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a(1)=1; for n > 1, a(n) = a(n-1)/gcd(a(n-1),n) + n + 1.
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%I #33 Sep 08 2022 08:46:21

%S 1,4,8,7,13,20,28,16,26,24,36,16,30,30,18,26,44,41,61,82,104,75,99,58,

%T 84,69,51,80,110,42,74,70,104,87,123,78,116,97,137,178,220,153,197,

%U 242,288,191,239,288,338,220,272,121,175,230,102,108,94,106,166

%N a(1)=1; for n > 1, a(n) = a(n-1)/gcd(a(n-1),n) + n + 1.

%C n is a lower bound on a(n), furthermore n+3 is a lower bound if n > 2. This can easily be proved by induction. It appears that both the average value and the upper bound grow either linearly or slightly faster than linearly.

%H Seiichi Manyama, <a href="/A309665/b309665.txt">Table of n, a(n) for n = 1..10000</a>

%e a(4) = a(3)/gcd(a(3),4) + 4 + 1 = 8/gcd(8,4) + 5 = 8/4 + 5 = 2 + 5 = 7.

%t a[1] = 1; a[n_] := a[n] = a[n-1]/GCD[a[n - 1], n] + n + 1; Array[a, 60] (* _Amiram Eldar_, Aug 14 2019 *)

%o (Magma) [n le 1 select 1 else Self(n-1)/Gcd(Floor(Self(n-1)),n) + n + 1 : n in [1..60]]; // _Marius A. Burtea_, Aug 11 2019

%Y Cf. A133058.

%K nonn,look

%O 1,2

%A _Dennis Reichard_, Aug 11 2019