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A298299
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a(n) is the smallest b > 0 such that b^(2n) + 1 has all prime divisors p == 1 (mod 2n).
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3
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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
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OFFSET
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1,1
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COMMENTS
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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)?
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)
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LINKS
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FORMULA
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a(n) = min{b > 1: for all prime p, if p | (b^(2n) + 1) then p == 1 (mod 2n)}.
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EXAMPLE
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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).
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PROG
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(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);
}
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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