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a(n) = a(n-1) + (a(n-2) AND n).
0

%I #7 Jan 23 2013 17:06:14

%S 0,1,1,2,2,2,4,6,6,6,8,10,18,26,28,38,54,54,72,90,90,106,124,126,150,

%T 174,192,202,202,210,220,238,238,270,304,306,338,370,372,406,438,438,

%U 472,506,514,554,556,598,630,646,696,698,746,794,828,846,902,910,912,922,938,962,1004,1006,1070

%N a(n) = a(n-1) + (a(n-2) AND n).

%F a(0)=0, a(1)=1, a(n) = a(n-1) + (a(n-2) AND n), where AND is the bitwise AND operator.

%t RecurrenceTable[{a[0]==0,a[1]==1,a[n]==a[n-1]+BitAnd[a[n-2],n]},a,{n,70}] (* _Harvey P. Dale_, Jan 23 2013 *)

%o (Python)

%o prpr, prev = 0,1

%o for n in range(2,99):

%o . current = prev + (prpr & n)

%o . print prpr,

%o . prpr, prev = prev, current

%Y Cf. A182537.

%K nonn,base,easy

%O 0,4

%A _Alex Ratushnyak_, May 04 2012