 2007.
J. H. Silverman, The Arithmetic of Elliptic Curves, Graduate Texts in Math., vol. 106, SpringerVerlag, Berlin and New York, 1986.
N. Koblitz, Introduction to Elliptic Curves and Modular Forms. New York: SpringerVerlag, 1993.


LINKS

Table of n, a(n) for n=1..19.
B. Mazur, The Structure of Error Terms in Number Theory and an Introduction to the SatoTate Conjecture
S. Fermigier, Collection of Links on Research Articles on Elliptic Curves and Related Topics


FORMULA

a(n) ~ (p(n+1) + 1)/5 as n > oo.
a(n) = (p+1  b(p))/5 where q*Prod(k=1 to oo, ((1  q^k)(1  q^11k))^2) = Sum(k=1 to oo, b(k)*q^k) and p is p(n) or p(n+1) according as n < 5 or n >= 5.


EXAMPLE

q*Prod(k=1 to oo, ((1  q^k)(1  q^11k))^2) = q  2q^2  ..., so a(1) = (p(1) + 1  b(p(1))/5 = (2 + 1  b(2))/5 = (3  (2))/5 = 1.


CROSSREFS

a(n) = A127310(n)/5. Cf. A000594, A127309.
Sequence in context: A113474 A089413 A159267 * A302929 A129229 A219029
Adjacent sequences: A127308 A127309 A127310 * A127312 A127313 A127314


KEYWORD

nonn


AUTHOR

Jonathan Sondow, Jan 12 2007


STATUS

approved



