Kival Ngaokrajang - cp's OEIS frontend

A275937 The number of distinct patterns of the smallest number of unit squares required to enclose n units of area, where corner contact is allowed.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 2, 4, 1

Offset: 0

Kival Ngaokrajang, Aug 12 2016

Inspired by A235382 and A261491. The rotations and/or reflections are excluded.

Kival Ngaokrajang, Illustration of initial terms

Cf. A235382, A261491, A100092.

A275113 a(n) is the minimal number of squares needed to enclose n squares with a wall so that there is a gap of at least one cell between the wall and the enclosed cells.

Original entry on oeis.org

12, 14, 15, 16, 16, 17, 18, 18, 19, 19, 20, 20, 20, 21, 21, 22

Offset: 1

Kival Ngaokrajang, Jul 17 2016

Inspired by beehive construction in which wax is used in the most efficient way. This problem is likened to construction of a fence around a house with minimum materials and maximum enclosed area. I conjectured that a specific house pattern shall be selected. See illustration in links.
If the conjecture in A261491 is true (i.e., A261491(n) is the number of squares required to enclose n squares without a gap), then a(n) = A261491(n) + 8. - Charlie Neder, Jul 11 2018
The conjecture in A261491 holds through a(16). - David Consiglio, Jr., Nov 10 2022

			     a(1) = 12:
     +--+--+--+
     | 1| 2| 3|
  +--+--+--+--+--+
  |12|        | 4|
  +--+  +--+  +--+
  |11|  | 1|  | 5|
  +--+  +--+  +--+
  |10|        | 6|
  +--+--+--+--+--+
     | 9| 8| 7|
     +--+--+--+
.
      a(2) = 14:
     +--+--+--+--+
     | 1| 2| 3| 4|
  +--+--+--+--+--+--+
  |14|           | 5|
  +--+  +--+--+  +--+
  |13|  | 1| 2|  | 6|
  +--+  +--+--+  +--+
  |12|           | 7|
  +--+--+--+--+--+--+
     |11|10| 9| 8|
     +--+--+--+--+
.
     a(3) = 15:
     +--+--+--+
     | 1| 2| 3|
  +--+--+--+--+--+
  |15|        | 4|
  +--+  +--+  +--+--+
  |14|  | 3|     | 5|
  +--+  +--+--+  +--+
  |13|  | 1| 2|  | 6|
  +--+  +--+--+  +--+
  |12|           | 7|
  +--+--+--+--+--+--+
     |11|10| 9| 8|
     +--+--+--+--+

David Consiglio, Jr., Python Program
David Consiglio, Jr., Illustration of a(11)
David Consiglio, Jr., Illustration of a(12)
David Consiglio, Jr., Illustration of a(13)
David Consiglio, Jr., Illustration of a(14)
David Consiglio, Jr., Illustration of a(15)
David Consiglio, Jr., Illustration of a(16)
Kival Ngaokrajang, Illustration of initial terms

Cf. A235382, A261491.

a(11)-a(16) from David Consiglio, Jr., Nov 10 2022

A273724 Place n equally-spaced points around a circle, labeled 0,1,2,...,n-1. For each i = 0..n-1 such that 3i != i mod n, draw an (undirected) chord from i to (3i mod n). Then a(n) is the total number of distinct chords.

Original entry on oeis.org

0, 0, 0, 2, 1, 4, 4, 6, 3, 8, 8, 10, 9, 12, 12, 14, 11, 16, 16, 18, 17, 20, 20, 22, 19, 24, 24, 26, 25, 28, 28, 30, 27, 32, 32, 34, 33, 36, 36, 38, 35, 40, 40, 42, 41, 44, 44, 46, 43, 48, 48, 50, 49, 52, 52, 54, 51, 56, 56, 58, 57, 60, 60, 62, 59, 64, 64, 66, 65, 68, 68, 70, 67, 72, 72, 74, 73, 76, 76, 78, 75, 80, 80

Offset: 0

Kival Ngaokrajang, May 28 2016

nonn

Brooke Logan and N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Kival Ngaokrajang, Illustration of initial terms
Burkard Polster, Times Tables, Mandelbrot and the Heart of Mathematics, Mathologer video (2015).

Cf. A117571 (if 3i is changed to 2i), A274462 (if 3i is changed to 4i).

a(n) = n-1 if n>0 is odd, n-2 if n == +-2 (mod 8), n-3 if n == 4 (mod 8), and n-5 if n == 0 (mod 8). These formulas are easily established by observing that the chord at i is missing if 2i == 0 mod n, and the chords starting at i and at 3i coincide if 8i == 0 mod n. The formulas then imply that the g.f. is 4+x^2/(1-x)^2-(4+x^2+2*x^4+x^6)/(1-x^8), which can be rewritten as (5*x^63*x^5+2*x^4+3*x^2-x+2)*x^3/((1-x)*(1-x^8)). (This g.f. was conjectured by Colin Barker.) - Brooke Logan and N. J. A. Sloane, Jun 23 2016

a(n) = a(n-1)+a(n-8)-a(n-9) for n>9. - Colin Barker, May 29 2016 (This follows from the above g.f. - Brooke Logan and N. J. A. Sloane)

Definition edited by N. J. A. Sloane, Jun 23 2016

A272721 The curvature (rounded down) of the n-th circle inscribed in the area related to the critical point of the Mandelbrot set at C = 1/4.

Original entry on oeis.org

13, 23, 35, 50, 66, 83, 103, 123, 146, 170, 196, 223

Offset: 1

Kival Ngaokrajang, May 05 2016

Inspired by "The dark side of the Mandelbrot set".

Consider the case C = 1/4, 0 <= x <= 1/2. Draw the inscribed circles in the area between the parabola y = x^2 + 1/4 and the line y = x. The radii of the circles are found using AutoCAD's "tan-tan-tan" function. See details and illustration in the links.

Burkard Polster, The dark side of the Mandelbrot set, Mathologer video (2016).
Kival Ngaokrajang, Illustration of initial terms.

A271509 List of 5-tuples: primitive integral pentagon sides in Cairo tiling.

Original entry on oeis.org

5, 5, 5, 5, 2, 13, 13, 13, 13, 14, 17, 17, 17, 17, 14, 25, 25, 25, 25, 34, 29, 29, 29, 29, 2, 37, 37, 37, 37, 46, 41, 41, 41, 41, 62, 53, 53, 53, 53, 34, 61, 61, 61, 61, 98, 65, 65, 65, 65, 94, 65, 65, 65, 65, 46, 73, 73, 73, 73, 14

Offset: 1

Kival Ngaokrajang, Apr 09 2016

Refer to Cairo tiling by Stick Cross Method (see details in the links). Each pentagon has four sides of equal length and one side which is either shorter or longer. All sides can be taken to have integral lengths related to primitive Pythagorean triples A103606.

If Pythagorean triple = (a, b, c), the 5-tuple is (s1, s2, s3, s4, s5) with s1 = s2 = s3 = s4 = c and s5 = 2*(b-a). See illustration in the links.

			List begins:
5, 5, 5, 5, 2,
13, 13, 13, 13, 14,
17, 17, 17, 17, 14,
25, 25, 25, 25, 34,
29, 29, 29, 29, 2,
...

David Bailey's World of Escher-like Tessellations, Stick Cross Method
Kival Ngaokrajang, Illustration of initial terms, Excel calculation sheet
Wikipedia, Cairo pentagonal tiling

Cf. A103606.

A270309 Irregular triangle read by rows: T(n,k) = ((n-k)+1)^2 if odd-n and odd-k; T(n,k) = k^2 if odd-n and even-k; T(n,k) = (n/2-(k/2-1/2))^2 if even-n and odd-k; T(n,k) = (k/2+1)^2 if even-n and even-k; where n >= 1, k = 1..2*n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 9, 4, 1, 1, 4, 9, 4, 1, 1, 4, 4, 1, 1, 4, 25, 4, 9, 16, 1, 1, 16, 9, 4, 25, 9, 1, 4, 4, 1, 9, 9, 1, 4, 4, 1, 9, 49, 4, 25, 16, 9, 36, 1, 1, 36, 9, 16, 25, 4, 49, 16, 1, 9, 4, 4, 9, 1, 16, 16, 1, 9, 4, 4, 9, 1, 16, 81, 4, 49, 16, 25, 36, 9, 64, 1, 1, 64, 9, 36, 25, 16, 49, 4, 81, 25, 1, 16, 4, 9, 9, 4, 16, 1, 25, 25, 1, 16, 4, 9, 9, 4, 16, 1, 25

Offset: 1

Kival Ngaokrajang, Mar 15 2016

Refer to A269845, but change to n+2 X n instead of n+1 X n.

There are triangles appearing along main diagonal. If the area of the smallest triangles are defined as 1, then the areas of all other triangles seem to be square numbers. Conjectures: (i) Even terms of row sum is A002492. (ii) Odd terms of row sum/2 is A100157. See illustration in links.

			Irregular triangle begins:
n\k  1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  ...
1    1,  1
2    1,  1,  1,  1
3    9,  4,  1,  1,  4,  9
4    4,  1,  1,  4,  4,  1,  1,  4
5   25,  4,  9, 16,  1,  1, 16,  9,  4, 25
6    9,  1,  4,  4,  1,  9,  9,  1,  4,  4,  1,  9
7   49,  4, 25, 16,  9, 36,  1,  1, 36,  9, 16, 25,  4, 49
8   16,  1,  9,  4,  4,  9,  1, 16, 16,  1,  9,  4,  4,  9,  1, 16
...

Kival Ngaokrajang, Illustration of initial terms, Row sum

Cf. A002492, A100157, A269845.

A269845 Irregular triangle read by rows: T(n,k) = (k/2+1/2)^2 if odd-k otherwise T(n,k) = (n-k/2)^2 where n >= 1, k = 0..2*n-1.

Original entry on oeis.org

1, 1, 4, 1, 1, 4, 9, 1, 4, 4, 1, 9, 16, 1, 9, 4, 4, 9, 1, 16, 25, 1, 16, 4, 9, 9, 4, 16, 1, 25, 36, 1, 25, 4, 16, 9, 9, 16, 4, 25, 1, 36, 49, 1, 36, 4, 25, 9, 16, 16, 9, 25, 4, 36, 1, 49, 64, 1, 49, 4, 36, 9, 25, 16, 16, 25, 9, 36, 4, 49, 1, 64, 81, 1, 64, 4, 49, 9, 36, 16, 25, 25, 16, 36, 9, 49, 4, 64, 1, 81, 100, 1, 81, 4, 64, 9, 49, 16, 36, 25, 25, 36, 16, 49

Offset: 1

Kival Ngaokrajang, Mar 06 2016

Inspired by A268317, but change to n+1 X n instead of Fib(n+1) X Fib(n).

There are triangles appearing along main diagonal. If the area of the smallest triangles are defined as 1, then the areas of all other triangles seem to be square numbers. Conjectures: (i) Odd terms of row sum/2 is A100157. (ii) Even terms of row sum/2 is A258582. See illustration in links.

			Irregular triangle begins:
n\k 0  1   2  3   4  5   6   7   8   9  10 11 12  13 14  15 ...
1   1, 1
2   4, 1,  1, 4
3   9, 1,  4, 4,  1, 9
4  16, 1,  9, 4,  4, 9,  1, 16
5  25, 1, 16, 4,  9, 9,  4, 16,  1, 25
6  36, 1, 25, 4, 16, 9,  9, 16,  4, 25, 1, 36
7  49, 1, 36, 4, 25, 9, 16, 16,  9, 25, 4, 36, 1, 49
8  64, 1, 49, 4, 36, 9, 25, 16, 16, 25, 9, 36, 4, 49, 1, 64
...

Kival Ngaokrajang, Illustration of initial terms, Row sum

Cf. A100157, A258582, A268317.

Table[If[OddQ@ k, (k/2 + 1/2)^2, (n - k/2)^2], {n, 8}, {k, 0, 2 n - 1}] // Flatten (* Michael De Vlieger, Apr 01 2016 *)

for (n = 1, 20, for (k = 0, 2*n-1, if (Mod(k,2)==0, t = (n-k/2)^2, t = (k/2+1/2)^2); print1(t, ", ")))

T(n,k) = (k/2+1/2)^2 if odd-k, T(n,k) = (n-k/2)^2 if even-k; n >= 1, k = 0..2*n-1.

A268292 a(n) is the total number of isolated 1's at the boundary between n-th and (n-1)-th iterations in the pattern of A267489.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 3, 5, 7, 9, 11, 14, 18, 22, 26, 30, 34, 39, 45, 51, 57, 63, 69, 76, 84, 92, 100, 108, 116, 125, 135, 145, 155, 165, 175, 186, 198, 210, 222, 234, 246, 259, 273, 287, 301, 315, 329, 344, 360, 376, 392, 408, 424, 441

Offset: 0

Kival Ngaokrajang, Jan 31 2016

Refer to pattern of A267489, The total number of isolated 1's is a(n) and A112421 when consider at the boundary between n-th and (n-1)-th iterations and at the boundary in the same iterations concatenate on horizontal respectively. See illustrations in the links.

Empirically, a(n+4) gives the number of solutions m where 0 < m < 2^n and A014682^n(m) < 3 and A014682^n(m+2^n) = A014682^n(m)+9. - Thomas Scheuerle, Apr 25 2021

Kival Ngaokrajang, Illustration of initial terms, Continuously concatenate pattern

Cf. A112421, A267489, A014682.

a = 3; d1 = 2; print1("0, 0, 0, 0, 0, 0, 0, 1, 3, ");
for (n = 3,100, d2 = 0; if (Mod(n,6)==1 || Mod(n,6)==2, d2 = 1); d1 = d1 + d2; a = a + d1; print1(a, ", "))

Empirical g.f.: x^7 / ((1-x)^3*(1-x+x^2)*(1+x+x^2)). - Colin Barker, Jan 31 2016

For n>0: a(n) = floor((n-3)^2/12) + floor((n-4)^2/12). - Hoang Xuan Thanh, Jun 02 2025

A268318 Irregular triangle read by rows: T(n,k) gives the row sums in the table Fib(n+1) X Fib(n), where k = 1..Fib(n+1), and 1's are assigned to cells on the longest diagonal path.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2

Offset: 0

Kival Ngaokrajang, Feb 01 2016

Inspired by sun flower spirals which come in  Fib(i) and Fib(i+1) numbers in opposite directions. The present Fib(n+1) X Fib(n) table has the following properties:
(i) Columns sum create the irregular triangle A268317.
(ii) Rows sum create the present irregular triangle.
(iii) The row sums of each of these irregular triangles is conjectured to be A000071.
(iv) The first differences of the sequence of half of the  voids (0's) are conjectured to give A191797.
See illustrations in the links of A268317.

			Irregular triangle begins:
0
1
1 1
1 2 1
1 2 1 2 1
1 2 1 2 2 1 2 1
1 2 1 2 2 1 2 1 2 2 1 2 1
1 2 1 2 2 1 2 1 2 2 1 2 2 1 2 1 2 2 1 2 1
...

Cf. A000071, A191797, A268317.

A268317 Irregular triangle read by rows: T(n,k) gives the columns sum in the table Fib(n+1) X Fib(n), where k = 1..Fib(n), and 1's are assigned to cells on the longest diagonal path.

Original entry on oeis.org

0, 1, 2, 2, 2, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 3, 2, 3, 2, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3

Offset: 0

Kival Ngaokrajang, Feb 01 2016

Inspired by sun flower spirals which come in Fib(i) and Fib(i+1) numbers in opposite directions. The present case of the Fib(n+1) X Fib(n) table has the following properties:
(i) Columns sum create the present irregular triangle.
(ii) Rows sum create the irregular triangle A268318.
(iii) The row sum of each of these irregular triangles is conjectured to be A000071.
(iv) The first differences of the sequence of half of the voids (0's) are conjectured to give A191797.
See illustrations in the links.

			Irregular triangle begins:
1
2
2  2
2  3  2
2  3  2  3  2
2  3  2  3  3  2  3  2
2  3  2  3  3  2  3  2  3  3  2  3  2
...

Kival Ngaokrajang, Illustration of initial terms, Sun flower spirals

Cf. A000071, A191797, A268318.

cp's OEIS Frontend

User: Kival Ngaokrajang

A275937 The number of distinct patterns of the smallest number of unit squares required to enclose n units of area, where corner contact is allowed.

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A275113 a(n) is the minimal number of squares needed to enclose n squares with a wall so that there is a gap of at least one cell between the wall and the enclosed cells.

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A273724 Place n equally-spaced points around a circle, labeled 0,1,2,...,n-1. For each i = 0..n-1 such that 3i != i mod n, draw an (undirected) chord from i to (3i mod n). Then a(n) is the total number of distinct chords.

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A272721 The curvature (rounded down) of the n-th circle inscribed in the area related to the critical point of the Mandelbrot set at C = 1/4.

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A271509 List of 5-tuples: primitive integral pentagon sides in Cairo tiling.

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A270309 Irregular triangle read by rows: T(n,k) = ((n-k)+1)^2 if odd-n and odd-k; T(n,k) = k^2 if odd-n and even-k; T(n,k) = (n/2-(k/2-1/2))^2 if even-n and odd-k; T(n,k) = (k/2+1)^2 if even-n and even-k; where n >= 1, k = 1..2*n.

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A269845 Irregular triangle read by rows: T(n,k) = (k/2+1/2)^2 if odd-k otherwise T(n,k) = (n-k/2)^2 where n >= 1, k = 0..2*n-1.

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A268292 a(n) is the total number of isolated 1's at the boundary between n-th and (n-1)-th iterations in the pattern of A267489.

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A268318 Irregular triangle read by rows: T(n,k) gives the row sums in the table Fib(n+1) X Fib(n), where k = 1..Fib(n+1), and 1's are assigned to cells on the longest diagonal path.

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A268317 Irregular triangle read by rows: T(n,k) gives the columns sum in the table Fib(n+1) X Fib(n), where k = 1..Fib(n), and 1's are assigned to cells on the longest diagonal path.

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