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a(n) = largest triangular number dividing n.
2

%I #13 Sep 03 2017 03:25:24

%S 1,1,3,1,1,6,1,1,3,10,1,6,1,1,15,1,1,6,1,10,21,1,1,6,1,1,3,28,1,15,1,

%T 1,3,1,1,36,1,1,3,10,1,21,1,1,45,1,1,6,1,10,3,1,1,6,55,28,3,1,1,15,1,

%U 1,21,1,1,66,1,1,3,10,1,36,1,1,15,1,1,78,1,10,3,1,1,28,1,1,3,1,1,45,91,1,3,1

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

%H Harvey P. Dale, <a href="/A115017/b115017.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = A083312(n) *(A083312(n) + 1)/2.

%e a(12)=6 because the triangular numbers dividing 12 are 1,3 and 6.

%p a:=proc(n) local P,j; P:={}: for j from 1 to n do if type(n/(j*(j+1)/2),integer)=true then P:=P union {j*(j+1)/2} else P:=P: fi od: P[nops(P)]; end: seq(a(n),n=1..105); # _Emeric Deutsch_, Mar 01 2006

%t With[{trnos=Accumulate[Range[100]]},Table[Last[Select[trnos,Divisible[ n,#]&]],{n,100}]] (* _Harvey P. Dale_, Nov 08 2011 *)

%Y Cf. A000217, A083312.

%K nonn

%O 1,3

%A _Leroy Quet_, Feb 23 2006

%E More terms from _Emeric Deutsch_, Mar 01 2006