The Overlord of the Über-Feral says: Welcome to my bijou bloguette. You can scroll down to sample more or simply:

• Read a Writerization at Random: RaWaR

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The Overlord of the Über-Feral says: Welcome to my bijou bloguette. You can scroll down to sample more or simply:

• Read a Writerization at Random: RaWaR

O.o.t.Ü.-F.: Not Just A Blog… A Key Lifestyle Commitment…

In maths, one thing leads to another. I wondered whether, in a spiral of integers, any number was equal to the digit-sum of the numbers on the route traced by moving to the origin first horizontally, then vertically. To illustrate the procedure, here is a 9×9 integer spiral containing 81 numbers:

| 65 | 64 | 63 | 62 | 61 | 60 | 59 | 58 | 57 | | 66 | 37 | 36 | 35 | 34 | 33 | 32 | 31 | 56 | | 67 | 38 | 17 | 16 | 15 | 14 | 13 | 30 | 55 | | 68 | 39 | 18 | 05 | 04 | 03 | 12 | 29 | 54 | | 69 | 40 | 19 | 06 |01| 02 | 11 | 28 | 53 | | 70 | 41 | 20 | 07 | 08 | 09 | 10 | 27 | 52 | | 71 | 42 | 21 | 22 | 23 | 24 | 25 | 26 | 51 | | 72 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 | | 73 | 74 | 75 | 76 | 77 | 78 | 79 | 80 | 81 |

Take the number 21, which is three places across and up from the bottom left corner of the spiral. The route to the origin contains the numbers 21, 22, 23, 8 and 1, because first you move right two places, then up two places. And 21 is what I call a route number, because 21 = 3 + 4 + 5 + 8 + 1 = digitsum(21) + digitsum(22) + digitsum(23) + digitsum(8) + digitsum(1). Beside the trivial case of 1, there are two more route numbers in the spiral:

58 = 13 + 14 + 6 + 7 + 7 + 6 + 4 + 1 = digitsum(58) + digitsum(59) + digitsum(60) + digitsum(61) + digitsum(34) + digitsum(15) + digitsum(4) + digitsum(1).

74 = 11 + 12 + 13 + 14 + 10 + 5 + 8 + 1 = digitsum(74) + digitsum(75) + digitsum(76) + digitsum(77) + digitsum(46) + digitsum(23) + digitsum(8) + digitsum(1).

Then I wondered about other possible routes to the origin. Think of the origin as one corner of a rectangle and the number being tested as the diagonal corner. Suppose that you always move away from the starting corner, that is, you always move up or right (or up and left, and so on, depending on where the corners lie). In a *x* by *y* rectangle, how many routes are there between the diagonal corners under those conditions?

It’s an interesting question, but first I’ve looked at the simpler case of an *n* by *n* square. You can encode each route as a binary number, with 0 representing a vertical move and 1 representing a horizontal move. The problem then becomes equivalent to finding the number of distinct ways you can arrange equal numbers of 1s and 0s. If you use this method, you’ll discover that there are two routes across the 2×2 square, corresponding to the binary numbers 01 and 10:

Across the 3×3 square, there are six routes, corresponding to the binary numbers 0011, 0101, 0110, 1001, 1010 and 1100:

Across the 4×4 square, there are twenty routes:

Across the 5×5 square, there are 70 routes:

Across the 6×6 and 7×7 squares, there are 252 and 924 routes:

After that, the routes quickly increase in number. This is the list for n = 1 to 14:

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600… (see A000984 at the Online Encyclopedia of Integer Sequences)

After that you can vary the conditions. What if you can move not just vertically and horizontally, but diagonally, i.e. vertically and horizontally at the same time? Now you can encode the route with a ternary number, or number in base 3, with 0 representing a vertical move, 1 a horizontal move and 2 a diagonal move. As before, there is one route across a 1×1 square, but there are three across a 2×2, corresponding to the ternary numbers 01, 2 and 10:

There are 13 routes across a 3×3 square, corresponding to the ternary numbers 0011, 201, 021, 22, 0101, 210, 1001, 120, 012, 102, 0110, 1010, 1100:

And what about cubes, hypercubes and higher?

“The recent election of Syriza in Greece offers a vibrant glimmer of hope for the future of social and economic democracy in Europe.” — from a letter to *The Guardian* by Judith Butler, Slavoj Žižek, Jacqueline Rose, *et al*.

Papyrocentric Performativity Presents:

• Sex/Dream Metaphors *– Extreme Metaphors: Selected Interviews with J.G. Ballard*, edited by Simon Sellars and Dan O’Hara (Fourth Estate 2014)

• DNAncient – *Neanderthal Man: In Search of Lost Genomes*, Svante Pääbo (Basic Books 2014)

• The Cult of Cthulhu – *H.P. Lovecraft: The Classic Horror Stories*, edited by Roger Luckhurst (Oxford University Press 2013)

• Rauc’ and Role – *Mortality*, Christopher Hitchens (Atlantic Books 2012)

• #BooksThatShouldNotBe — Tip-top Transgressive Texts…

Or Read a Review at Random: RaRaR

“A cloud of incense is worth a thousand sermons.” — Nicolás Gómez Dávila (1913-94).

It’s such a simple thing: repeatedly doubling a number: 1, 2, 4, 8, 16, 32, 61, 128… And yet it yields such riches, reminiscent of DNA or a literary text:

2^0 = 1

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

2^7 = 128

2^8 = 256

2^9 = 512

2^10 = 1024

2^20 = 1048576

2^30 = 1073741824

2^40 = 1099511627776

2^50 = 1125899906842624

2^60 = 1152921504606846976

2^70 = 1180591620717411303424

2^80 = 1208925819614629174706176

2^90 = 1237940039285380274899124224

2^100 = 1267650600228229401496703205376

2^200 = 1606938044258990275541962092341162602522202993782792835301376

Although, by Benford’s law*, 1 is the commonest leading digit, do all numbers eventually appear as the leading digits of some power of 2? I conjecture that they do. indeed, I conjecture that they do infinitely often. If the function first(n) returns the power of 2 whose leading digits are the same as the digits of n, then:

first(1) = 2^0 = 1

first(2) = 2^1 = 2

first(3) = 2^5 = 32

first(4) = 2^2 = 4

first(5) = 2^9 = 512

first(6) = 2^6 = 64

first(7) = 2^46 = 70368744177664

first(8) = 2^3 = 8

first(9) = 2^53 = 9007199254740992

first(10) = 2^10 = 1024

And I conjecture that this is true of all bases except bases that are powers of 2, like 2, 4, 8, 16 and so on. A related question is whether the leading digits of any 2^n are the same as the digits of n. Yes:

2^6 = 64

2^10 = 1024

2^1542 = 1.54259995… * 10^464

2^77075 = 7.70754024… * 10^23201

2^113939 = 1.13939932… * 10^34299

2^1122772 = 1.12277217… * 10^337988

That looks like a look of calculation, but there’s a simple way to cut it down: restrict the leading digits. Eventually they will lose accuracy, because the missing digits are generating carries. With four leading digits, this happens:

1: 0001

2: 0002

4: 0004

8: 0008

16: 0016

32: 0032

64: 0064

128: 0128

256: 0256

512: 0512

1024: 1024

2048: 2048

4096: 4096

8192: 8192

16384: 1638…

32768: 3276…

65536: 6552…

But working with only fifteen leading digits, you can find that 1122772 = the leading digits of 2^1122772, which has 337989 digits when calculated in full.

Previously pre-posted (please peruse):

*Not Zipf’s law, as I originally said.

Glittering skeins of piping birds

Fleet through the web of all our words.

*“Flying words”.

When we are conscious of being conscious, what are we consciousness-conscious with? If consciousness is a process in the brain, the process has become aware of itself, but how does it do so? And what purpose does consciousness-of-consciousness serve? Is it an artefact or an instrument? Is it an illusion? A sight or sound or smell is consciousness of a thing-in-itself, but that doesn’t apply here. We aren’t conscious of the thing-in-itself: the brain and its electro-chemistry. We’re conscious of the glitter on the swinging sword, but not the sword or the swing.

We can also be conscious of being conscious of being conscious, but beyond that my head begins to spin. Which brings me to an interesting reminder of how long the puzzle of consciousness has existed in its present form: how do we get from matter to mind? As far as I can see, science understands the material substrate of consciousness – the brain – in greater and greater detail, but is utterly unable to explain how objective matter becomes subjective consciousness. We have not moved an inch towards understanding how quanta become qualia since this was published in 1871:

Were our minds and senses so expanded, strengthened, and illuminated, as to enable us to see and feel the very molecules of the brain; were we capable of following all their motions, all their groupings, all their electric discharges, if such there be; and were we intimately acquainted with the corresponding states of thought and feeling, we should be as far as ever from the solution of the problem, “How are these physical processes connected with the facts of consciousness?” The chasm between the two classes of phenomena would still remain intellectually impassable.

Let the consciousness of love, for example, be associated with a right-handed spiral motion of the molecules of the brain, and the consciousness of hate with a left-handed spiral motion. We should then know, when we love, that the motion is in one direction, and, when we hate, that the motion is in the other; but the “Why?” would remain as unanswerable as before. — John Tyndall,

Fragments of Science(1871), viâ Rational Buddhism.

Elsewhere other-posted:

• Double Bubble

• The Brain in Pain

• The Brain in Train

• This Mortal Doyle

“After a million years or so, those screens are about to be removed, and once they have gone, then, for the first time, men will really know what it is to be alive.” — *Extreme Metaphors: Collected Interviews with J.G. Ballard, 1967-2008*, ed. Simon Sellars and Dan O’Hara (2012).

“A fertile imagination is better than any drug.” — *Ibid*.

Elsewhere other-posted:

• Vermilion Glands — review of *The Inner Man: The Life of J.G. Ballard* (W&N 2011)