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a(1) = 1; a(n+1) = -Sum_{d|n} d*a(d).
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%I #11 Nov 13 2018 08:37:14

%S 1,-1,1,-4,17,-86,514,-3599,28809,-259285,2592766,-28520427,342245654,

%T -4449193503,62288705445,-934330581764,14949289337033,

%U -254137918729562,4574482536873349,-86915168200593632,1738303364014465422,-36504370644303777464,803096154174654583783,-18471211546017055427010

%N a(1) = 1; a(n+1) = -Sum_{d|n} d*a(d).

%H Seiichi Manyama, <a href="/A321384/b321384.txt">Table of n, a(n) for n = 1..450</a>

%F L.g.f.: log(Product_{n>=1} (1 - x^n)^a(n)) = Sum_{n>=1} a(n+1)*x^n/n.

%F a(n) ~ -(-1)^n * c * (n-1)!, where c = 0.7144978951771230847588633755835851845867260778566988217176856019246992... - _Vaclav Kotesovec_, Nov 09 2018

%t a[1] = 1; a[n_] := a[n] = -Sum[d a[d], {d, Divisors[n - 1]}]; Table[a[n], {n, 24}]

%o (PARI) a(n) = if (n==1, 1, -sumdiv(n-1, d, d*a(d))); \\ _Michel Marcus_, Nov 09 2018

%Y Cf. A281487, A318372.

%K sign

%O 1,4

%A _Ilya Gutkovskiy_, Nov 08 2018