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a(n) = phi(2^prime(n) - 1)/prime(n); a(0) = 0 by convention.
2

%I #3 Mar 30 2012 17:30:26

%S 0,1,2,6,18,176,630,7710,27594,356960,18407808,69273666,3697909056,

%T 53630700752,204064589160,2992477516800,169917983040000,

%U 9770466930024800,37800705069076950,2202596295934991760

%N a(n) = phi(2^prime(n) - 1)/prime(n); a(0) = 0 by convention.

%p with numtheory; A056743 := proc(n) phi( 2^ithprime(n) - 1 )/ithprime(n); end;

%t Phi( A001348) / A000040. Table[EulerPhi[(2^Prime[n] - 1)]/Prime[n], {n, 1, 25}]

%Y Cf. A000020, A064535.

%K nonn

%O 0,3

%A _Robert G. Wilson v_, Aug 14 2000