From Lee Sallows:
“The above three strips of ten numbers have an intriguing property. They record how many times each of the decimal digits (shown at left) occur in the other two strips. Hence the 6 in the left-hand strip identifies the number of 0’s in strips B and C, while the 2 in the centre strip counts the number of 3’s present in strips A and C. Moreover, the same property holds for every number in all three strips.”
See Reciprocation.
(Thanks, Lee.)
From Lee Sallows:
A feature common to many geomagic squares is that the set of shapes they employ reveal an atomic structure. That is, they are built up from repeated copies of a single unit shape. Examples of this are piece sets composed of polyominoes, the unit shape then being a (relatively small) square.
For the would-be geomagic square constructor, a key advantage of the atomic property is that the shapes concerned are each describable in terms of the positions of their constituent atoms. Or, to put it another way, they can be represented by a set of numbers. Hence, unlike non-atomic shapes, they are readily amenable to analysis and manipulation by computer.
Take, for example, an algorithm able to identify and list each of the different ways in which a given planar shape can be tiled by some specified set of smaller shapes. Such a program might be challenging to write, but provided the pieces concerned are composed of repeated units, implementation ought to be straightforward. But could the same be said in the case of non-atomic pieces? Without a set of numbers to describe piece shapes, how are they to be represented in a digital computer?
This is worth noting since, as inspection will show, the shapes employed in the square above are plainly non-atomic. In line with this I can confirm that the only computer program involved in deriving this solution was a vector graphics editor used to create the drawing seen above.
(Thanks, Lee.)
From Lee Sallows:
As the reader can check, the English number names less than “twenty” are composed using 16 different letters of the alphabet. We assign a distinct integral value to each of these as follows:
E F G H I L N O R S T U V W X Z 3 9 6 1 -4 0 5 -7 -6 -1 2 8 -3 7 11 10
The result is the following run of so called “perfect” numbers:
Z+E+R+O = 10 + 3 – 6 – 7 = 0 O+N+E = –7 + 5 + 3 = 1 T+W+O = 2 + 7 – 7 = 2 T+H+R+E+E = 2 + 1 – 6 + 3 + 3 = 3 F+O+U+R = 9 – 7 + 8 – 6 = 4 F+I+V+E = 9 – 4 – 3 + 3 = 5 S+I+X = –1 – 4 + 11 = 6 S+E+V+E+N = –1 + 3 – 3 + 3 + 5 = 7 E+I+G+H+T = 3 – 4 + 6 + 1 + 2 = 8 N+I+N+E = 5 – 4 + 5 + 3 = 9 T+E+N = 2 + 3 + 5 = 10 E+L+E+V+E+N = 3 + 0 + 3 – 3 + 3 + 5 = 11 T+W+E+L+V+E = 2 + 7 + 3 + 0 – 3 + 3 = 12
The above is due to a computer program in which nested Do-loops try out all possible values in systematically incremented steps. The above solution is one of two sets coming in second place to the minimal (lowest set of values) solution seen here:
E F G H I L N O R S T U V W X Z –2 –6 0 –7 7 9 2 1 4 3 10 5 6 –9 –4 –3
But why does the list above stop at twelve? Given that 3 + 10 = 13, and assuming that THREE, TEN and THIRTEEN are all perfect, we have T+H+I+R+T+E+E+N = T+H+R+E+E + T+E+N. But cancelling common letters on both sides of this equation yields E = I, which is to say E and I must share the same value, contrary to our requirement above that the letters be assigned distinct values. Thus, irrespective of letter values selected, if it includes THREE and TEN, no unbroken run of perfect numbers can exceed TWELVE. This might be decribed as a formal proof that THIRTEEN is unlucky.
But not all situations call for an unbroken series of perfect numbers. Sixteen distinct numbers occur in the following, eight positive, eight negative. This lends itself to display on a checkerboard:
Choose any number on the board. Call out the letters that spell its name, adding up their associated numbers when on white squares, subtracting when on black. Their sum is the number you selected.
(Thanks, Lee.)
From Lee Sallows:
(Thanks, Lee!)
From Lee Sallows:
(Thanks, Lee!)
From Lee Sallows:
(Thanks, Lee!)
In 2015 British computer scientist Chris Patuzzo produced a self-enumerating pangram — a sentence that itemizes its own contents — that records its totals as percentages:
This sentence is dedicated to Lee Sallows and to within one decimal place four point five percent of the letters in this sentence are a’s, zero point one percent are b’s, four point three percent are c’s, zero point nine percent are d’s, twenty point one percent are e’s, one point five percent are f’s, zero point four percent are g’s, one point five percent are h’s, six point eight percent are i’s, zero point one percent are j’s, zero point one percent are k’s, one point one percent are l’s, zero point three percent are m’s, twelve point one percent are n’s, eight point one percent are o’s, seven point three percent are p’s, zero point one percent are q’s, nine point nine percent are r’s, five point six percent are s’s, nine point nine percent are t’s, zero point seven percent are u’s, one point four percent are v’s, zero point seven percent are w’s, zero point five percent are x’s, zero point three percent are y’s and one point six percent are z’s.
The next challenge was to extend the precision beyond one decimal place. Impressively, Matthias Belz produced this specimen in 2017:
Rounded to five decimal places, two point six five two five two percent of the letters of this sentence are a’s, zero point zero eight eight four two percent are b’s, two point six five two five two percent are c’s, zero point four four two zero nine percent are d’s, nineteen point eight zero five four eight percent are e’s, three point four four eight two eight percent are f’s, one point seven six eight three five percent are g’s, two point nine one seven seven seven percent are h’s, seven point eight six nine one four percent are i’s, zero point zero eight eight four two percent are j’s, zero point zero eight eight four two percent are k’s, zero point three five three six seven percent are l’s, zero point one seven six eight three percent are m’s, ten point two five six four one percent are n’s, eight point nine three zero one five percent are o’s, four point seven seven four five four percent are p’s, zero point zero eight eight four two percent are q’s, nine point five four nine zero seven percent are r’s, four point nine five one three seven percent are s’s, nine point six three seven four nine percent are t’s, two point zero three three six zero percent are u’s, two point seven four zero nine four percent are v’s, one point six seven nine nine three percent are w’s, zero point nine seven two five nine percent are x’s, zero point zero eight eight four two percent are y’s and one point nine four five one eight percent are z’s.
These numbers are still rounded, so later that year he surpassed that with an instance giving precisely accurate values:
Exactly three point eight seven five percent of the letters of this autogram are a’s, zero point one two five percent are b’s, three point five percent are c’s, zero point two five percent are d’s, twenty-one point two five percent are e’s, three point seven five percent are f’s, zero point three seven five percent are g’s, one point five percent are h’s, seven point two five percent are i’s, zero point one two five percent are j’s, zero point one two five percent are k’s, zero point three seven five percent are l’s, zero point two five percent are m’s, nine point seven five percent are n’s, seven point five percent are o’s, six point five percent are p’s, zero point one two five percent are q’s, nine point three seven five percent are r’s, five point one two five percent are s’s, ten percent are t’s, zero point three seven five percent are u’s, four point six two five percent are v’s, one point five percent are w’s, zero point five percent are x’s, zero point three seven five percent are y’s and one point five percent are z’s.
From Lee Sallows:
(Thanks, Lee!)
From Lee Sallows:
(Thanks, Lee!)
From Lee Sallows:
(Thanks, Lee!)