The Spanish text on the poster, …rendería saber lo que puede provocar un ataque de asma, says something like “…would [le]arn to know what can provoke an attack of asthma”.
Previously pre-posted
• Moz Mit Mog #1
Monthly Archives: February 2022
The Viscount of Bi-Count
Today is 22/2/22 and, as I hoped on 2/2/22, I can say more about an interesting little palindromic-pattern problem. For each set of integers <= 1[0]1 in base 10, I looked at the count of palindromes exactly divisible by 1, 2, 3, 4, 5, 6, 7, 8 and 9. For example, 2, 4, 6 and 8 are the 4 palindromes divisible by 2 that are less than 11, so countdiv(2) = 4 for pal <= 11; 3, 6 and 9 are the 3 palindromes divisible by 3, so countdiv(3) = 3; and so on. Here are the counts — and some interesting patterns — for palindromes <= (powers-of-10 + 1) up to 1,000,000,000,001:
count for palindromes <= 101 (prime)
countdiv(1) = 19
countdiv(2) = 8
countdiv(3) = 6
countdiv(4) = 4
countdiv(5) = 2
countdiv(6) = 2
countdiv(7) = 2
countdiv(8) = 2
countdiv(9) = 2
count for palindromes <= 1001 = 7 * 11 * 13
countdiv(1) = 109
countdiv(2) = 48
countdiv(3) = 36
countdiv(4) = 24
countdiv(5) = 12
countdiv(6) = 15
countdiv(7) = 15
countdiv(8) = 12
countdiv(9) = 12
count for palindromes <= 10001 = 73 * 137
countdiv(1) = 199
countdiv(2) = 88
countdiv(3) = 66
countdiv(4) = 44
countdiv(5) = 22
countdiv(6) = 28
countdiv(7) = 32
countdiv(8) = 22
countdiv(9) = 22
count for palindromes <= 100001 = 11 * 9091
countdiv(1) = 1099
countdiv(2) = 488
countdiv(3) = 366
countdiv(4) = 244
countdiv(5) = 122
countdiv(6) = 161
countdiv(7) = 163
countdiv(8) = 122
countdiv(9) = 122
count for palindromes <= 1000001 = 101 * 9901
countdiv(1) = 1999
countdiv(2) = 888
countdiv(3) = 666
countdiv(4) = 444
countdiv(5) = 222
countdiv(6) = 294
countdiv(7) = 303
countdiv(8) = 222
countdiv(9) = 222
count for palindromes <= 10000001 = 11 * 909091
countdiv(1) = 10999
countdiv(2) = 4888
countdiv(3) = 3666
countdiv(4) = 2444
countdiv(5) = 1222
countdiv(6) = 1627
countdiv(7) = 1588
countdiv(8) = 1222
countdiv(9) = 1222
count for palindromes <= 100000001 = 17 * 5882353
countdiv(1) = 19999
countdiv(2) = 8888
countdiv(3) = 6666
countdiv(4) = 4444
countdiv(5) = 2222
countdiv(6) = 2960
countdiv(7) = 2878
countdiv(8) = 2222
countdiv(9) = 2222
count for palindromes <= 1000000001 = 7 * 11 * 13 * 19 * 52579
countdiv(1) = 109999
countdiv(2) = 48888
countdiv(3) = 36666
countdiv(4) = 24444
countdiv(5) = 12222
countdiv(6) = 16293
countdiv(7) = 15734
countdiv(8) = 12222
countdiv(9) = 12222
count for palindromes <= 10000000001 = 101 * 3541 * 27961
countdiv(1) = 199999
countdiv(2) = 88888
countdiv(3) = 66666
countdiv(4) = 44444
countdiv(5) = 22222
countdiv(6) = 29626
countdiv(7) = 28783
countdiv(8) = 22222
countdiv(9) = 22222
count for palindromes <= 100000000001 = 11^2 * 23 * 4093 * 8779
countdiv(1) = 1099999
countdiv(2) = 488888
countdiv(3) = 366666
countdiv(4) = 244444
countdiv(5) = 122222
countdiv(6) = 162959
countdiv(7) = 157361
countdiv(8) = 122222
countdiv(9) = 122222
count for palindromes <= 1000000000001 = 73 * 137 * 99990001
countdiv(1) = 1999999
countdiv(2) = 888888
countdiv(3) = 666666
countdiv(4) = 444444
countdiv(5) = 222222
countdiv(6) = 296292
countdiv(7) = 286461
countdiv(8) = 222222
countdiv(9) = 222222
As you can see, the counts for some numbers alternate between rep-digits (all digits the same) and nearly rep-digits. For example, the counts for palindromes exactly divisible by 5, 8 and 9 are alternately all 2s or 1 followed by all 2s. And you get counts of 2, 12, 22, 122, 222, 1222, 2222 in other even bases greater than base 2 when the counts are represented in that base. Here’s base 8:
count for palindromes <= 101 in b8 = 65 in b10 = 5 * 13
countdiv(1) = 17 in b8 (15 in b10)
countdiv(2) = 6
countdiv(3) = 11 in b8 (9)
countdiv(4) = 2
countdiv(5) = 3
countdiv(6) = 4
countdiv(7) = 2
count for palindromes <= 1001 in b8 = 513 in b10 = 3^3 * 19
countdiv(1) = 107 in b8 (71 in b10)
countdiv(2) = 36 in b8 (30)
countdiv(3) = 34 in b8 (28)
countdiv(4) = 12 in b8 (10)
countdiv(5) = 20 in b8 (16)
countdiv(6) = 14 in b8 (12)
countdiv(7) = 12 in b8 (10)
count for palindromes <= 10001 in b8 = 4097 in b10 = 17 * 241
countdiv(1) = 177 in b8 (127 in b10)
countdiv(2) = 66 in b8 (54)
countdiv(3) = 123 in b8 (83)
countdiv(4) = 22 in b8 (18)
countdiv(5) = 34 in b8 (28)
countdiv(6) = 44 in b8 (36)
countdiv(7) = 22 in b8 (18)
count for palindromes <= 100001 in b8 = 32769 in b10 = 3^2 * 11 * 331
countdiv(1) = 1077 in b8 (575 in b10)
countdiv(2) = 366 in b8 (246)
countdiv(3) = 352 in b8 (234)
countdiv(4) = 122 in b8 (82)
countdiv(5) = 164 in b8 (116)
countdiv(6) = 144 in b8 (100)
countdiv(7) = 122 in b8 (82)
count for palindromes <= 1000001 in b8 = 262145 in b10 = 5 * 13 * 37 * 109
countdiv(1) = 1777 in b8 (1023 in b10)
countdiv(2) = 666 in b8 (438)
countdiv(3) = 1251 in b8 (681)
countdiv(4) = 222 in b8 (146)
countdiv(5) = 316 in b8 (206)
countdiv(6) = 444 in b8 (292)
countdiv(7) = 222 in b8 (146)
count for palindromes <= 10000001 in b8 = 2097153 in b10 = 3^2 * 43 * 5419
countdiv(1) = 10777 in b8 (4607 in b10)
countdiv(2) = 3666 in b8 (1974)
countdiv(3) = 3524 in b8 (1876)
countdiv(4) = 1222 in b8 (658)
countdiv(5) = 1645 in b8 (933)
countdiv(6) = 1444 in b8 (804)
countdiv(7) = 1222 in b8 (658)
count for palindromes <= 100000001 in b8 = 16777217 in b10 = 97 * 257 * 673
countdiv(1) = 17777 in b8 (8191 in b10)
countdiv(2) = 6666 in b8 (3510)
countdiv(3) = 12523 in b8 (5459)
countdiv(4) = 2222 in b8 (1170)
countdiv(5) = 3164 in b8 (1652)
countdiv(6) = 4444 in b8 (2340)
countdiv(7) = 2222 in b8 (1170)
The counts for 4-palindromes and 7-palindromes in base 8 run: 1, 12, 22, 122, 222, 1222, 2222…, just like the counts for 5-palindromes, 8-palindromes and 9-palindromes in base 10. Here’s base 14:
count for palindromes <= 101 in b14 = 197 in b10 (prime)
countdiv(1) = 1D in b14 (27 in b10)
countdiv(2) = C in b14 (12)
countdiv(3) = 13 in b14 (17)
countdiv(4) = 6
countdiv(5) = 11 in b14 (15)
countdiv(6) = 8
countdiv(7) = 2
countdiv(8) = 2
countdiv(9) = 5
countdiv(A) = 7
countdiv(B) = 2
countdiv(C) = 4
countdiv(D) = 2
count for palindromes <= 1001 in b14 = 2745 in b10 = 3^2 * 5 * 61
countdiv(1) = 10D in b14 (209 in b10)
countdiv(2) = 6C in b14 (96)
countdiv(3) = 58 in b14 (78)
countdiv(4) = 36 in b14 (48)
countdiv(5) = 3A in b14 (52)
countdiv(6) = 28 in b14 (36)
countdiv(7) = 12 in b14 (16)
countdiv(8) = 19 in b14 (23)
countdiv(9) = 1C in b14 (26)
countdiv(A) = 19 in b14 (23)
countdiv(B) = 14 in b14 (18)
countdiv(C) = 14 in b14 (18)
countdiv(D) = 12 in b14 (16)
count for palindromes <= 10001 in b14 = 38417 in b10 = 41 * 937
countdiv(1) = 1DD in b14 (391 in b10)
countdiv(2) = CC in b14 (180)
countdiv(3) = 147 in b14 (259)
countdiv(4) = 66 in b14 (90)
countdiv(5) = 129 in b14 (233)
countdiv(6) = 88 in b14 (120)
countdiv(7) = 22 in b14 (30)
countdiv(8) = 31 in b14 (43)
countdiv(9) = 66 in b14 (90)
countdiv(A) = 79 in b14 (107)
countdiv(B) = 26 in b14 (34)
countdiv(C) = 44 in b14 (60)
countdiv(D) = 22 in b14 (30)
count for palindromes <= 100001 in b14 = 537825 in b10 = 3 * 5^2 * 71 * 101
countdiv(1) = 10DD in b14 (2939 in b10)
countdiv(2) = 6CC in b14 (1356)
countdiv(3) = 594 in b14 (1110)
countdiv(4) = 366 in b14 (678)
countdiv(5) = 3B2 in b14 (744)
countdiv(6) = 288 in b14 (512)
countdiv(7) = 122 in b14 (226)
countdiv(8) = 1A1 in b14 (337)
countdiv(9) = 1CA in b14 (374)
countdiv(A) = 1A7 in b14 (343)
countdiv(B) = 150 in b14 (266)
countdiv(C) = 144 in b14 (256)
countdiv(D) = 122 in b14 (226)
count for palindromes <= 1000001 in b14 = 7529537 in b10 = 37 * 197 * 1033
countdiv(1) = 1DDD in b14 (5487 in b10)
countdiv(2) = CCC in b14 (2532)
countdiv(3) = 1493 in b14 (3657)
countdiv(4) = 666 in b14 (1266)
countdiv(5) = 12B1 in b14 (3291)
countdiv(6) = 888 in b14 (1688)
countdiv(7) = 222 in b14 (422)
countdiv(8) = 331 in b14 (631)
countdiv(9) = 63A in b14 (1228)
countdiv(A) = 7A7 in b14 (1519)
countdiv(B) = 278 in b14 (498)
countdiv(C) = 444 in b14 (844)
countdiv(D) = 222 in b14 (422)
count for palindromes <= 10000001 in b14 = 105413505 in b10 = 3 * 5 * 7027567
countdiv(1) = 10DDD in b14 (41159 in b10)
countdiv(2) = 6CCC in b14 (18996)
countdiv(3) = 5948 in b14 (15548)
countdiv(4) = 3666 in b14 (9498)
countdiv(5) = 3B2A in b14 (10426)
countdiv(6) = 2888 in b14 (7176)
countdiv(7) = 1222 in b14 (3166)
countdiv(8) = 1A31 in b14 (4747)
countdiv(9) = 1C6D in b14 (5193)
countdiv(A) = 1A79 in b14 (4811)
countdiv(B) = 1513 in b14 (3741)
countdiv(C) = 1444 in b14 (3588)
countdiv(D) = 1222 in b14 (3166)
count for palindromes <= 100000001 in b14 = 1475789057 in b10 = 17 * 5393 * 16097
countdiv(1) = 1DDDD in b14 (76831 in b10)
countdiv(2) = CCCC in b14 (35460)
countdiv(3) = 14947 in b14 (51219)
countdiv(4) = 6666 in b14 (17730)
countdiv(5) = 12B29 in b14 (46097)
countdiv(6) = 8888 in b14 (23640)
countdiv(7) = 2222 in b14 (5910)
countdiv(8) = 3331 in b14 (8863)
countdiv(9) = 631D in b14 (17079)
countdiv(A) = 7A79 in b14 (21275)
countdiv(B) = 278B in b14 (6983)
countdiv(C) = 4444 in b14 (11820)
countdiv(D) = 2222 in b14 (5910)
Now 7-palindromes and D-palindromes (D = 13 in base 10) are following the [1]2222… pattern. What explains it? If you’re good at math, you won’t need telling. But I’m not good at maths, so I’m going to tell myself and other members of the not-good-at-math community what’s going on. Let’s go back to base 10 and the counts for 5-palindromes, that is, palindromes exactly divisible by 5. In base 10, the only integers exactly divisible by 5 have to end in either 5 or 0. But a palindrome can’t end in 0, because then the leading digit would have to be 0 too. Therefore only palindromes ending in 5 are exactly divisible by 5 in base 10. And if the palindromes end in 5, they have to start with 5 too.
Once we know that, we can easily calculate, for a given number of digits, how many 5-palindromes there are. Take 5-palindromes with three digits. If the three-digit 5-palindromes end and start with 5, we have to consider only the middle digit, which can obviously range from 0 to 9: 505, 515, 525, 535, 545, 555, 565, 575, 585 and 595. So there are 10 3-digit 5-palindromes. We add that count to the count for the single one-digit 5-palindrome, 5, and the single two-digit 5-palindrome, 55. So the cumulative count for 5-palindromes < 1001 is: 10 + 1 + 1 = 12.
Now look at four-digit 5-palindromes. They start and end with 5, therefore we have to consider only the middle two digits. And those middle digits have to be identical: 5005, 5115, 5225, 5335, 5445, 5555, 5665, 5775, 5885, 5995. So there are also 10 four-digit 5-palindromes and count of 5-palindromes < 10001 is: 10 + 10 + 1 + 1 = 22.
Now look at five-digit 5-palindromes. Again we have consider only the middle digits, because the first and fifth digits have to be 5. The second digit of a five-digit 5-palindrome has to be the same as the fourth digit: 50005, 51715, 52425, 53135, and so on. And the second and fourth digits can obviously range from 0 to 9. And so can the third and middle digit of the 5-palindromes. But the third digit doesn’t have to be the same as the second and fourth digits: 50005, 50105, 50205, and so on. Therefore the number of five-digit 5-palindromes is 10 * 10 = 100. And the count of 5-palindromes < 100001 is: 100 + 10 + 10 + 1 + 1 = 122.
Now look at six-digit 5-palindromes. The second digit of a six-digit 5-palindrome has to be same as the fifth digit and the third digit has to be the same as the fourth digit. So once you have the second and third digits, you automatically have the fourth and fifth digits: 500005, 523325, 587785, and so on. Clearly, the second and third digits range from 00 to 99 (i.e., 00, 01, 02 … 97, 98, 99), so there must be 100 six-digit 6-palindromes. And the count of 5-palindromes < 1000001 is: 100 + 100 + 10 + 10 + 1 + 1 = 222.
It should be clear, then, that the count of 5-palindromes for an odd number of digits, d, will be always the same as the count of 5-palindromes for the even number of digits d+1. There is 1 one-digit 5-palindrome, namely 5, and 1 two-digit 5-palindrome, namely 55. There are 10 three-digit 5-palindromes, 505 to 595, and 10 four-digit 5-palindromes, 5005 to 5995. Now, the count of 5-palindromes with an odd number of digits, d, will be equal to 10^(d\2), where d\2 = (d-1)/2. And the count for 5-palindromes with the even number of digits d+1 will be the same, 10^(d\2). Therefore the count for both sets of 5-palindromes, d-digit palindromes and (d+1)-digit palindromes, will be 2 * 10^(d\2). And that’s why the cumulative count of 5-palindromes looks the way it does in base 10: 1, 2, 12, 22, 122, 222, 1222, 2222, 12222, 22222…
The same reasoning applies in other even bases greater than base 2. When a palindrome divisible by a particular number has to start and end with the same digit, s, in base b, the middle digits will dictate a count of b^(d\2) for both d-digit s-palindromes and (d+1)-digit s-palindromes. And you’ll get the same cumulative count for s-palindromes in that base: 1, 2, 12, 22, 122, 222, 1222, 2222, 12222, 22222…
Some other patterns in the palindrome-counts can be explained by extending the reasoning given above. For example, if an s-palindrome can begin and end with two possible numbers, you’ll get cumulative counts of 2, 4, 24, 44, 244, 444, 2444, 4444, 24444, 44444 and so on. If the s-palindrome can end with three possible numbers, you’ll get cumulative counts of 3, 6, 36, 66, 366, 666, 3666, 6666, 36666, 66666 and so on.
Post-Performative Post-Scriptum
The discussion above is of very simple mathematics, but that’s the only kind I can cope with. All the same, I’m pleased that I managed to work out why the count of 5-palindromes behaves like that in base 10. So I’ve decided to award myself a title. Remember that the count for 5-palindromes of length d and d+1 is 2 * 10^(d\2), where d is an odd number. And you could say that 2 * 10^(d\2) is a bi-count of 10^(d\2). So I’m calling myself the Viscount of Bi-Count.
BB’s Butterfly
Illustration of a swallowtail butterfly (Papilio machaon) by Denys Watkins-Pitchford (1905-90), who used the pen-name of BB for books like The Little Grey Men (1942)
Isolingatto
Basque is one. Etruscan is another. Sumerian is a third.
What are they? Well, they’re all languages, but they’re more than that. John Donne said that no man is an island, entire of itself. That isn’t true of languages. Basque, Etruscan and Sumerian are all what might be called language-islands, entire of themselves and unrelated to any other language in the world, alive or dead.
I find that a powerful idea in all sorts of ways. Living or dead, language isolates (as linguists call them) are weird and wonderful things. But in some ways they’re at their weirdest and most wonderful when they’re poised between life and death. Many people down the millennia have been the last living speaker of a once widely spoken language. Often that language will still have had far-thrown and flourishing relatives, so its imminent death won’t throw all its treasures of phonology and syntax and lexicon into oblivion.
But sometimes the last living speaker will be of a language isolate. And when the speaker dies, an entire linguistic world will die with them. That kind of tragedy reminds me of one of Clark Ashton Smith’s most memorable and moving stories: the brief but brain-ballistic “Sadastor”. It’s the weird tale of a huge planet “far-fissured with enormous chasms, and covered from pole to pole with the never-ebbing tides of the desert sand.” The planet is called Sadastor and is “without moon or satellite, an abomination and a token of doom to fairer and younger worlds.” I won’t describe what the demon Charnadis discovers on Sadastor, because you should read the story for yourself if you haven’t already.
If you do read it, or you already know it, you’ll understand why it reminds me of the last living speaker of a language isolate. And here is such a speaker and such a language:
Gyani Maiya Sen, a 75-year-old woman from western Nepal, can perhaps be forgiven for feeling that the weight of the world rests on her shoulders. She is the only person still alive in Nepal who fluently speaks the Kusunda language. The unknown origins and mysterious sentence structures of Kusunda have long baffled linguists.
As such, she has become a star attraction for campaigners eager to preserve her dying tongue. Madhav Prasad Pokharel, a professor of linguistics at Nepal’s Tribhuwan University, has spent a decade researching the vanishing Kusunda tribe.
Professor Pokharel describes Kusunda as a “language isolate”, not related to any common language of the world. “There are about 20 language families in the world,” he said, “among them are the Indo-European, Sino-Tibetan and Austro-Asiatic group of languages.
“Kusunda stands out because it is not phonologically, morphologically, syntactically and lexically related to any other languages of the world.” – Nepal’s mystery language on the verge of extinction, BBC News, 13v12
Wikitionary has a word-list for Kusunda and I looked there for something that also reminds me of a language-isolate. What do you call one of the most beautiful, mysterious and solitary animals in the world? In Kusunda, you call it myaqo getse. And what does that mean? Well, in English it means “cat”. But in Kusunda itself it means “leopard-child”, from myaq or myaχ, “leopard”, and getse, “offspring, child.” So I suppose you could also translate myaqo getse as “leopardling”.
Panthera pardus fuscus, the Indian Leopard
And I’ve chosen to try and express the theme of this blog-post in Italian as isolingatto, a portmanteau of isola, “island”, lingua, “language”, and gatto, “cat”. But isolingatto could almost be the past participle of the verb isolingare, meaning “to speak in a language isolate”. That is, isolingatto could mean “spoken in a language isolate” or “spoken from a language-island”.
A Little Light Night Music
Cover of Albinö Rhino’s Upholder (2016)
I don’t like the music, but I do like the cover.
Domination. Exclusion. Inequality.
Sudoku has conquered the world, but I think futoshiki is more fun — more concentrated, more compact and quicker. When complete, a 5×5 futoshiki will be a Roman square in which every row and column contains the numbers 1 to 5, but no number is repeated in any row or column. You have to work out the missing numbers using logic and the “inequality” signs that show whether one square contains a number more than or less than a number in a neighboring square — futōshiki, 不等式, means “inequality” in Japanese (fu- is the negative prefix). Here’s an example of a futoshiki puzzle:
Futoshiki puzzle, with 4 in square (3,2) and 3 in square (1,1)
If you identify the squares by row and column, 4 is in (3,2) and 3 is in (1,1). And you can say, for example, that the empty square (3,4) dominates the empty square (3,3) or that (3,3) is dominated by (3,4). I’ll describe one route (not the best or most efficient) to completing the puzzle. Let’s start by considering the general rules that 1 cannot appear in any square that dominates another square and that 5 cannot appear in any square dominated by another square.
If you extend that logic, you’ll see that 4 cannot appear in any square that is at the end of what you might call a chain of dominations, where one square dominates a second square that in turn dominates a third square. Therefore, in the puzzle above, 4 cannot appear in squares (2,1) and (4,1) of column 1. And it can’t appear in square (3,1), because that would mean two 4s in the same row. This leaves one place for 4 to appear: square (5,1). And if 4 is there, 5 has to be in square (3,1):
Now look at row 3. Two of the remaining three empty squares are dominators: (3,4) dominates (3,2) and (3,5) dominates (2,5). 1 cannot appear in a dominating square, so 1 has to be in the dominated square (3,3):
The next step I’ll take is a bit more complicated. In row 2, the number 4 cannot be in (2,1) and (2,4). It can’t be in (2,1) because that would mean 4 was greater than itself. And it can’t be in (2,4), because (2,4) is dominated by (3,4) and (3,4) can’t contain 5, the only number that dominates 4. Therefore 4 must be in either (2,3) or (2,4). But so must 5. Therefore (2,3) contains either 4 or 5 and (2,4) contains either 4 or 5. That means that the numbers [1,2,3] must be in the other three squares of row 2. Now, 3 can’t be in square (2,1), because the chain of dominations is too long. And 3 can’t be in (2,5), because (2,5) is dominated by (3,5), which contains either 2 or 3. Therefore 3 must be in (2,2):
Now consider column 2. Square (4,2) cannot contain 1 or 5, because it’s both dominated and dominating. And if it can’t contain 1 or 5, there’s only one number it can contain: 2. And it immediately follows that (4,1) must contain 1, the only number less than 2. And if four squares of column 1 now contain the numbers [1,3,4,5], the remaining empty square (2,1) must contain 2:
Now consider row 2. Squares (2,3) and (2,4) contain either 4 or 5, therefore (2,5) must contain 1:
Now consider row 5. The number 1 is logically excluded from three squares: from (5,3), because there’s a 1 in (3,3); from (5,4), because (5,4) dominates (5,5); and from (5,5), because there’s a 1 in (2,5). Therefore 1 must be in (5,2). And if 1 is in (5,2), the number 5 must be in (1,2):
Now 1 pops up in row 1 because it can only be in (1,4):
And 5 pops up in row 4 because it can only be in (4,5):
Once 5 is in (4,5), the number 4 must be in (4,4) and the number 3 in (4,3):
The 4 of (4,4) immediately collapses the ambiguity of (2,4), which must contain 5. Therefore (2,3) contains 4:
Next, 5 pops up in (5,3):
And 3 must be in (5,4), dominating 2 in (5,5):
With 3 in (5,4) and 2 in (5,5), the ambiguity of (3,4) and (3,5) collapses:
And the square is completed like this:
⬇
Here’s an animated version of the steps to completion:
Futoshiki puzzle animated
Reading the Roons
In terms of core issues around maximal engagement with keyly committed core components of the counter-cultural community, one of the saddest, sorriest and sighfullest sights among them is that of the talented lad from the wrong side of the tracks who betrays his class by turning himself into a Guardian-reader, in terms of core cultural assumptions and behaviour.
Northampton’s Alan Moore has done it.
London’s Stewart Home has done it.
Huddersfield’s John Coulthart has done it.
How do I know?
[Readers’ Advisory: If you are easily disturbed, distressed and/or disgusted, please stop reading NOW.]
I know because
[I mean it. Stop reading or you may well regret it.]
I know because each of these talented lads from the wrong side of the tracks now bears the Mark of the Beast, metaphorically speaking.
[Last chance.]
Each of them has, on multiple occasions and without the minimalest micro-metric of shame or irony, deployed the key Guardianista phrase “in terms of”.
• For proof of Alan Moore’s deplorable delinquency, please see here.
• For proof of Stewart Home’s dep-del, please see here.
• For proof of John Coulthart’s dep-del, please see in the same place as you possibly saw or are-about-to-see Stewart Home’s, i.e. here.
So. After seeing and lamenting those horrific examples of class-betrayal, I thought I was hermeneutically hardened and would never again experience sadness, sorrow or sighfullness at the sight of a talented lad etc.
I was wrong.
As I learned when I read this interview in The Mail on Sunday:
There was a lot of negativity in terms of my mum getting frustrated with us as kids, messing around all the time, smashing things in the house and my nan lived in the same road, a few houses down. […] In terms of therapy, I have spoken to a few different people. I have never done a period of time where I have done two years with someone and it has been ongoing. […] Everything I am asking of those players in terms of hard work, honesty, trust, commitment…if I was just to turn round and say “I have had an offer, I’m off”, I honestly couldn’t do that to the players and the staff. — Wayne Rooney reveals his secret two-day drinking binges etc
Oh, Wayne, Wayne, Wayne. How could you do it? But I think we can easily guess where he was infected: it was during his therapy-sessions.
Elsewhere other-accessible
• Ex-Term-In-Ate! — interrogating issues around why “in terms of” is so teratographically toxic…
• All posts interrogating issues around “in terms of”…
• All posts interrogating issues around the Guardian-reading community and its affiliates…
To Wit: 2/2
Today is 2/2/22, so here’s a little problem about palindromes. For each set of integers <= 1[0…]1 in base 10, find the cumulative count of palindromes exactly divisible by 1, 2, 3, 4, 5, 6, 7, 8 and 9. For example, 2, 4, 6 and 8 are the 4 palindromes exactly divisible by 2 that are less than 11, so countdiv(2) = 4 for pal <= 11; 3, 6 and 9 are the 3 palindromes exactly divisible by 3, so countdiv(3) = 3; and so on:
count for palindromes <= 11 (prime)
countdiv(1) = 10
countdiv(2) = 4
countdiv(3) = 3
countdiv(4) = 2
countdiv(5) = 1
countdiv(6) = 1
countdiv(7) = 1
countdiv(8) = 1
countdiv(9) = 1
count for palindromes <= 101 (prime)
countdiv(1) = 19
countdiv(2) = 8
countdiv(3) = 6
countdiv(4) = 4
countdiv(5) = 2
countdiv(6) = 2
countdiv(7) = 2
countdiv(8) = 2
countdiv(9) = 2
count for palindromes <= 1001 = 7 * 11 * 13
countdiv(1) = 109
countdiv(2) = 48
countdiv(3) = 36
countdiv(4) = 24
countdiv(5) = 12
countdiv(6) = 15
countdiv(7) = 15
countdiv(8) = 12
countdiv(9) = 12
Some interesting patterns appear as the ceiling-palindromes reach 10001, 100001, 1000001… And one particular pattern doesn’t always disappear when you try the same problem in other bases. I hope to say more on 22/2/22.
Overlord In Terms of Core Issues Around Maximal Engagement with Key Notions of the Über-Feral 
