OFFSET
1,1
COMMENTS
a(n) == 1 (mod 4), n >= 1. This is because 4*k+1, k>=0, not a square, can only have an even number of odd primes of the type 3 (mod 4) with odd exponents in the prime number factorization. The squarefree part of 4*k+1 has then an even number (maybe 0) of primes of the type 3 (mod 4). Examples:
a(4) = 21 = 3*7, a(6) = 33 = 3*11.
D(n) = A077425(n) are the 1 (mod 4) discriminants of indefinite binary quadratic forms (they are the odd numbers from A079896). sqrt(D(n)) becomes then, up to an integer factor, sqrt(a(n)), which defines a real quadratic number field Q(sqrt(a(n))) with a basis <1, omega(a(n))> for the ring of integers in this field, where omega(a(n)) = (1 + sqrt(a(n))/2. Example: sqrt(D(9)) = sqrt(45) = 3*sqrt(a(9)) = 3*sqrt(5), with omega(5) = (1 + sqrt(5))/2 (the golden section) for Q(sqrt(5)) = Q(omega(5)).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
MATHEMATICA
SquareFreePart[n_] := Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ FactorInteger[n]); SquareFreePart /@ (4*Range[65] + 1) // DeleteCases[#, 1] & (* Jean-François Alcover, Jun 14 2013 *)
PROG
(PARI) [core(n) | n <- vector(100, n, 4*n+1), !issquare(n)] \\ Charles R Greathouse IV, Mar 11 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jun 14 2013
STATUS
approved