A218543 Number of times when an odd number is encountered, when going from 2^(n+1)-1 to (2^n)-1 using the iterative process described in A071542.

0, 1, 1, 2, 3, 6, 9, 18, 31, 54, 93, 167, 306, 574, 1088, 2081, 3998, 7696, 14792, 28335, 54049, 102742, 194948, 369955, 703335, 1340834, 2563781, 4915378, 9444799, 18180238, 35047841, 67660623, 130806130, 253252243, 491034479, 953404380, 1853513715, 3607440034

Offset

0,4

Comments

Ratio a(n)/A213709(n) develops as: 0, 1, 0.5, 0.666..., 0.6, 0.666..., 0.529..., 0.6, 0.574..., 0.551..., 0.520..., 0.506..., 0.498..., 0.499..., 0.503..., 0.511..., 0.521..., 0.531..., 0.539..., 0.545..., 0.547..., 0.546..., 0.542..., 0.536..., 0.531..., 0.525..., 0.520..., 0.516..., 0.512..., 0.508..., 0.504..., 0.501..., 0.498..., 0.497..., 0.495..., 0.495..., 0.495..., 0.495..., 0.495..., 0.496..., 0.496..., 0.497..., 0.497..., 0.498..., 0.498..., 0.498..., 0.497..., 0.497...

Ratio a(n)/A218542(n) develops as follows from n>=2 onward:

1, 2, 1.5, 2, 1.125, 1.5, 1.348..., 1.227..., 1.081..., 1.025..., 0.994..., 0.997..., 1.013..., 1.045..., 1.086..., 1.132..., 1.172..., 1.198..., 1.208..., 1.201..., 1.182..., 1.157..., 1.131..., 1.107..., 1.085..., 1.065..., 1.047..., 1.031..., 1.016..., 1.004..., 0.994..., 0.986..., 0.981..., 0.979..., 0.978..., 0.979..., 0.981..., 0.983..., 0.986..., 0.988..., 0.989..., 0.990..., 0.991..., 0.991..., 0.989..., 0.987...

Observation: A179016 seems to alternatively slightly favor the odd numbers and then again the even numbers, at least for the terms computed so far.

Please plot this sequence against A218542 in the "ratio mode" (given as a link) to see how smoothly (almost "continuously") the ratios given above develop.

What is the reason for that smoothness? (Conjecture: The distribution of "tendrils", i.e. finite subtrees in the beanstalk and its almost fractal nature? Cf: A218787.)

Links

Antti Karttunen, Table of n, a(n) for n = 0..47

OEIS Server, Sequence plotted together with A218542 showing how their ratio develops.

Formula

a(n) = Sum_{i=A218600(n) .. (A218600(n+1)-1)} A213729(i)

Example

(2^0)-1 (0) is reached from (2^1)-1 (1) with one step by subtracting A000120(1) from 1. Zero is not an odd number, so a(0)=0.
(2^1)-1 (1) is reached from (2^2)-1 (3) with one step by subtracting A000120(3) from 3. One is an odd number, so a(1)=1.
(2^2)-1 (3) is reached from (2^3)-1 (7) with two steps by first subtracting A000120(7) from 7 -> 4, and then subtracting A000120(4) from 4 -> 3. Four is not an odd number, but three is, so a(2)=1.

Prog

(Scheme with memoizing definec-macro): (definec (A218543 n) (if (zero? n) 0 (let loop ((i (- (expt 2 (1+ n)) n 2)) (s 0)) (cond ((pow2? (1+ i)) (+ s (modulo i 2))) (else (loop (- i (A000120 i)) (+ s (modulo i 2))))))))
(define (pow2? n) (and (> n 0) (zero? (A004198bi n (- n 1))))) ;; A004198 is bitwise AND
;; Or with a summing-function add:
(define (A218543v2 n) (add A213729 (A218600 n) (-1+ (A218600 (1+ n)))))
(define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))

Crossrefs

a(n) = A213709(n)-A218542(n). Cf. A213733, A218787, A218789.

Analogous sequence for factorial number system: A219663.

Sequence in context: A059966 A095718 A038751 * A266925 A018518 A091326

Adjacent sequences: A218540 A218541 A218542 * A218544 A218545 A218546

Keyword

nonn

Author

Antti Karttunen, Nov 02 2012

Status

approved