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Sum of previous terms divided by their distance from n
0

%I #2 Mar 31 2012 10:25:22

%S 1,3,6,11,20,33,54,87,142,226,360,571,907,1438,2278,3606,5708,9032,

%T 14293,22614,35778,56604,89550,141670,224122,354560,560907,887348,

%U 1403768,2220731,3513149,5557726,8792197,13909055,22003808

%N Sum of previous terms divided by their distance from n

%F a(n) = n + floor(a(n-1)/1) + floor(a(n-2)/2) + ... + floor(a(1)/(n-1))

%e a(1)=1 by definition

%e a(4)=4+a(3)/1+a(2)/2+a(1)/3=4+floor(6/1)+floor(3/2)+floor(1/3)=11

%K nonn

%O 2,2

%A _Dominick Cancilla_, Aug 09 2010