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A108618
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A quaternion-generated sequence calculated using the rules given in the comment box with initial seed x = .5'i + .5'j + .5'k + .5e; version: "tes".
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16
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1, 2, -1, -2, -3, -6, -6, 1, 4, 3, 0, -5, -10, -8, 3, 8, 5, -2, -9, -12, -6, 7, 16, 10, -9, -18, -11, 4, 15, 14, -2, -16, -20, -3, 14, 17, 6, -12, -24, -11, 10, 21, 14, -8, -22, -20, 3, 20, 17, -2, -21, -24, -6, 19, 28, 10, -21, -36, -18, 19, 40, 22, -21, -42, -23, 16, 39, 26, -14, -40, -32, 9, 38, 29, -8, -39, -36, 2, 36, 38, -1, -38
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| Set y = x = .5'i + .5'j + .5'k + .5e Define a(0) = 1 (this is twice the coefficient of the unit e in x), then "loop" steps 1-5, below. a(n) is given by twice the coefficient of e (the unit) in y from step 4 inside of the n-th loop. Step 1 (Loop 1): Calculate x*y Result: x*y = .5'i + .5'j + .5'k - .5e Step 2 (Loop 1): Add the fractional parts of the real coefficient basis vectors of x*y (i.e. 'i, 'j, 'k, e) Result: .5 + .5 + .5 - .5 = 1 = s Step 3 (Loop 1): Calculate x + x*y + se Result .5'i + .5'j + .5'k + .5e + (.5'i + .5'j + .5'k - .5e) + se = 'i + 'j + 'k + e. Step 4 (Loop 1): Set y equal to the result from Step 3. Result: y = 'i + 'j + 'k + e; thus a(1) = 2*1 = 2 Step 5 (Loop 1): Return to Step 1 Step 1 (Loop 2): Result: x*y = 'i + 'j + 'k - e Step 2 (Loop 2): Result: s = 0 Step 3 (Loop 2): 1.5'i + 1.5'j + 1.5'k -.5e Step 4 (Loop 2): y = 1.5'i + 1.5'j + 1.5'k -.5e; thus a(2) = 2*(-.5) = -1 **Loop 1** + 'i + 'j + 'k + e **Loop 2** + 1.5'i + 1.5'j + 1.5'k - .5e **Loop 3** + 'i + 'j + 'k - e **Loop 4** + .5'i + .5'j + .5'k - 1.5e **Loop 5** - 3e **Loop 6** - 'i - 'j - 'k - 3e **Loop 7** - 1.5'i - 1.5'j - 1.5'k + .5e **Loop 8** + 2e **Loop 9** + 1.5'i + 1.5'j + 1.5'k + 1.5e **Loop 10** + 2'i + 2'j + 2'k **Loop 11** + 1.5'i + 1.5'j + 1.5'k - 2.5e **Loop 12** - 5e
Notice the horizontal line segments in the graph of (a(n)) against the natural numbers. These may be referred to as "Gerald's diamonds" (after Gerald McGarvey, who pointed them out shortly after this sequence was submitted). It could be an interesting task to find the approximate area of these diamonds and compare to the approximate area of the other diamonds.
Contribution from Benoit Jubin (benoit_jubin(AT)yahoo.fr), Aug 12 2009: (Start)
Define the function f on the integers to be the odd function such that for n>=0, f(2n)=0 and f(2n+1)=1. Define the sequences a and b by
a(0)=b(0)=0,
a(n+1) = 1 + (a(n)-3b(n))/2 + f((a(n)-3b(n))/2) + 3 f((a(n)+b(n))/2),
b(n+1) = 1 + (a(n)+b(n))/2.
Then (with an offset shifted by 1), a=A108618 and b=A108619. (End)
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LINKS
| C. Dement, Table of n, a(n) for n = 0..10000
C. Dement, Plot of A108618 against A108619 (patch on)
C. Dement, Plot of A108618 against A108619 (patch off)
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CROSSREFS
| Cf. A108619, A108620, A108621.
Sequence in context: A198123 A106576 A128474 * A097719 A056493 A001371
Adjacent sequences: A108615 A108616 A108617 * A108619 A108620 A108621
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KEYWORD
| sign
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AUTHOR
| Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Jun 12 2005
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