OFFSET
1,1
COMMENTS
For p = 1 (mod 2^k), the 2^(k-1)-adic, 2^(k-2)-adic, etc. Gauss sums g(d,p) are real. This comes from symmetry: both x and -x (mod p) are residues. This sequence is the set the primes p = 1 (mod 8) for which g(4,p) > g(2,p) =1 [if we divide each Gauss sum by sqrt(p)]. So for a prime, for which 2^d || (p-1), we can assign a sequence of + or - signs, the signs of (g(2^f,p) - g(2^(f-1),p)) f=2,3,...,(d-1)
MATHEMATICA
d = 3; Select[Prime[Range[5000]], Mod[#, 2^(d)] == 1 && Re[Sum[Exp[k^(2^(d - 1))*2*Pi*I/#] - Exp[k^(2^(d - 2))*2*Pi*I/#], {k, 0, # - 1}]/Sqrt[#]] > 0 &]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zoltan Reti, Nov 23 2025
STATUS
approved
