A274721 a(n) is the least k such that A051903(k^2+1) = n.

1, 7, 57, 182, 2057, 1068, 32318, 110443, 280182, 3626068, 23157318, 120813568, 123327057, 1097376068, 11109655182, 49925501068, 407838170807, 355101282318, 3459595983307, 15613890344818, 365855836217682, 110981321985443, 2273204469030182, 9647724486047943

Offset

1,2

Comments

Least k such that the largest exponent of a prime in the factorization of k^2+1 is n.

Conjecture: for each n > 1, a(n) = A034939(n) or 5^n - A034939(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

Cf. A034939, A051903.

Sequence in context: A304691 A218428 A365102 * A180817 A201438 A202510

Adjacent sequences: A274718 A274719 A274720 * A274722 A274723 A274724

Keyword

nonn

Author

Robert Israel, Jul 14 2016

Status

approved