A034939 -id:A034939 - cp's OEIS frontend

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

Robert Israel, Jul 14 2016

nonn

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.

			1^2 + 1 = 2.
7^2 + 1 = 2*5^2.
57^2 + 1 = 2*5^3*13.
182^2 + 1 = 5^4 * 53.

Robert Israel, Table of n, a(n) for n = 1..109

Cf. A034939, A051903.

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]);

A218720 a(n) is smallest number such that a(n)^2 + 1 is divisible by 101^n.

Original entry on oeis.org

0, 10, 515, 296344, 35764191, 1108900220, 316411915250, 47023298541694, 3156215819652023, 310872228812491206, 28124944860980892220, 3783840171259076226254, 208193145695151069244665, 19364218657938636320485082, 663491749602035014400202724

Offset: 0

Michel Lagneau, Nov 04 2012

nonn

			a(3) = 296344 because 296344^2+1 = 101 ^ 3 * 85237.

Cf. A002522, A049532, A034939, A218709, A218710, A218712, A218713, A218714, A218715, A218716, A218717, A218718, A218719.

b=10;n101=101;jo=Join[{0,b},Table[n101=101*n101;b=PowerMod[b, 101,n101];b=Min[b,n101-b],{99}]]

cp's OEIS Frontend

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

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