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Number of 2's in partition of n-th Ulam number into sum of two distinct Ulam numbers.
1

%I #2 Mar 31 2012 13:50:44

%S 0,1,1,1,2,3,4,5,6,7,10,11,14,15,18,18,20,21,24,26,27,29,31,33,37,38,

%T 39,40,43,47,50,53,55,56,59,66,67,68,69,72,75,78,79,83,84,90,91,92,93,

%U 96,98,99,104,107,118,120,121,123,128,129,136,137,138,141,146,149,153,154

%N Number of 2'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 2's in (1+2)+(2+(2+(1+(1+2)))) is 4, so a(7) = 4.

%Y Cf. A002858.

%K easy,nonn

%O 1,5

%A _Naohiro Nomoto_, Feb 22 2003