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a(n) = least k such that 3^k mod k = 2^n.
2

%I #15 Mar 31 2012 14:12:21

%S 2,2929,41459,2352527,144937,1055,1829903,7316185805,114491,

%T 3146746271,5028467,20299,69609309001,129433,15307006153,2149705,

%U 66469,559182815,18429503,4529951,7094711,83591212702535,1251548749,38088889

%N a(n) = least k such that 3^k mod k = 2^n.

%F a(n) = A078457(2^n).

%e a(1) = A128149(3) = 2929.

%e a(2) = A128150(3) = 41459.

%Y Cf. A078457 = least k such that the remainder when 3^k is divided by k is n.

%Y Cf. A036236, A128149, A128150.

%K hard,nonn

%O 0,1

%A _Alexander Adamchuk_, Feb 16 2007

%E a(7)-a(9) from A078457. _Max Alekseyev_, Mar 11 2009

%E Extended by _Max Alekseyev_, Mar 15 2009

%E a(20) from _Hagen von Eitzen_, Aug 01 2009

%E a(21)-a(23) from _Max Alekseyev_, Feb 13 2012