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a(n+1) = k(a(n), n), where k(m, n) = (m*n) mod (m+n) and with a(1) = 5.
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%I #34 Sep 30 2019 09:21:17

%S 5,5,3,3,5,5,8,11,12,3,4,14,12,6,4,3,10,8,14,2,18,27,6,22,22,33,32,38,

%T 8,10,20,8,16,38,68,11,20,56,60,63,48,10,4,31,14,40,34,59,50,74,104,

%U 34,48,19,4,43,32,44,2,57,27,63,31,73,14,41,31,19,74,101,59,29,68,29,86,10,72,31,20,95,75,147,146,210,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0

%N a(n+1) = k(a(n), n), where k(m, n) = (m*n) mod (m+n) and with a(1) = 5.

%F a(k) = 0 for k >= 85. - _Jinyuan Wang_, Aug 26 2019

%F 5 is one of the first starting values that produces rather long sequence until collapses to zero.

%t a[1] = 5; a[n_] := a[n] = Mod[a[n - 1] * (n - 1), a[n - 1] + n - 1]; Array[a, 100] (* _Amiram Eldar_, Aug 29 2019 *)

%o (Python)

%o def k(a,b):

%o return (a*b)%(a+b)

%o a = [5]

%o for n in range(1, 100):

%o a.append(k(a[-1], n))

%o print(a)

%K nonn,easy

%O 1,1

%A _Stefan Gog_, Aug 23 2019