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%I #17 Jul 21 2021 09:18:36
%S 1,3,2,3,5,3,3,5,3,3,7,3,3,7,3,3,7,5,3,7,3,3,8,3,5,7,3,5,7,3,3,11,5,3,
%T 7,3,3,7,7,3,9,3,5,7,3,7,7,5,3,11,3,3,11,3,3,7,3,5,11,7,5,7,4,3,7,3,7,
%U 11,3,3,7,7,5,11,3,3,11,5,3,7,7,3,11,3,5
%N a(n) is the number of proper divisors of A053742(n) ending with 5.
%F a(n) = A346392(A053742(n)).
%e a(10) = 7 since there are 7 proper divisors of A053742(10) = 525 ending with 5: 5, 15, 25, 35, 75, 105 and 175.
%t a[n_]:=Length[Drop[Select[Divisors[25+50n], (Mod[#,10]==5&)], -1]]; Array[a, 90, 0]
%Y Cf. A032741, A053742, A346389 (ending with 6), A346392.
%K nonn,base
%O 0,2
%A _Stefano Spezia_, Jul 15 2021
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