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Least k >= 1 such that k^n + n is semiprime, or 0 if no such k exists.
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%I #25 Jun 19 2023 17:07:32

%S 3,2,1,3,1,7,3,1,1,11,2,7,1,1,7,3,5,23,4,1,1,3,2,1,1,21,14,11,12,7,16,

%T 1,1,1,26,37,1,1,4,21,6,31,4,25,1,71,14,1,10,1,10,371,36,1,3,1,1,185,

%U 2,43,1,49,104,1,18,205,70,1,2,33,38,541,1,105,8,1,24,395,30,3,1,71,20,1,1,1

%N Least k >= 1 such that k^n + n is semiprime, or 0 if no such k exists.

%C There exist values of n for which k^n + n is never prime (cf. A072883). Do there exist values of n for which k^n + n is never semiprime?

%C Compare with A361803, the equivalent sequence for k^n - n, where a generalized factorization (effectively a polynomial factorization) into 3 factors is given to show that k^n - n is never semiprime for certain n. - _Peter Munn_, Jun 19 2023

%H Sean A. Irvine, <a href="/A130827/b130827.txt">Table of n, a(n) for n = 1..100</a>.

%e a(1)=3 because 1^1 + 1 = 2 (prime) and 2^1 + 1 = 3 (prime) but 3^1 + 1 = 4 = 2*2 (semiprime).

%e a(2)=2 because 1^2 + 2 = 3 (prime) but 2^2 + 2 = 6 = 2*3 (semiprime).

%e a(3)=1 because 1^3 + 3 = 4 = 2*2 (semiprime).

%e a(4)=3 because 1^4 + 4 = 5 (prime) and 2^4 + 4 = 20 = 2^2 * 5 but 3^4 + 4 = 85 = 5*17 (semiprime).

%e a(5)=1 because 1^5 + 5 = 6 = 2*3 (semiprime).

%o (PARI) a(n) = my(k=1); while (bigomega(k^n+n)!=2, k++); k; \\ _Michel Marcus_, Jun 19 2023

%Y Cf. A097792 (n such that x^n+n is reducible), A072883 (least k >= 1 such that k^n+n is prime, or 0 if no such k exists).

%Y Cf. A361803.

%K nonn

%O 1,1

%A _Zak Seidov_, Aug 18 2007

%E More terms from _Sean A. Irvine_, Oct 20 2009