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%I #15 Sep 14 2021 04:43:05
%S 0,0,0,0,1,1,1,1,1,1,1,1,2,1,2,2,2,1,1,2,1,1,2,2,1,2,1,2,1,2,3,3,1,3,
%T 1,3,1,3,3,1,3,1,3,1,1,3,3,3,3,1,1,3,1,3,1,3,1,3,1,1,3,1,3,3,1,1,3,1,
%U 2,1,1,3,2,3,4,3,4,2,1,4,4,1,1,3,4,3
%N a(n) = prime(n) mod floor(prime(n) / n).
%H Simon Strandgaard, <a href="/A347342/a347342.png">Plot of 1000 terms</a>
%F a(n) = A000040(n) mod A038605(n).
%e a(1) = 2 mod floor( 2 / 1) = 2 mod 2 = 0,
%e a(2) = 3 mod floor( 3 / 2) = 3 mod 1 = 0,
%e a(3) = 5 mod floor( 5 / 3) = 5 mod 1 = 0,
%e a(4) = 7 mod floor( 7 / 4) = 7 mod 1 = 0,
%e a(5) = 11 mod floor(11 / 5) = 11 mod 2 = 1.
%t A347342[n_] := Mod[Prime[n] , Floor[Prime[n]/n]]; Table[A347342[n], {n, 1, 86}] (* _Robert P. P. McKone_, Aug 27 2021 *)
%o (PARI) a(n) = prime(n) % (prime(n) \ n);
%o (Ruby) require 'prime'
%o values = []
%o Prime.first(30).each_with_index do |prime,i|
%o values << prime % (prime/(i+1))
%o end
%o p values
%Y Cf. A000040, A038605.
%K nonn
%O 1,13
%A _Simon Strandgaard_, Aug 27 2021
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