%I #18 Jul 07 2023 21:10:54
%S 6,9,12,10,11,9,11,23,25,25,22,18,19,15,14,38,35,24,28,25,27,21,17,38,
%T 44,27,43,16,16,23,42,35,37,30,29,14,23,62,41,51,28,26,24,19,50,29,39,
%U 25,62,36,29
%N Total number of prime factors of n-th cyclic number.
%C A004042(n) factorized with Dario Alpern's ECM.
%C Extended using factors of 10^(A001913(n)-1)-1, see Kamada link.
%H Dario A. Alpern, <a href="https://www.alpertron.com.ar/ECM.HTM">Factorization using the Elliptic Curve Method</a>.
%H Makoto Kamada, <a href="https://stdkmd.net/nrr/repunit/">Factorizations of 11...11 (Repunit)</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CyclicNumber.html">Cyclic Number</a>
%F a(n) = A001222(A004042(n+1)).
%e A004042(2) = 142857 = 37*13*11*3^3, therefore a(1) = A087021(1)+2 = 6.
%Y Cf. A001913, A087020-A087026.
%K nonn,more,hard
%O 1,1
%A _Reinhard Zumkeller_, Jul 30 2003
%E a(3) corrected, a(12)-a(42) added by _Ray Chandler_, Nov 16 2011
%E a(43)-a(51) from _Max Alekseyev_, May 13 2022
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