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%I #12 Aug 07 2019 01:44:29
%S 2,11,29,92,113,223,295,333,397,1076
%N a(n) is the smallest k such that 2^k-1 has n primitive prime factors.
%C A prime factor of 2^n-1 is called primitive if it does not divide 2^r-1 for any r<n. Equivalently, p is a primitive prime factor of 2^n-1 if ord(2,p)=n. See A086251 for the number of primitive prime factors in 2^n-1.
%C No more terms < 673. (2^673-1 is the first that is not completely factored in the linked reference.) - _David Wasserman_, Feb 22 2005
%C 2^1207-1 is now the first not completely factored number of the form 2^k-1. - _Hugo Pfoertner_, Aug 06 2019
%D J. Brillhart et al., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 3rd edition, 2002.
%H J. Brillhart et al., Factorizations of b^n +- 1 <a href="http://dx.doi.org/10.1090/conm/022">Available on-line</a>
%H factordb.com, <a href="http://factordb.com/index.php?query=2%5E1076-1">Factors of 2^1076-1</a>.
%H factordb.com, <a href="http://factordb.com/index.php?query=2%5En-1&use=n&n=1201&VP=on&VC=on&EV=on&OD=on&FF=on&CF=on&U=on&C=on&perpage=20&format=1&sent=Show">Status of 2^n-1 for n>1200</a>.
%e a(2) = 11 because 2^11-1 = 23*89, both 23 and 89 have order 11 and the numbers 2^r-1 have only 0 or 1 primitive prime factors.
%Y Cf. A086251.
%K hard,more,nonn
%O 1,1
%A _T. D. Noe_, Jul 14 2003
%E More terms from _David Wasserman_, Feb 22 2005
%E a(10) from _Hugo Pfoertner_, Aug 06 2019
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