

A236183


Primes p such that for all primes q dividing p1 every residue mod p is the sum of two qth powers mod p.


1



2, 3, 5, 13, 17, 19, 37, 73, 97, 101, 109, 151, 163, 181, 193, 197, 211, 241, 251, 257, 271, 281, 337, 379, 397, 401, 421, 433, 449, 487, 491, 541, 577, 601, 631, 641, 661, 673, 701, 727, 751, 757, 769, 811, 881, 883, 991, 1009
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OFFSET

1,1


COMMENTS

Alternative definition: increasing list of primes p such that for ALL primes q, every residue mod p is the sum of two qth powers mod p (if q does not divides p1 then every residue mod p is a qth power, so only the case q divides p1 is not trivial).
Related to the conjecture:
For every prime q there are only finitely many primes p such that not every residue mod p is the sum of two qth powers mod p.


LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 1..400


EXAMPLE

p=7 is not in the list because q=3 divides p1 and the only residues mod 7 which are the sum of two cubic residues mod 7 are: 0,1,1,2,2.
To check that p=13 is in the sequence, since 2 and 3 are the only primes dividing p1=12, we only need to see that every residue mod 13 is both the sum of two quadratic residues mod 13 and the sum of two cubic residues mod 13.


PROG

(Sage)
for p in prime_range(a, b):
c=1
C=GF(p)
u=combinations_with_replacement(C, 2)
v=[x for x in u]
for q in prime_divisors(p1):
w=(k[0]^q+k[1]^q for k in v)
s=set(w)
l=len(s)
if l!=p:
c=0
break
if c==1:
print(p)
(PARI) is(p)=if(!isprime(p), return(0)); my(f=factor(p1)[, 1], v, u); for(i=1, #f, u=vector(p); v=vector(p, j, j^f[i]%p); for(j=1, p, for(k=j, p, u[(v[j]+v[k])%p+1]=1)); if(!vecmin(u), return(0))); 1 \\ Charles R Greathouse IV, Jan 23 2014


CROSSREFS

Sequence in context: A259188 A173971 A095083 * A093077 A249016 A241123
Adjacent sequences: A236180 A236181 A236182 * A236184 A236185 A236186


KEYWORD

nonn


AUTHOR

Esteban Arreaga Ambéliz, Jan 19 2014


STATUS

approved



