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%I #12 Jan 17 2019 17:20:01
%S 1,2,4,3,8,12,6,4,16,24,24,24,12,18,8,5,32,48,48,48,48,72,48,40,24,36,
%T 36,36,16,24,10,6,64,96,96,96,96,144,96,80,96,144,144,144,96,144,80,
%U 60,48,72,72,72,72,108,72,60,32,48,48,48,20,30,12,7,128,192,192,192,192,288,192,160,192,288,288,288,192,288,160,120
%N a(n) = A323505(n) / A246660(n).
%H Antti Karttunen, <a href="/A323506/b323506.txt">Table of n, a(n) for n = 0..16383</a>
%F a(n) = A323505(n) / A246660(n).
%F For n > 1, a(2n) = 2*a(n).
%e This sequence can be represented as a binary tree, as both A323505 and A246660 have similar tree structures:
%e 1
%e |
%e ...................2....................
%e 4 3
%e 8......../ \........12 6........./ \.......4
%e / \ / \ / \ / \
%e / \ / \ / \ / \
%e / \ / \ / \ / \
%e 16 24 24 24 12 18 8 5
%e 32 48 48 48 48 72 48 40 24 36 36 36 16 24 10 6
%e etc.
%o (PARI)
%o A001511(n) = (1+valuation(n,2));
%o A036987(n) = !bitand(n,1+n);
%o A323505(n) = if(!n,1,if(!(n%2), 2*A323505(n/2), (A001511(n+1)+1-A036987(n))*A323505((n-1)/2)));
%o A246660(n) = { my(i=0, p=1); while(n>0, if(n%2, i++; p = p * i, i = 0); n = n\2); p; };
%o A323506(n) = (A323505(n)/A246660(n));
%Y Cf. A246660, A323505.
%K nonn
%O 0,2
%A _Antti Karttunen_, Jan 16 2019
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