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A098294
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Smallest exponent of 2 which gives a power of 2 which is equal or bigger than (3/2)^n, n=0,1,...
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3
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0, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 10, 11, 12, 12, 13, 13, 14, 15, 15, 16, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 23, 24, 24, 25, 26, 26, 27, 27, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 34, 35, 36, 36, 37, 37, 38, 39, 39, 40, 40, 41, 41, 42, 43, 43, 44
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Stacking perfect fifths (the frequency ratio of a fifth is 3/2) this sequence determines into which octave the n-th fifth falls. For example, the third fifth, (3/2)^3, falls into the second octave, which means that it lies in the interval [2^1,2^2)=[2,4). The k-th octave comprises ratios in the interval [2^(k-1),2^k), k=1,2,...
Related to the initial number of sequential even terms in an "ideal" sequence under iteration of the 3x+1 Problem on a positive odd value m, where the piecewise function f is given by f(2*m)=m, f(2*m+1)=6*m+4, to ensure f^{<a href="https://oeis.org/A122437">A122437(n)</a>} (m)<m, where n>1 is the number of odds in the sequence (including m) and floor(1+n*(log3/log2))is the number of evens. An "ideal" sequence minimizes the effects of f(2*m+1) by following a certain order of even or odd terms along with the rules of the function. A representation of such sequences in terms of parity sequences for values n>=2 follows:
n=2, (o,e,e,o,e,e)
n=3, (o,e,e,o,e,o,e,e)
n=4, (o,e,e,e,o,e,o,e,o,e,e)
n=5, (o,e,e,e,o,e,o,e,o,e,o,e,e)
n=6, (o,e,e,e,e,o,e,o,e,o,e,o,e,o,e,e)
n=7, (o,e,e,e,e,e,o,e,o,e,o,e,o,e,o,e,o,e,e)
The pattern is clear, and the formula for the initial number of sequential even terms in each sequence is given by a(n) = floor(1+n*(log3/log2))-n for n>1, where the sum of the number of even and odd terms is given by <a href="https://oeis.org/A122437">A122437(n)</a> for n>1. Of course, most values m do not have sequences following this pattern of iteration under f. Also, the reason for placing an extra even term at the end of such sequences is to mitigate to some degree the effects of the possibility that the last odd term is only "slightly" larger than m, i.e., (3*m+1) / 4 < m for all m > 1. [Jeffrey R. Goodwin, Aug 25 2011]
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LINKS
| Pythagorean Scale.
Eric Weisstein's World of Music, Pythagorean Scale
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FORMULA
| 2^a(n) >= (3/2)^n but 2^(a(n) - 1)< (3/2)^n, n >= 0.
a(n)= ceiling(tau*n) with tau := ln(3)/ln(2) - 1 = .584962501..., n>=0.
a(n) = floor(1 + n * ln(3)/ln(2)) - n, n >= 1, [from Mike Winkler (mike.winkler(at)gmx.de), Dec 31 2010]
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EXAMPLE
| a(0)=0 because 2^0=1 = (3/2)^0 but 2^(-1)= 1/2 < 1. a(11)=7 because
2^7=128 > 86.497.. = (3/2)^11 but 2^6=64 < (3/2)^11.
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CROSSREFS
| Cf. A098295, A020914, A020857
Sequence in context: A055038 A085268 A194979 * A195119 A077467 A005378
Adjacent sequences: A098291 A098292 A098293 * A098295 A098296 A098297
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KEYWORD
| nonn,easy
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AUTHOR
| Wolfdieter Lang (wolfdieter.lang_AT_physik_DOT_uni-karlsruhe_DOT_de), Oct 18 2004
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