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

1,1

Comments

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)

Links

Table of n, a(n) for n=1..44.

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

Formula

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

Example

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).

Maple

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

Mathematica

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 *)

Prog

(PARI) 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

Crossrefs

Cf. A298299.

Sequence in context: A332730 A248955 A071053 * A094439 A122037 A363796

Adjacent sequences: A298395 A298396 A298397 * A298399 A298400 A298401

Keyword

nonn,hard,more

Author

Thomas Ordowski, Jan 18 2018

Extensions

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

Status

approved