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A299119 Positive solution to 2^(n-1) = (1/n) * Sum_{d|n} a(d) * a(n/d).

%I A299119 #12 Mar 07 2018 10:17:16
%S A299119 1,2,6,14,40,84,224,484,1134,2480,5632,12036,26624,56896,122640,
%T A299119 261078,557056,1176876,2490368,5237360,11008704,23057408,48234496,
%U A299119 100635144,209714400,436154368,905962860,1878931264,3892314112,8052800160,16642998272,34359209436
%N A299119 Positive solution to 2^(n-1) = (1/n) * Sum_{d|n} a(d) * a(n/d).
%C A299119 For prime p, a(p) = 2^(p-2)*p. - _Jon E. Schoenfield_, Feb 03 2018
%H A299119 Alois P. Heinz, <a href="/A299119/b299119.txt">Table of n, a(n) for n = 1..1000</a>
%p A299119 with(numtheory):
%p A299119 a:= proc(n) option remember; `if`(n=1, 1, n*2^(n-2)-
%p A299119        add(a(d)*a(n/d), d=divisors(n) minus {1, n})/2)
%p A299119     end:
%p A299119 seq(a(n), n=1..35);  # _Alois P. Heinz_, Mar 07 2018
%t A299119 nn=50;
%t A299119 sys=Table[2^(n-1)*n==Sum[a[d]*a[n/d],{d,Divisors[n]}],{n,nn}];
%t A299119 Array[a,nn]/.Solve[sys,Array[a,nn]][[2]]
%Y A299119 Cf. A000005, A000010, A000740, A001787, A018804, A023900, A029935, A034691, A046643, A059966, A228369, A257098, A296302, A298971, A299149, A299151.
%K A299119 nonn
%O A299119 1,2
%A A299119 _Gus Wiseman_, Feb 03 2018