A227408 - OEIS

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

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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)