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Product_{n>=1} (1 + x^n)^a(n) = 1 + Sum_{n>=1} phi(n) * x^n, where phi = A000010.
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%I #5 Oct 27 2019 13:44:09

%S 1,1,1,1,2,-2,4,-3,4,-7,14,-21,30,-38,50,-79,128,-190,286,-419,598,

%T -895,1386,-2121,3178,-4733,7122,-10796,16414,-25011,38056,-57722,

%U 87568,-133308,203618,-311318,475536,-726069,1109718,-1698185,2601166,-3987305,6114666,-9378656,14389676

%N Product_{n>=1} (1 + x^n)^a(n) = 1 + Sum_{n>=1} phi(n) * x^n, where phi = A000010.

%C Inverse weigh transform of A000010.

%p b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,

%p add(binomial(a(i), j)*b(n-i*j, i-1), j=0..n/i)))

%p end:

%p a:= proc(n) option remember; numtheory[phi](n)-b(n, n-1) end:

%p seq(a(n), n=1..45); # _Alois P. Heinz_, Oct 27 2019

%t b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[Binomial[a[i], j] b[n - i j, i - 1], {j, 0, n/i}]]]; a[n_] := a[n] = EulerPhi[n] - b[n, n - 1]; Array[a, 45]

%Y Cf. A000010, A299069, A320778.

%K sign

%O 1,5

%A _Ilya Gutkovskiy_, Oct 27 2019