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%I #40 Apr 14 2021 01:17:05
%S 1,11,11,111,101,111,111,1111,1001,1111,1011,1111,1101,1111,1111,
%T 11111,10001,11011,10011,11111,10101,11111,10111,11111,11001,11111,
%U 11011,11111,11101,11111,11111,111111,100001,110011,100011,111111,100101,110111,100111,111111,101001,111111,101011,111111,101101,111111,101111,111111,110001,111011,110011,111111
%N The sum of the divisors of n in base-2 lunar arithmetic.
%C More precisely, in base-2 lunar arithmetic, the lunar sum of the lunar divisors of the n-th nonzero binary number.
%C Theorem: a(n) = binary representation of n iff n is odd.
%H D. Applegate, M. LeBrun and N. J. A. Sloane, <a href="http://arxiv.org/abs/1107.1130">Dismal Arithmetic</a> [Note: we have now changed the name from "dismal arithmetic" to "lunar arithmetic" - the old name was too depressing]
%H N. J. A. Sloane, <a href="/A188548/a188548.txt">Table giving n (written in base 10), n (written in base 2), a(n) (written in base 2), a(n) (written in base 10)</a>
%H <a href="/index/Di#dismal">Index entries for sequences related to dismal (or lunar) arithmetic</a>
%e The 4th binary number is 100 which has lunar divisors 1, 10, 100, whose lunar sum is 111, so a(4)=111.
%e The 5th binary number is 101 which has lunar divisors 1 and 101, whose lunar sum is 101, so a(5)=101.
%e It might be tempting to conjecture that if n is even then a(n) = 111...111, but a(18)=11011 shows that this is false (see A190149).
%Y Cf. A067399 (number of divisors), A190149, A190632.
%K nonn,base
%O 1,2
%A _N. J. A. Sloane_, Apr 04 2011
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