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.

0, 22, 25, 38, 41, 70, 73, 134, 137, 237, 243, 262, 265, 365, 371, 429, 435, 461, 467, 492, 494, 498, 501, 518, 521, 621, 627, 685, 691, 717, 723, 748, 750, 754, 757, 813, 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

Offset: 1

Author

Andres M. Torres, Jul 10 2013

Keywords

• nonn
• base

Comments

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.

Examples

			0  = r(s(0)) = s(r(0))  = r(0)  = s(0)  = 0.
22 = r(s(22))= s(r(22)) = r(19) = s(25) = 22.
25 = r(s(25))= s(r(25)) = r(22) = s(28) = 25.
38 = r(s(38))= s(r(38)) = r(35) = s(41) = 38.
		

Links

• Andres M. Torres, Table of n, a(n) for n = 1..10000

Crossrefs

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

Programs

PARI

npbc(n) = n + hammingweight(n)
nmbc(n) = n - hammingweight(n)
isok(n) = (n == npbc(nmbc(n))) && (n == nmbc(npbc(n))) \\ Michel Marcus, Aug 08 2013

Formula

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

Extensions

Offset changed from 0 to 1 by Michel Marcus, Aug 08 2013