OFFSET
1,1
COMMENTS
These numbers are the odd D candidates for the (generalized) Pell equation x^2 - D*y^2 = +8 which could have proper solutions (x, y) with x and y both odd (and gcd(x, y) = 1).
Proof: Put x =2*X + 1, y = 2*Y + 1; then 8*(T(X) - D*T(Y)) = 8 - 1 + D = 7 + D, with the triangular numbers T = A000217. Hence, D == -7 (mod 8) == +1 (mod 8). Only nonsquare numbers D are considered for the Pell equation (square D leads to a factorization with only one solution: D = 1, (x, y) = (3, 1)). Consider a prime factor p == 3 or 5 (mod 8) (A007520 or A007521) of D. Then x^2 == 8 (mod p). Because the Legendre symbol (8/p) = (2*2^2/p) = (2/p) == (-1)^(p^2-1)/8 (see, e.g., Nagell, eq. (3), p. 138) this becomes -1 for these primes p, and therefore a candidate for D cannot have any prime factors 3 or 5 (mod 8).
However, not all of these candidates admit solutions. For the exceptions see A264348.
The remaining Ds (that admit solutions) are given in A263012.
REFERENCES
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10000
MATHEMATICA
Select[8 Range@ 154 + 1, Or[PrimeQ@ #, CompositeQ@ # && AllTrue[Union@ Mod[First /@ FactorInteger@ #, 8], ! MemberQ[{3, 5}, #] &]] && ! IntegerQ@ Sqrt@ # &] (* Michael De Vlieger, Dec 11 2015, Version 10 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Nov 17 2015
STATUS
approved