A298299 -id:A298299 - cp's OEIS frontend

A298398 a(n) is the smallest odd b > 1 such that (b^(2n) + 1)/2 has all prime divisors p == 1 (mod 2n).

3, 3, 5, 3, 9, 5, 15, 3, 199, 3, 45, 13, 25, 13, 181, 3, 35, 71, 39, 9, 545, 21, 45, 5, 101, 5, 1405, 13, 59, 107, 61, 3, 5369, 13, 7069, 305, 221, 39, 131, 3, 165, 169, 85, 43

Offset: 1

Thomas Ordowski, Jan 18 2018

Conjecture: a(n) exists for every n. This is implied by the generalized Bunyakovsky conjecture (Schinzel's hypothesis H).
The number (a(n)^(2n) + 1)/2 has all divisors d == 1 (mod 2n).
Thus, here is the congruence a(n)^(2n) == 1 (mod 2n).
If n is a power of 2, then a(n) = 3.
From Kevin P. Thompson, Mar 13 2022: (Start)
Additional terms: a(47) = 95, a(48) = 19, a(50) = 851, a(51) = 6425, a(52) = 47, a(56) = 29, a(57) = 571.
a(45) >= 2746511 (a C169 + C276 remain to be factored to verify b=2746511).
a(46) >= 275 (a C182 remains to be factored to verify b=275).
a(49) >= 979 (a C234 remains to be factored to verify b=979).
a(53) >= 425 (a C195 remains to be factored to verify b=425).
a(54) >= 1457 (a C164 remains to be factored to verify b=1457).
a(55) >= 10361 (a C307 remains to be factored to verify b=10361).
(End)

			a(5) = 9 since (9^10 + 1)/2 = 41 * 42521761, 41 = 1 (mod 5*2) and 42521761 = 1 (mod 5*2), so all divisors d == 1 (mod 10).

Kevin P. Thompson, Factorizations to support known terms of a(n) for n = 1..57

Cf. A298299.

g:= proc(t)
   convert(select(type,map(s -> s[1], ifactors(t,easy)[2]),integer),set);
end proc:
F:= proc(n) local s,t,b,C,B,k,bb,Cb, easyf; uses numtheory;
  t:= 2^padic:-ordp(n,2);
  s:= n/t;
  C:= unapply({seq(numtheory:-cyclotomic(m,-b^(2*t)),m=numtheory:-divisors(s) minus {1}), (b^(2*t)+1)/2},b);
   B:= select(t -> C(t) mod (2*n) = {1}, [seq(b,b=1..2*n-1,2)]);
   for k from 0 do
     for bb in B do
       b:= k*2*n+bb;
       if b < 2 then next fi;
       Cb:= remove(isprime,C(b));
       if Cb = {} then return b fi;
       easyf:= map(g, Cb) mod (2*n);
       if not (`union`(op(easyf)) subset {1}) then next fi;
       if andmap(c -> factorset(c) mod (2*n) = {1}, Cb) then return b fi;
     od
   od
end proc:
map(F, [$1..26]); # Robert Israel, Jan 18 2018

Array[Block[{b = 3}, While[Union@ Mod[FactorInteger[(b^(2 #) + 1)/2][[All, 1]], 2 #] != {1}, b += 2]; b] &, 20] (* Michael De Vlieger, Jan 20 2018 *)
f[n_] := Block[{b = 3}, Label[init]; While[ PowerMod[b, 2n, 2n] != 1, b += 2]; d = First@# & /@ FactorInteger[(b^(2n) +1)/2]; If[ Union@ Mod[d, 2n] != {1}, b += 2; Goto[init]]; b]; Array[f, 30] (* Robert G. Wilson v, Jan 22 2018 *)

isok(b, n) = {pf = factor((b^(2*n) + 1)/2)[, 1]; for (j=1, #pf, if (lift(Mod(pf[j], 2*n)) != 1, return (0));); return(1);}
a(n) = {my(b = 3); while (!isok(b, n), b += 2); b;} \\ Michel Marcus, Jan 19 2018

a(n) = min{b > 1: b is odd and for all prime p, if p | (b^(2n) + 1)/2 then p == 1 (mod 2n)}. - Kevin P. Thompson, Mar 14 2022

a(9)-a(30) from Robert Israel, Jan 18 2018
a(20) corrected by Michel Marcus, Jan 19 2018
a(31)-a(44) from Kevin P. Thompson, Mar 13 2022

A352597 a(n) is the smallest k > 1 such that k^n + 1 has all prime divisors p == 1 (mod n).

Original entry on oeis.org

2, 2, 6, 2, 10, 6, 28, 2, 18, 10, 22, 6, 52, 14, 60, 2, 102, 36, 190, 20, 756, 66, 46, 18, 2550, 26, 2970, 28, 58, 120, 310, 2, 330, 170, 11550, 6, 148, 570, 156, 140, 82, 2184, 172, 88, 3040020, 184, 282, 42, 7252, 110, 7548, 312, 106, 1440, 41800, 42, 11172

Offset: 1

Kevin P. Thompson, Mar 21 2022

Equivalently, a(n) is the smallest k > 1 such that for all divisors d of k^n + 1, d == 1 (mod n).
A298299 is a subsequence.
All terms in this sequence are even since for odd k the expression k^n + 1 is divisible by 2 which is not congruent to 1 (mod n) for any n > 1.
If n is odd, a(n)^n + 1 is divisible by a(n) + 1. Therefore, a(n) + 1 == 1 (mod n) and so n | a(n) for odd n.
Theorem: a(n) = 2 if and only if n is a power of 2.

			a(3) = 6 since 6^3 + 1 = 217 = 7 * 31 and both factors are congruent to 1 (mod 3).

Cf. A298076, A298299 (bisection), A298310, A298398.

isok(k,n) = my(f=factor(k^n+1)); for (i=1, #f~, if (Mod(f[i,1], n) != 1, return(0))); return(1);
a(n) = my(k=2); while (!isok(k, n), k+=2); k; \\ Michel Marcus, Mar 22 2022

a(2n) = A298299(n).

cp's OEIS Frontend

A298398 a(n) is the smallest odd b > 1 such that (b^(2n) + 1)/2 has all prime divisors p == 1 (mod 2n).

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