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%I #19 Nov 20 2020 05:35:01
%S 3,7,5,31,9,127,17,73,33,2047,65,8191,129,1057,257,131071,513,524287,
%T 1025,16513,2049,8388607,4097,1082401,8193,262657,16385,536870911,
%U 32769,2147483647,65537,4196353,131073,270549121,262145,137438953471,524289
%N If b(n) = the largest proper divisor of n, then a(n) = (2^n - 1)/(2^b(n) - 1).
%H Alois P. Heinz, <a href="/A161818/b161818.txt">Table of n, a(n) for n = 2..1000</a>
%e a(6) = (2^6 - 1)/(2^3 - 1) = 63/7 = 9. - _Emeric Deutsch_, Jun 26 2009
%p with(numtheory): a := proc (n) options operator, arrow: (2^n-1)/(2^divisors(n)[tau(n)-1]-1) end proc: seq(a(n), n = 2 .. 40); # _Emeric Deutsch_, Jun 26 2009
%t b[n_] := Divisors[n][[-2]];
%t a[n_] := (2^n - 1)/(2^b[n] - 1);
%t a /@ Range[2, 40] (* _Jean-François Alcover_, Nov 20 2020 *)
%o (PARI) a(n) = my(d=divisors(n)); (2^n-1)/(2^d[#d-1]-1); \\ _Michel Marcus_, Nov 20 2020
%Y Cf. A032742.
%K nonn
%O 2,1
%A _Leroy Quet_, Jun 20 2009
%E Extended by _Emeric Deutsch_, Jun 26 2009
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