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%I #2 Mar 31 2012 13:50:44
%S 1,0,1,2,2,2,3,3,4,4,6,6,8,8,11,12,13,15,14,17,18,19,20,21,23,23,24,
%T 26,28,32,31,32,35,36,37,43,43,44,44,45,47,50,51,53,53,56,56,57,57,61,
%U 62,62,65,68,73,76,77,78,83,83,84,84,87,88,90,92,94,94,97,98,98,101,101
%N Number of 1's in partition of n-th Ulam number into sum of two distinct Ulam numbers.
%e For n=7: A002858(7) = 11 = 3+8 = (1+2)+(2+6) = (1+2)+(2+(2+4)) = (1+2)+(2+(2+(1+3))) = (1+2)+(2+(2+(1+(1+2)))). Number of 1's in (1+2)+(2+(2+(1+(1+2)))) is 3, so a(7) = 3.
%Y Cf. A002858.
%K easy,nonn
%O 1,4
%A _Naohiro Nomoto_, Feb 22 2003