Yury Kazakov has authored 2 sequences.
Original entry on oeis.org
8, 11, 14, 15, 18, 22, 23, 29, 32, 35, 38, 40, 41, 45, 47, 51, 53, 54, 55, 58, 59, 62, 66, 68, 69, 71, 74, 77, 78, 80, 83, 87, 88, 92, 95, 96, 98, 99, 105, 106, 107, 113, 115, 116, 118, 119, 123, 125, 126, 128, 130, 131, 134, 135, 137, 138, 141, 143, 149, 150, 153, 154, 155
Offset: 1
Yury Kazakov, S. P. Obukhov, Sean Sun, and N. A. Shikhova, May 03 2024
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import math
L=20 #tr are terms A372477 below L**2+1
numb=set()
Lmax=math.trunc((1+2*math.sqrt(3*L**2+1))/3)+1
tr=set()
tr.add(1)
for n in range(2,Lmax):
for k in range(0,n):
p1=n*n+k*k-k*n
p2=p1+k-n
if p1<=L**2:
tr.add(p1)
if p2<=L**2:
tr.add(p2)
for k in range (1, L**2):
numb.add(k)
print(sorted(numb-tr))
A372477
Areas of alternating equilateral and non-equilateral triangles that make up a three-leaf tiling over a regular triangular grid.
Original entry on oeis.org
1, 2, 3, 4, 5, 6, 7, 9, 10, 12, 13, 16, 17, 19, 20, 21, 24, 25, 26, 27, 28, 30, 31, 33, 34, 36, 37, 39, 42, 43, 44, 46, 48, 49, 50, 52, 56, 57, 60, 61, 63, 64, 65, 67, 70, 72, 73, 75, 76, 79, 81, 82, 84, 85, 86, 89, 90, 91, 93, 94, 97, 100
Offset: 1
Yury Kazakov, S. P. Obukhov, Sean Sun, and N. A. Shikhova, May 02 2024
For L=4:
Number Layer n = 1, Min Layer 1, [1, 1]
Number Layer n = 2, Min Layer 2, [4, 2, 3, 2, 4]
Number Layer n = 3, Min Layer 5, [9, 6, 7, 5, 7, 6, 9]
Number Layer n = 4, Min Layer 10, [16, 12, 13, 10, 12, 10, 13, 12, 16]
Number Layer n = 5, Min Layer 16, [25, 20, 21, 17, 19, 16, 19, 17, 21, 20, 25]
Number of terms below L^2+1=17 is 12.
In increasing order, without duplicates: [1, 2, 3, 4, 5, 6, 7, 9, 10, 12, 13, 16].
Terms below 17 are a(1)=1, a(2)=2, ..., a(11)=13, a(12)=16.
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=== Alternative layout idea ===
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The table below lists the numbers in layer n for n = 1..5. For each layer n >= 2, the table shows a pair of rows; the upper and lower rows in each pair list the triangle areas computed using the above formulas for b(n,k) and c(n,k), respectively.
min.
--+-----+-------+-------+-------+-------+-------+-------+ number
n | b/c | k = 0 | 1 | 2 | 3 | 4 | 5 | in layer
==+=====+=======+=======+=======+=======+=======+=======+==========
1 | - | 1 1 | | | | | | 1
--+-----+-------+-------+-------+-------+-------+-------+----------
2 | c | 2 | 2 | | | | | 2
| b | 4 | 3 | 4 | | | |
--+-----+-------+-------+-------+-------+-------+-------+----------
3 | c | 6 | 5 | 6 | | | | 5
| b | 9 | 7 | 7 | 9 | | |
--+-----+-------+-------+-------+-------+-------+-------+----------
4 | c | 12 | 10 | 10 | 12 | | | 10
| b | 16 | 13 | 12 | 13 | 16 | |
--+-----+-------+-------+-------+-------+-------+-------+----------
5 | c | 20 | 17 | 16 | 17 | 20 | | 16
| b | 25 | 21 | 19 | 19 | 21 | 25 |
--+-----+-------+-------+-------+-------+-------+-------+----------
-
import math
L=10 #generates terms below L**2+1
Lmax=math.trunc((1+2*math.sqrt(3*L**2+1))/3)+1
tr=set()
tr.add(1)
for n in range(2,Lmax):
for k in range(0,n):
p1=n*n+k*k-k*n
p2=p1+k-n
if p1<=L**2:
tr.add(p1)
if p2<=L**2:
tr.add(p2)
print('Number terms below', L**2+1, 'is', len(tr))
print(sorted(tr))
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