A255064 Number of times an odious number is encountered when iterating from 2^(n+1)-2 to (2^n)-2 with the map x -> x - (number of runs in binary representation of x).

0, 1, 1, 1, 3, 4, 9, 15, 29, 45, 94, 155, 318, 548, 1088, 1976, 3812, 7115, 13617, 25733, 49247, 93739, 179691, 343816, 660735, 1270112, 2448975, 4727786, 9146539, 17717760, 34366228, 66718749, 129619199, 251958752, 489959621, 953155315, 1854898028

Offset: 0

Author

Antti Karttunen, Feb 14 2015

Keywords

• nonn

Examples

			For n=0 we count the odious numbers (A000069) found in range A255056(0..0), and A255056(0) = 0 is not an odious number, thus a(0) = 0.
For n=1 we count the odious numbers in range A255056(1..1), and A255056(1) = 2 is an odious number, thus a(1) = 1.
For n=2 we look at the numbers in range A255056(2..3), i.e. 4 and 6 and while 4 is an odious number, 6 is not, thus a(2) = 1.
For n=5 we look at the numbers in range A255056(12..20) which are (32, 36, 42, 46, 50, 54, 58, 60, 62), or if we take them in the order the come when iterating A236840 (as in A255066(12..20): 62, 60, 58, 54, 50, 46, 42, 36, 32), that is, we start iterating with map m(n) = A236840(n) from the initial value (2^(5+1))-2 = 62. Thus we get m(62) = 60, m(60) = 58, m(58) = 54, m(54) = 50, m(50) = 46, m(46) = 42, m(42) = 36 and finally m(36) = 32 which is (2^5). Of the nine numbers encountered, only 62, 50, 42 and 32 are odious numbers, thus a(5) = 4.
		

Crossrefs

Cf. A000069, A005811, A010060, A236840, A254114, A255056, A255063, A255066, A255071.

Similar sequences: A255126, A218543.

Programs

PARI

\\ Use the Pari-code given in A255063.

Scheme

(define (A255064 n) (if (zero? n) n (let loop ((i (- (expt 2 (+ 1 n)) 4)) (s (modulo n 2))) (cond ((pow2? (+ 2 i)) s) (else (loop (- i (A005811 i)) (+ s (A010060 i))))))))

Scheme

(define (A255064 n) (add A254114 (A255062 n) (A255061 (+ 1 n))))

Scheme

(define (A255064 n) (add (COMPOSE A010060 A255066) (A255062 n) (A255061 (+ 1 n))))

Formula

a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A254114(k).

a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A010060(A255066(k)).

Other identities. For all n >= 1:

a(n) = A255071(n) - A255063(n).