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 A227408 Set of all n,  where n = r(s(n)) = s(r(n)), given that r(n) = n+bitcount(n),  s(n) = n-bitcount(n), and bitcount(n) is the count of binary 1's in n. 3

%I

%S 0,22,25,38,41,70,73,134,137,237,243,262,265,365,371,429,435,461,467,

%T 492,494,498,501,518,521,621,627,685,691,717,723,748,750,754,757,813,

%U 819,845,851,876,878,882,885,909,915,940,942,946,949,972,974,978,981,988,995,1002,1009,1030,1033,1133,1139,1197,1203,1229

%N Set of all n, where n = r(s(n)) = s(r(n)), given that r(n) = n+bitcount(n), s(n) = n-bitcount(n), and bitcount(n) is the count of binary 1's in n.

%C This is a simple sequence where the nesting of functions r(n), and s(n), are grouped in a special way: n = r(s(n)) = s(r(n)), and those three values must be equal.

%H Andres M. Torres, <a href="/A227408/b227408.txt">Table of n, a(n) for n = 1..10000</a>

%F Find all n, such that: n = r(s(n)) = s(r(n)), where r(n) = n+bitcount(n) and s(n) = n-bitcount(n)

%e 0 = r(s(0)) = s(r(0)) = r(0) = s(0) = 0.

%e 22 = r(s(22))= s(r(22)) = r(19) = s(25) = 22.

%e 25 = r(s(25))= s(r(25)) = r(22) = s(28) = 25.

%e 38 = r(s(38))= s(r(38)) = r(35) = s(41) = 38.

%o (PARI) npbc(n) = n + hammingweight(n)

%o nmbc(n) = n - hammingweight(n)

%o isok(n) = (n == npbc(nmbc(n))) && (n == nmbc(npbc(n))) \\ _Michel Marcus_, Aug 08 2013

%Y Cf. A055938, A010061, A010062, A227359, A227361.

%K nonn,base

%O 1,2

%A _Andres M. Torres_, Jul 10 2013

%E Offset changed from 0 to 1 by _Michel Marcus_, Aug 08 2013

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Last modified June 19 06:47 EDT 2019. Contains 324218 sequences. (Running on oeis4.)