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

Author

Thomas Ordowski, Jan 18 2018

Keywords

• nonn
• hard
• more

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)

Examples

			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).
		

Links

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

Crossrefs

Cf. A298299.

Programs

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 *)

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

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

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