A227361 If n is even, then a(n) = n + bitsum(n), else a(n) = n - bitsum(n), where bitsum(n) is the count of binary 1's in n, A000120.

0, 0, 3, 1, 5, 3, 8, 4, 9, 7, 12, 8, 14, 10, 17, 11, 17, 15, 20, 16, 22, 18, 25, 19, 26, 22, 29, 23, 31, 25, 34, 26, 33, 31, 36, 32, 38, 34, 41, 35, 42, 38, 45, 39, 47, 41, 50, 42, 50, 46, 53, 47, 55, 49, 58, 50, 59, 53, 62, 54, 64, 56, 67, 57, 65, 63, 68, 64, 70, 66, 73, 67, 74, 70, 77, 71, 79, 73, 82, 74, 82, 78, 85, 79, 87, 81, 90, 82, 91, 85

Offset: 0

Author

Andres M. Torres, Jul 08 2013

Keywords

• nonn
• easy
• base

Comments

I gathered together some interesting statistics for this seq A227361:

Within the first 200000001 members of this sequence, only 70 were repeated 4 times, and then only when n > 2 million. None were repeated 5 times.

The first value to become repeated 3 times is 50, occurring at indexes (n=) 46, 48, and 55.

The first value to become repeated 4 times is 2097170, occurring at indexes (n=) 2097150, 2097166, 2097168, and 2097175.

The total count of those only occurring once is 96226727, or about 48.11 %.

Total count of those repeated 2 times is 45055158.

Total count of those repeated 3 times is 4554221, or about 2.28 %.

Total count of those repeated 4 times is 70 (extremely low).

Repeatedly applying this BitStoneA(v) function to values in a recursive (nested) style has shown that only 21 starting values shall become zero. These are those values: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 23, 27. All other values shall cycle forever in small loops.

274877906962 is the smallest number that occurs 5 times. - Donovan Johnson, Jul 27 2013

Examples

			a(0) = 0 because 0 is even, so 0 + bitsum(0) = 0.
a(1) = 0 because 1 is odd, so 1 - bitsum(1) = 0.
a(2) = 3 because 2 is even, so 2 + bitsum(2) = 3.
a(3) = 1 because 3 is odd, so 3 - bitsum(3) = 1.
		

Links

• Andres M. Torres, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000120, A055938, A010061, A010062.

Programs

Blitz3D

;; Each a(n) is generated simply as follows:  a(n) = BitStoneA(n)
Function BitStoneA(n)
         If  (n Mod 2)              ;; if is odd
                 Return n-bitsum(n)
         Else                       ;; if is even
                 Return n+bitsum(n)
         End If
End Function
;; --- Or, If n is even, then return n+A000120(n), else return n-A000120(n), where A000120(n) = bitsum(n)

Mathematica

Table[n + (-1)^n DigitCount[n, 2, 1], {n, 0, 127}] (* Alonso del Arte, Jul 08 2013 *)

PARI

a(n)=n+(-1)^(n%2)*hammingweight(n) \\ Charles R Greathouse IV, Jul 09 2013

Formula

a(n) = n + (-1)^n * Sum_{j = 1 .. floor(log_2(n)) + 1} (floor(n/2^j + 1/2) - floor(n/2^j)). - Alonso del Arte, Jul 08 2013, based on one of Hieronymus Fischer's formulas for A000120.