OFFSET
1,1
COMMENTS
If we consider the polynomial (1-lambda(t)+lambda(t)^2)^3-(j(t)/256)*lambda(t)^2*(1-lambda(t))^2 = 0 for D <> 5 (mod 8) where D is the fundamental discriminant of the imaginary quadratic field, then this polynomial factorizes into a product of three quadratic polynomials over Q(j(t)). From this follows that 2*A000003(n) gives us the highest exponent of a minimal polynomial with a root in lambda(I*sqrt(n)). This leads to the relation that a(n) is also given by 2*gcd(2, n+1)*A000003(n) for n > 1 because lambda^* is related to lambda by square root. By using the substitution lambda(I*sqrt(4*n)) = ((1-sqrt(1-lambda(I*sqrt(n))^2))/(1+sqrt(1-lambda(I*sqrt(n))^2)))^2 a somehow nicer expression which holds for all n can be derived: gcd(2, n+1)*A000003(4*n). - Thomas Scheuerle, Oct 30 2025
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Elliptic Lambda Function
FORMULA
a(n) = gcd(2, n+1)*A000003(4*n). - Thomas Scheuerle, Oct 30 2025
EXAMPLE
k_3 = (1/4)*sqrt(2)*(sqrt(3)-1) this is a root of 1 - 16*x^2 + 16*x^4 (row 3 from A389848), so a(3) = 4.
MATHEMATICA
Table[Exponent[MinimalPolynomial[RootApproximant[N[Sqrt[ModularLambda[I*Sqrt[n]]], 100*n]], x], x], {n, 1, 50}] (* Vaclav Kotesovec, Apr 07 2018 *)
PROG
(PARI)
a(n) = poldegree(factor(numerator(subst(polclass(-(n*4)), 'x, 256*(1-x^2+x^4)^3/(x^4*(1-x^2)^2))))[, 1][1]); \\ Thomas Scheuerle, Oct 30 2025
a(n) = qfbclassno(-16*n)*gcd(2, n+1); \\ Thomas Scheuerle, Oct 30 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Eric W. Weisstein, May 29 2003
EXTENSIONS
a(29)-a(100) from Vaclav Kotesovec, Apr 07 2018
STATUS
approved
