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a(n) = n/(largest triangular number dividing n).
1

%I #9 Oct 11 2019 14:11:28

%S 0,1,2,1,4,5,1,7,8,3,1,11,2,13,14,1,16,17,3,19,2,1,22,23,4,25,26,9,1,

%T 29,2,31,32,11,34,35,1,37,38,13,4,41,2,43,44,1,46,47,8,49,5,17,52,53,

%U 9,1,2,19,58,59,4,61,62,3,64,65,1,67,68,23,7,71,2,73,74,5,76,77,1,79,8,27,82

%N a(n) = n/(largest triangular number dividing n).

%F a(n) = n/(A115017(n))

%t f[n_] := Max[Select[Divisors[n], IntegerQ[(8# + 1)^(1/2)] & ]]; Table[n/f[n], {n, 0, 82}] (* _Ray Chandler_, Aug 29 2006 *)

%Y Cf. A000217, A083312, A115017.

%K nonn

%O 0,3

%A _Leroy Quet_, Aug 24 2006

%E Extended by _Ray Chandler_, Aug 29 2006