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

1,2

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.

Links

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

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)

Example

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.

Prog

(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

Crossrefs

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

Sequence in context: A178423 A108632 A045096 * A303304 A234540 A352158

Adjacent sequences: A227405 A227406 A227407 * A227409 A227410 A227411

Keyword

nonn,base

Author

Andres M. Torres, Jul 10 2013

Extensions

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

Status

approved