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a(n) = number of k with 0 < 2*k < n-1 such that a(n-k) AND a(n-2*k) = a(n-k) (where AND denotes the bitwise AND operator).
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%I #12 May 10 2019 18:10:05

%S 0,0,1,0,1,1,3,0,2,1,2,1,3,3,4,0,3,2,4,2,4,3,5,1,3,3,7,2,4,4,7,0,4,3,

%T 6,5,7,4,8,2,6,4,12,1,10,5,7,2,7,1,9,3,5,6,9,4,7,3,7,3,11,5,8,3,8,4,

%U 10,3,11,6,11,1,9,4,11,8,10,8,13,2,11,7,15

%N a(n) = number of k with 0 < 2*k < n-1 such that a(n-k) AND a(n-2*k) = a(n-k) (where AND denotes the bitwise AND operator).

%C This sequence has similarities with A317420.

%H Rémy Sigrist, <a href="/A317922/b317922.txt">Table of n, a(n) for n = 1..50000</a>

%H Rémy Sigrist, <a href="/A317922/a317922.png">Scatterplot of the first 10000000 terms</a>

%H Rémy Sigrist, <a href="/A317922/a317922_1.png">Colored scatterplot of the first 10000000 terms</a> (where the color is function of the 2-adic valuation of n (A001511))

%H Rémy Sigrist, <a href="/A317922/a317922_2.png">Colored scatterplot of the first 10000000 terms</a> (where the color is function of A000265(n) mod 16)

%H Rémy Sigrist, <a href="/A317922/a317922.txt">C++ program for A317922</a>

%e For n = 5:

%e - a(5-1) AND a(5-2) = 0 AND 1 = 0 = a(5-1),

%e - a(5-2) AND a(5-4) = 1 AND 0 = 0 <> a(5-2),

%e - hence a(5) = 1.

%o (C++) See Links section.

%Y Cf. A000265, A001511, A317420.

%K nonn,base

%O 1,7

%A _Rémy Sigrist_, Aug 11 2018