OFFSET
1,1
COMMENTS
a(n) == 1, 5 (mod 16).
Conjecture: Consider the smallest prime p of the form k^2+1 such that p is congruent to A002496(n) modulo q, q prime of the form m^2+1 > A002496(n). Then q = A002496(n+1).
Corollary: For any pair (A002496(n), A002496(n+1)), there exist two integers m, k such that A002496(m) = A002496(n) + k*A002496(n+1), m>n+1 and n=1,2,3,...
Examples (see A352582):
LINKS
Michel Lagneau, a(n),n=1..90
MAPLE
with(numtheory):T:=array(1..30000):k:=0:
nn:=500000:
for m from 1 to nn do:
if isprime(m^2+1)
then
k:=k+1:T[k]:=m^2+1:
else
fi:
od:
for n from 1 to 32 do:
ii:=0:r:=T[n]:q:=T[n+1]:
for i from 1 to k while(ii=0) do:
p:=T[i]:r1:=irem(p, q):
if r1=r and p>q
then
ii:=1: printf(`%d, `, p)
else
fi:
od:
od:
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Mar 20 2022
STATUS
approved