OFFSET
1,2
COMMENTS
Least k such that the largest exponent of a prime in the factorization of k^2+1 is n.
For any n > 1, -1 has two square roots mod 5^n; at least one of these is not a square root of -1 mod 5^(n+1). If v is this number, v < 5^n so v^2 < 25^n. v^2+1 might be divisible by p^(n+1) for p = 13 or 17, or a square root of -1 mod 13^n or 17^n might be smaller than v, but that seems very unlikely. Thus the conjecture.
LINKS
Robert Israel, Table of n, a(n) for n = 1..109
EXAMPLE
1^2 + 1 = 2.
7^2 + 1 = 2*5^2.
57^2 + 1 = 2*5^3*13.
182^2 + 1 = 5^4 * 53.
MAPLE
F:= proc(n) local v, p, w;
v:= numtheory:-msqrt(-1, 5^n);
v:= min(v, 5^n-v);
if max(seq(t[2], t=ifactors(v^2+1)[2])) > n then
v:= 5^n - v;
if max(seq(t[2], t=ifactors(v^2+1)[2])) > n then
error "neither %d nor %d works", 5^n-v, v fi
fi;
for p from 13 by 4 while p^n <= v^2+1 do
if isprime(p) then
w:= numtheory:-msqrt(-1, p^n);
w:= min(w, p^n-w);
if w < v then
if max(seq(t[2], t=ifactors(w^2+1)[2])) = n then
v:= w;
fi
fi
fi
od;
v
end proc:
F(1):= 1:
map(F, [$1..100]);
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert Israel, Jul 14 2016
STATUS
approved