A298299 a(n) is the smallest b > 0 such that b^(2n) + 1 has all prime divisors p == 1 (mod 2n).

2, 2, 6, 2, 10, 6, 14, 2, 36, 20, 66, 18, 26, 28, 120, 2, 170, 6, 570, 140, 2184, 88, 184, 42, 110, 312, 1440, 42, 116, 9060, 124, 2, 7656, 34, 13650, 132, 74, 228, 7800, 40, 1066, 4158, 430, 132, 6283590, 46, 94, 12, 1246, 1960

Offset: 1

Author

Krzysztof Ziemak and Thomas Ordowski, Jan 16 2018

Keywords

• nonn
• hard
• more

Comments

All the terms are even.

The number a(n)^(2n) + 1 has all divisors d == 1 (mod 2n).

Conjecture: a(n) exists for every n. This is implied by the generalized Bunyakovsky conjecture (Schinzel's hypothesis H).

Theorem: a(n) = 2 if and only if n is a power of 2.

Note: rad(2n) divides rad(a(n)), where rad(m) = A007947(m).

Even numbers 2n such that a(n) = rad(2n) are powers of two and 6, 10, 12, 14, 26, 36, ... Are there infinitely many such numbers?

We have a(n) = 2n = 2, 6, 10, 14, 20, 26, 28, ...

Problem: are there infinitely many even numbers m <> 2^k such that the number m^m + 1 has all divisors d == 1 (mod m)?

From Kevin P. Thompson, Mar 13 2022: (Start)

Additional terms: a(46) = 46, a(47) = 94, a(48) = 12, a(49) = 1246, a(50) = 1960, a(52) = 208, a(53) = 636, a(55) = 17600, a(56) = 476.

a(45) > 1000000 (sequence A298398 likewise has a very large value for n=45).

a(51) >= 16524 (a C241 remains to be factored to verify b=16524).

a(54) >= 6864 (a C201 remains to be factored to verify b=6864).

(End)

Examples

			a(5) = 10, because 10^10 + 1 = 10000000001 = 101*3541*27961 and all the prime factors p == 1 (mod 2*5), so all divisors d == 1 (mod 10).
		

Links

• Kevin P. Thompson, Factorizations to support known terms of a(n) for n = 1..56

Crossrefs

Cf. A007947, A298076 (see PARI subroutines used for a(48)), A298398.

Programs

PARI

find_a_ORDOWSKI2(n=2, a=1, B_START=2, LIM=10^11,DEBUG=1)={
  my(B,FF,LL);
  my(fn="_THOMAS_ORDOWSKI_b_a_n.txt");
  LL=R2('b,a,n);   \\ R(b,a,n)=(b^n+a)
  FF=factor(LL);
  if(DEBUG==1,
    print(FF);
    print(LL);
  );
  if(Mod(n,2)==0,  \\ n-EVEN
    B=FIND_BASE(n,BSTART=B_START,LIM,STEP=2,FF);
  );
  if(B>0,
     return([n,B,[subst(FF,'b,B)]]);
  );
  return(0);
}

Formula

a(n) = min{b > 1: for all prime p, if p | (b^(2n) + 1) then p == 1 (mod 2n)}.

Extensions

a(31)-a(44) from Kevin P. Thompson, Mar 13 2022

a(45)-a(50) from Daniel Suteu, Jul 01 2022