OFFSET
1,2
COMMENTS
Equivalently, numbers n such that there are no positive integers m such that m^r is of the form x^2 + n*y^2 and m is not of the form x^2 + n*y^2, for any odd integer r >= 3.
Because if m is not of the form x^2 + n*y^2 and m^(r-2) is not of the form x^2 + n*y^2, but m^r = a^2 + n*b^2, then m^(r+2) = (m*a)^2+n*(m*b)^2, m^(r+4) = (m^2*a)^2+n*(m^2*b)^2, etc. are also of the form x^2 + n*y^2. Then m^(3*r-6) = (m^(r-2))^3 = (m^(r-3)*a)^2 + n*(m^(r-3)*b)^2 is of the form x^2 + n*y^2 and m^(r-2) is not of the form x^2 + n*y^2.
All squarefree convenient numbers (A000926) congruent to {1, 2, 3, 5, 6} mod 8 are members of this sequence.
It appears that for a given value of n, if there is an m such that m^3 is of the form x^2 + n*y^2 and m is not of the form x^2 + n*y^2, then there exists some m < n that satisfies this condition.
Is this sequence finite?
Is 66045 the largest term of this sequence?
LINKS
V. Raman, Table of n, a(n) for n = 1..172
EXAMPLE
n = 11 is not a member of this sequence because for m = 23, 23 is not of form x^2 + 11*y^2, but 23^3 = 12167 = 54^2 + 11*29^2.
PROG
(PARI) for(n=1, 100000, flag=0; for(m=1, n, a=0; b=0; for(x=0, ceil(sqrt(m/n)), if(issquare(m-n*x^2), a=1; break)); if(a==0, for(y=0, ceil(sqrt(m^3/n)), if(issquare(m^3-n*y^2), b=1; break))); if(a==0&&b==1, flag=1)); if(flag==0, print1(n", ")))
CROSSREFS
KEYWORD
nonn
AUTHOR
V. Raman, Nov 25 2013
STATUS
approved