OFFSET
1,1
COMMENTS
If we define f(n) to be the number of primes (counted with multiplicity) of the form 4k + 3 that divide n, then with this sequence f(a(n)) is always odd. For example, 95 is divisible by 17 and 99 is divisible by 3 (twice) and 11. - Alonso del Arte, Jan 13 2016
With the Jan 05 2004 Jovovic comment on A078703: The number of 1 and -1 (mod 4) divisors of a(n) are identical. Proof: each number 3 (mod 4) is trivially not a sum of two squares. The number of solutions of n as a sum of two squares is r_2(n) = 4*(d_1(n) - d_3(n)), where d_k(n) is the number of k (mod 4) divisors of n. See e.g., Grosswald, pp. 15-16 for the proof of Jacobi. - Wolfdieter Lang, Jul 29 2016
REFERENCES
E. Grosswald, Representations of Integers as Sums of Squares. Springer-Verlag, NY, 1985.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
EXAMPLE
27 = 4 * 6 + 3 = 3^3.
35 = 4 * 8 + 3 = 5 * 7.
a(8) = 75 with 2*A078703(19) = 6 divisors [1, 3, 5, 15, 25, 75], which are 1, -1, 1, -1, 1, -1 (mod 4). - Wolfdieter Lang, Jul 29 2016
MATHEMATICA
Select[Range[1000], !PrimeQ[#] && IntegerQ[(# - 3)/4] &] (* Harvey P. Dale, Aug 16 2013 *)
Select[4Range[100] - 1, Not[PrimeQ[#]] &] (* Alonso del Arte, Jan 13 2016 *)
PROG
(PARI) lista(nn) = for(n=1, nn, if(!isprime(k=4*n+3), print1(k, ", "))); \\ Altug Alkan, Jan 17 2016
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Labos Elemer, Feb 24 2004
STATUS
approved