|
|
A116605
|
|
Smallest prime p such that p == 1 (mod prime(n)) and not p == 1 (mod k) for 2 < k < prime(n).
|
|
2
|
|
|
3, 7, 11, 239, 23, 443, 647, 1103, 47, 59, 2543, 3923, 83, 9203, 6299, 107, 7907, 8663, 11927, 14627, 12119, 15959, 167, 179, 20759, 20807, 23279, 23327, 28559, 227, 37847, 263, 43019, 54767, 53939, 54059, 54323, 54443, 66467, 347, 359, 69143, 383
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
a(n) > 2*prime(n) for n > 1.
a(n) = 2*prime(n)+1 if prime(n) is in A005384. Otherwise, a(n) > 2*prime(n)^2+1 for n > 1. - Robert Israel, Mar 29 2017
|
|
LINKS
|
|
|
EXAMPLE
|
a(1) = 3 since prime(1) = 2 and 3 == 1 (mod 2).
a(4) = 239 since prime(4) = 7, 239 == 1 (mod 7) and for each of the primes q smaller than 239 with q == 1 (mod 7) there is a k (2 < k < 7) such that q == 1 (mod k): 29 == 1 (mod 4), 43 == 1 (mod 6), 71 == 1 (mod 5), 113 == 1 (mod 4), 127 == 1 (mod 3), 197 == 1 (mod 4), 211 == 1 (mod 5), whereas 239 == 2 (mod 3), 3 (mod 4), 4 (mod 5), 5
(mod 6).
|
|
MAPLE
|
V:= {seq(4*i+2, i=1..10^5)}: A[1]:= 3:
for n from 2 do
pn:= ithprime(n);
R:= select(t -> t mod pn = 0, V);
found:= false;
for r in R do
if isprime(r+1) then
found:= true;
A[n]:= r+1;
break
fi
od;
if not found then break fi;
V:= V minus R;
od:
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|