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More on convex shapes of hexagons
I continue looking into convex shapes of
hexagons and looking into the conjecture that the sequence deviates at most
one from the sequence A216522. With respect to
this sequence and the sequence A135711
the following are true:
A216522(n) = ceiling(sqrt(12n + 9)) - 3
A135711(n) = 2.ceiling(sqrt(12n - 3))
A135711(n) = 2.A216522(n-1) + 6
A216522(n) = A135711(n+1)/2 - 3
(It is important to note that the sequence A135711 is one based where the
sequence A216522 is zero based.) If we take the sequence A216522 as a reference
then the perimeter should be defined as the distance between the hexagons on
the outside where the distance between the center of two hexagon is defined as
one. The table below gives for a given perimeter (for which k is
positive) the number of hexagons is can enclose. So for k is one,
seven hexagons can be enclosed with a perimeter of six.
perimeter: sides: #triangles: #hexagons:
6k k, k, k, k, k, k 6k² 3k² + 3k + 1
6k + 1 k+1, k-1, k+1, k, k, k 6k² + 2k - 1 3k² + 4k + 1
6k + 2 k+1, k, k, k+1, k, k 6k² + 4k 3k² + 5k + 2
6k + 3 k+1, k , k+1, k, k+1, k 6k² + 6k + 1 3k² + 6k + 3
6k + 4 k+1, k+1, k, k+1, k+1, k 6k² + 8k + 2 3k² + 7k + 4
6k + 5 k, k+2, k, k+1, k+1, k+1 6k² + 10k + 3 3k² + 8k + 5
If we use a zero based variant of sequence A216522, the formulea becomes:
ceiling(sqrt(12n - 3)) - 3 for the number of
perimeters. Lets denote this number with p, then the equations with
this formulea are:
p + 3 - 1 < sqrt(12n - 3) ≤ p + 3
(p + 2)² < 12n - 3 ≤ (p + 3)²
p² + 4p + 4 < 12n - 3 ≤ p² + 6p + 9
p² + 4p + 7 < 12n ≤ p² + 6p + 12
Now if we fill in the values for the perimers and matching number of hexagons,
we get:
6k: 36k² + 24k + 7 < 36k² + 36k + 12 ≤ 36k² + 36k + 12
6k + 1: 36k² + 36k + 12 < 36k² + 48k + 12 ≤ 36k² + 48k + 19
6k + 2: 36k² + 48k + 19 < 36k² + 60k + 24 ≤ 36k² + 60k + 28
6k + 3: 36k² + 60k + 28 < 36k² + 72k + 36 ≤ 36k² + 72k + 39
6k + 4: 36k² + 72k + 39 < 36k² + 84k + 48 ≤ 36k² + 84k + 52
6k + 5: 36k² + 84k + 52 < 36k² + 96k + 60 ≤ 36k² + 96k + 67
It is quite obvious that for all positive k the above equations are
true. This means that for all those minimal perimeter for the given number of
hexagons is equal to the given formulea. So, there is not a value, above
which the perimeter might be always one larger than given by the formulea.
This seems to support the conjecture that it is at most one higher than the
value given by the formulea. It should be noted that the number of hexagons
for which there is a simple solution for the minimal perimeter appear further
and further apart leaving more and more room for complicated cases.
Today, is the last day of a two week long regional heat wave. In the past two
weeks the maximum temperature was at least 25° Celsius. There were eight
days on which the temperature was 30° or higher of which one day higher than 35° with a new 37.9° record for the day.
There was a total of four days with temperature
records for the day of the year. There was also one night that the minimum
temperature was 21.6°, which is also rather exceptional. The national heat wave was only eleven days, but nevertheless a record number
of days. In the South of the country there was even a 'super heat wave' with
four days with temperatures of 35° or higher.
Graduation Show at Rietveld Academie
I traveled to Amsterdam to attend the Graduation Show 2026 at Gerrit Rietveld Academie. I found the works of the following students
noteworthy, which is very subjective and often based on the first impression,
in the order I encountered them:
- Isabel Heatley
- Niloofar Salehi
- Anne-Sophie de Lange
- Clément Lobjoy
with If a Tree Falls in a City...
- Elektra Tatalia-Aloupi
- Astrid Hemmingsen
- Nynne Pors with
Frem og tilbage har aldring vaeret lige langt
- Jonathan Doekwijt with
Four Patitions, Sky Palace, Amphora, and Untitled.
- Kateryna Lymar with
All lines trying to become a circle
- Piet Woudt with Oh, unknown nothingness.
- Anouk Albrecht with
installation that used clothing, steel, oak flooring, daylight lamp. (The
clothing had images that looked like they were made with a laser cutter.)
- Ian Marquez Garcia
- Mabel Blokzijl with
When I Look at You, I See Myself II *
- Miro Jacob with Empty Set
- Loren Xu with No Likes!
- Oliver Matzner with Ghost Ritual Machine +
- Nim Smit with I am famous in Amsterdam
- Sohju Kim with loaded screens and We Love That Echo +
- Alberte Rydahl.
(At 13:13, I took one of her booklets.)
- Maren Weertman with Apricot
trees exist, apricot trees exist
- Lily Vogt
- Freja Oline
Østergaard and Josefine Saietz with Mending the gap between
us
- Freja Oline
Østergaard with we all carry a butterfly inside
- Amin Mohammadaliei
- Alara Bora with Moult
- Bowie van de Loo
- Roya Hes
- Gabin Kim
- Liepa Pagareckaitė
- Kishan Graanoogst with Purpaduas-Aphroditi and Blueprint +
- Iris Vermeulen with
I Am Daniel Claus
- Marjoleine Timmer
- Saskia van der Kam
- Caroline Girardot-Bijnen with
X-ray 7.04
- Minky Kim with
I stiched my time, care, exhaustion, and I repeat +
- Aleksy Domke with The
Mathematical Inscription of the world jomel002.github.io
- Cassiopée
Charles-Messance with FM 120Km/h (installation self made music
instruments producing interesting resonances) *
- Feliz Aaliyah Sánchez Suitrago
- Stefania Escalona Millano
Afterwards, I went to Stedelijk Museum Amsterdam. There I first saw the exhibtion Kho Liang Ie, which contained the following works from his collection:
- Wetmatige beweging (Systematic Movement), Peter Struycken, 1964.
- Homage to Minakshi Sundaram Pillai, Joseph Ongenae, 1956.
- Vierkant met sektor (Square with sector), Ad Dekkers, 1968.
- Pawukon (calendar with important religious, social, and econmic days,
based on arithmetic) from Bali, date unknown.
- Homage to Kho Liang Ie, Sheila Hicks, 1975.
- Koh Liang Ie Collection, Kho Liang Ie and Ger van Vliet, 1963-1964.
From some of the other exhibtions:
- Windmill, Piet Mondriaan, ca.1917.
- Vegatable Gardens on Montmartre, Vincent van Gogh, 1887.
- Bal Tabarin, Jan Sluiters, 1907.
- Seated Woman, Leo Gestel, 1912.
- Reliëf Nr.16, Henryk Stażewski, 1964.
- Composition, Bart van der Leck, 1918-1920.
- Composition with Blue, Yellow, Red, Black, and Grey, Piet Mondriaan, 1948.
- Lozenge Composition with two Lines, Piet Mondriaan, 1931.
- Composition No.IV, with Red, Blue, and Yellow, Piet Mondriaan, 1929.
- Old Oaks in Surrey, Jan Toorop, 1890.
- no hobbies, mid 40s, German artist, Berlin, Germany, Reba Mayburry, 2026.
- Untitles, Hamiski Farah, 2026.
- Young Girl's Dream, Sophie Calle, 1992.
- Nude, Lying, Leo Gestel, 1910.
- Komputerstrukturen IV (Computer Structure IV), Peter Struycken, 1969.
- Montauk IV, Willem de Kooning, 1969.
- 40% Blue, 40% Red, 10% Green, 10% Orange (chartres), François Morellet, 1960.
I also saw the exhibition Danh Vo - πνεῦμα (Ἔλισσα).
At FOAM, I saw the exhibitions Foam Talent 2026 and Martin Parr - Very Modern and Rather Ugly.
Graduation Show at KABK
Today, I went to the Graduation Show 2026 at KABK. I found the works of the following students noteworthy, which is
very subjective and often based on my first impression, in the order I
encountered them:
In the train home, I finished reading the booklet
Wildsam Field Guides: Los Angeles, which I started reading on March 22
after I bought it on February 18. Initial I read
some pages during commercial breaks. In the past days, I finished reading it
in the train. Although it has many pages with fun and mundane facts, it also
has some serious content giving you a good impression of the city.
Flowers again
Just like last year and some years before, our
magnolia has some flowers again. The three plants that I planted about
a month ago are still doing well. They still rather
small, but I understand that that is normal as they focus on growing roots
first. I regularly give them water with the waste water from rinsing
vegatables.
Cyber Ægg
In the afternoon, I went to TkkrLab to help
with the 'sweatshop' for assembling the Cyber Ægg badge for BornHack. I first unpacked some PCBs, next I helped with attaching the
E-Ink display, because that was the bottleneck in the whole assembly line. I
did the same in 2017 for the SHA2017 badge. I felt it was a little bit more difficutl this time. I did
a bit over thirty before I left.
I went into the city and at photo gallery Objektief, I saw the exhibition Meesterwerken 2026 with students finishing some the
photography training. I saw the presentations by:
At Concordia, I saw the exhibition AKI Finals '26 with works by the
following graduates:
I also saw the exhibition Expositie Boswinkel with amateur art from the
neighborhood with that same name. I was rather surprised by the quality as the
neighborhood is not known as one of the best neighborhoods in Enschede.
Some interesting relationships
I found some interesting relationships with respect to convex shapes of hexagons. I realized that the each hexagon on a
triangluar grid can be viewed as the intersection of two triangles. I also
realized that there is a simple relationship with respect to the sizes of the
hexagon. Let the sizes of the hexagon, be denoted by a, b,
c, d, e, and f, then the following equations are
true:
a + b = d + e
b + c = e + f
c + d = f + a
Note that the third equation can be derived from the first two. If n is
the sum of all sides, the following equation can be derived:
d = n - a - 2b - 2c
e = -n + 2a + 3b + 2c
f = n - 2a - 2b - c
Now it is also possible to calculate the number of enclosed triangles from the
values of a, b, c, and n with the formulea:
n(4(a + c) + 6b) - n² - 4(a² + c²) - 8b² - 10b(a + c) - 6ac
The formulea shows that a and c can be interchanged, which is
what one would expect. This formulea can also be used to find generic
formuleas on k of the form 6k² + nk + l,
where l is equal to the above formulea for enclosed triangles. I wonder
if there is an efficient method to discover for giving values for n and
l there is a solution for some values of a, b, and
c or not. We could for example, turn the formulea in a quadratic
equation of b, which results in (if I am not mistaken):
-8b² + (6n - 10(a + c))b - n² + 4n(a + c) - 4(a² + c²) - 6ac - l = 0
The determinant of the quadratic equation on b is (if I am not
mistaken):
-28a² - 28c² + (128n - 120)a + (128n - 120)c + 8ac + 4n² - 32l
For there to be a integer solution for b, the determinant must be a
square number.
Remark July 13: I was mistaken with respect to last formulea.
I was mistaken
The formulea that I presented yesterday for the
determinant (related to convex shapes of
hexagons) is incorrect. The correct formulea, written slightly different,
is:
-28(a² + c²) + 8ac + 8n(a + c) + 4n² - 32l
Taking out a factor of four, we get:
-7(a² + c²) + 2ac + 2n(a + c) + n² - 8l
So, how do we find integer values for a and c for given n
and l such that the above formulea returns a square? Even if l is
a negative number, there are only a limited number of pairs to be evaluated,
due to the negative factor with the quadratic terms. The values for n
are always positive. It looks like all positive values are within a certain
circle or elipse. The function is symmetric around the line
a = b, The highest value probably lays on that line.
If we replace a and b we get the formulea:
-12x² + 4nx + n² - 8l
From the first derivative we know that the highest point is at
x = n/6. This might be used to determine the center of
the circle. I wrote some code to investigate values for which the formulea
returns squares for various values of n and l and discovered that
only for some multiples of 3 for n and multiples of -6 for l
there are no squares. The others have many squares and it seems that the number
increases with smaller values for l.
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