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a(0)=1, a(n) = a(n-1) + a(2*n AND n), where AND is the bitwise AND operator.
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%I #15 Aug 06 2025 17:41:17

%S 1,2,3,6,7,8,15,30,31,32,33,36,67,98,165,330,331,332,333,336,337,338,

%T 345,360,691,1022,1353,1686,2377,3068,5445,10890,10891,10892,10893,

%U 10896,10897,10898,10905,10920,10921,10922,10923,10926,10957,10988,11055,11220,22111,33002

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

%H Robert Israel, <a href="/A215488/b215488.txt">Table of n, a(n) for n = 0..10000</a>

%p f:= proc(n) option remember;

%p procname(n-1) + procname(Bits:-And(2*n,n))

%p end proc:

%p f(0):= 1:

%p seq(f(i),i=0..100); # _Robert Israel_, Dec 29 2016

%t A215488[n_] := A215488[n] = If[n == 0, 1, A215488[n-1] + A215488[BitAnd[2*n, n]]];

%t Array[A215488, 50, 0] (* _Paolo Xausa_, Aug 06 2025 *)

%o (Python)

%o a = [1]*1000

%o for n in range(1,777):

%o print(a[n-1], end=', ')

%o a[n]= a[n-1] + a[2*n & n]

%Y Cf. A213370.

%K nonn,base,easy

%O 0,2

%A _Alex Ratushnyak_, Aug 13 2012