login
Number of representations of A005243(n) as sum of consecutive earlier terms in A005243.
5

%I #3 Mar 30 2012 18:50:54

%S 0,0,1,1,1,1,1,2,2,1,1,1,1,1,1,2,2,2,1,1,1,1,4,1,2,1,3,1,2,1,2,2,3,1,

%T 2,3,1,2,3,2,1,1,2,1,1,1,2,3,1,1,1,1,1,3,2,1,1,1,3,2,2,1,2,1,1,1,3,1,

%U 1,4,1,2,1,3,1,2,1,2,1,1,2,2,1,3,1,3,4,2,1,3,1,2,2,2,3,1,1,1,2,2,1,3,1,1,4

%N Number of representations of A005243(n) as sum of consecutive earlier terms in A005243.

%C A118165(n) = Min{m: a(m) = n};

%C for n>2: a(n) > 0 by definition of A005243.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HofstadterSequences.html">Hofstadter Sequences</a>

%e A005243(33) = 54 = 29+25 = Sum(A005243[17:18]) =

%e 19+18+17 = Sum(A005243[11:13]) = 14+11+10+8+6+5 =

%e Sum(A005243[4:9]), therefore a(33) = 3.

%K nonn

%O 1,8

%A _Reinhard Zumkeller_, Apr 13 2006