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a(n) = denominator of b(n), where b(1)=1, b(n+1) = sum{k|n} 1/b(k).
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%I #12 May 02 2018 19:16:14

%S 1,1,1,2,3,8,22,71,93,638,9144,26821,2373690,11013191,950468551,

%T 23819080360,4860600739719,18887557655957,23748158395676,

%U 102452785866447,1153981434332852712,722919835385951370578

%N a(n) = denominator of b(n), where b(1)=1, b(n+1) = sum{k|n} 1/b(k).

%H G. C. Greubel, <a href="/A127940/b127940.txt">Table of n, a(n) for n = 1..68</a>

%e {b(n)}:1,1,2,3/2,8/3,11/8,71/22,93/71,...So b(7) = sum{k|6} 1/b(k) = 1/b(1) + 1/ b(2) + 1/b(3) + 1/b(6) = 1 + 1 + 1/2 + 8/11 = 71/22.

%t f[l_List] := Block[{n = Length[l], d = Divisors[n]},Append[l, Sum[1/l[[d[[i]]]], {i, Length[d]}]]];Denominator[Nest[f, {1}, 22]] (* _Ray Chandler_, Feb 09 2007 *)

%Y Cf. A127939 (numerators).

%K frac,nonn

%O 1,4

%A _Leroy Quet_, Feb 08 2007

%E Extended by _Ray Chandler_, Feb 09 2007