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A218543
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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.
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11
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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
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OFFSET
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0,4
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COMMENTS
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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.)
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LINKS
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FORMULA
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EXAMPLE
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(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.
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PROG
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(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 (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))
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CROSSREFS
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Analogous sequence for factorial number system: A219663.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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