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a(n) is the smallest b > 0 such that b^(2n) + 1 has all prime divisors p == 1 (mod 2n).
3

%I #46 Jul 01 2022 21:36:51

%S 2,2,6,2,10,6,14,2,36,20,66,18,26,28,120,2,170,6,570,140,2184,88,184,

%T 42,110,312,1440,42,116,9060,124,2,7656,34,13650,132,74,228,7800,40,

%U 1066,4158,430,132,6283590,46,94,12,1246,1960

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

%C All the terms are even.

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

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

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

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

%C 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?

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

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

%C From _Kevin P. Thompson_, Mar 13 2022: (Start)

%C 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.

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

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

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

%C (End)

%H Kevin P. Thompson, <a href="/A298299/a298299.txt">Factorizations to support known terms of a(n) for n = 1..56</a>

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

%e 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).

%o (PARI)

%o find_a_ORDOWSKI2(n=2, a=1, B_START=2, LIM=10^11,DEBUG=1)={

%o my(B,FF,LL);

%o my(fn="_THOMAS_ORDOWSKI_b_a_n.txt");

%o LL=R2('b,a,n); \\ R(b,a,n)=(b^n+a)

%o FF=factor(LL);

%o if(DEBUG==1,

%o print(FF);

%o print(LL);

%o );

%o if(Mod(n,2)==0, \\ n-EVEN

%o B=FIND_BASE(n,BSTART=B_START,LIM,STEP=2,FF);

%o );

%o if(B>0,

%o return([n,B,[subst(FF,'b,B)]]);

%o );

%o return(0);

%o }

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

%K nonn,hard,more

%O 1,1

%A _Krzysztof Ziemak_ and _Thomas Ordowski_, Jan 16 2018

%E a(31)-a(44) from _Kevin P. Thompson_, Mar 13 2022

%E a(45)-a(50) from _Daniel Suteu_, Jul 01 2022