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%I #6 Feb 22 2019 01:45:11
%S 1,2,2,3,3,4,3,6,3,5,4,5,5,9,3,4,9,3,6,10,6,7,6,11,5,11,10,5,10,9,3,
%T 12,6,10,9,5,6,13,9,6,11,6,10,16,4,4,6,9,6,11,8,4,10,10,5,13,10,7,11,
%U 6,6,21,4,23,8,6,8,15,15,7,12,7,8,19,8,13,14,5,6,20,6,10,13,12,7,9,9,6,21
%N Number of prime factors with multiplicity of 1 + (n^(n+1)).
%C As also noticed by _T. D. Noe_, for odd n: 2 | a(n), for even n: (n+1)^2 | a(n). Coincidentally, a(74) includes 13 multidigit prime factors all of which end with the digit 1. There is no upper limit to this sequence, which rapidly becomes slow to compute. The derived sequences of n such that a(n) = k for any constant k > 2 do not yet appear in the OEIS. For instance, a(n) = 3 for n = 4, 5, 7, 9, 15, 18, 31, ... Is each such derived sequence finite?
%F a(1) = 1. For n>1, a(n) = A001222(A110567(n)) = A001222(1 + A007778(n)) = A001222(1 + (n^(n+1))).
%e a(1) = 1 because 1+1^2 = 2 is prime (and the only such prime).
%e a(2) = 2 because 1 + 2^3 = 9 = 3^2 which has (with multiplicity) two prime factors.
%e a(3) = 2 because 1 + 3^4 = 82 = 2 * 41 (the last such semiprime?).
%e a(4) = 3 because 1 + 4^5 = 1025 = 5^2 * 41 which has (with multiplicity) 3 prime factors.
%e a(8) = 6 because 1 + 8^9 = 134217729 = 3^4 * 19 * 87211.
%e a(14) = 9 because 1 + 14^15 = 155568095557812225 = 3^2 * 5^2 * 61 * 71 * 101 * 811 * 1948981.
%e a(1000) > 52.
%Y Cf. A001222, A007778, A110567.
%K easy,nonn
%O 1,2
%A _Jonathan Vos Post_, Sep 14 2005
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