|
%I #21 Jul 07 2023 19:02:36
%S 4,8,9,8,10,8,10,21,23,19,19,15,16,12,11,33,31,19,24,22,24,18,14,33,
%T 39,23,36,13,13,19,36,32,29,27,25,11,20,56,37,46,25,22,21,16,47,25,33,
%U 22,55,32,25
%N Number of distinct 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="https://mathworld.wolfram.com/CyclicNumber.html">Cyclic Number</a>
%F a(n) = A001221(A004042(n+1)).
%F For n>1, let p = A001913(n). If p is a base-10 Wieferich prime, then a(n) = A102347(p-1) + 2; otherwise a(n) = A102347(p-1) + 1. Also, we have A102347(p-1) = A102347((p-1)/2) + A119704((p-1)/2). - _Max Alekseyev_, Apr 26 2022
%e A004042(2) = 142857 = 37*13*11*3^3, therefore a(1) =
%e #{3,11,13,37} = 4.
%Y Cf. A001913, A087021-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
|
