cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 21-29 of 29 results.

A105521 Sums of area and perimeter of primitive Pythagorean triples.

Original entry on oeis.org

18, 60, 100, 140, 270, 280, 294, 462, 648, 728, 756, 1078, 1080, 1210, 1496, 1530, 1584, 1768, 2028, 2090, 2574, 2772, 2860, 2990, 3150, 3588, 3910, 4550, 4624, 4680, 4950, 5434, 5670, 5984, 6498, 6960, 7140, 7548, 8330, 8398, 8432, 8436, 8820, 9568, 10098
Offset: 1

Views

Author

Alexandre Wajnberg, May 02 2005

Keywords

Crossrefs

Extensions

Corrected and extended by Harvey P. Dale, Oct 27 2018

A198464 Consider triples a<=b

Original entry on oeis.org

4, 8, 13, 19, 9, 13, 26, 17, 34, 43, 14, 18, 53, 32, 64, 38, 76, 19, 23, 30, 89, 103, 59, 118, 24, 28, 42, 67, 134, 151, 35, 43, 169, 29, 33, 63, 94, 188, 104, 208, 47, 58, 229, 31, 34, 38, 84, 251, 56, 137, 274, 149, 298, 39, 43, 108, 323, 52, 71, 88, 349
Offset: 1

Views

Author

Charlie Marion, Nov 26 2011

Keywords

Comments

See A198453.

Examples

			2*5 + 3*6 = 4*7
3*6 + 7*10 = 8*11
4*7 + 12*15 = 13*16
5*8 + 18*21 = 19*22
6*9 + 6*9 = 9*12
6*9 + 11*14 = 13*16
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

Crossrefs

A198465 Consider triples a<=b

Original entry on oeis.org

3, 3, 4, 4, 6, 7, 5, 10, 11, 6, 7, 9, 6, 15, 16, 7, 10, 12, 7, 21, 22, 8, 28, 29, 9, 11, 14, 9, 36, 37, 10, 14, 17, 10, 22, 24, 10, 45, 46, 11, 27, 29, 11, 55, 56, 12, 15, 19, 12, 21, 24, 12, 66, 67, 13, 18, 22, 13, 25, 28, 13, 78, 79, 14, 45, 47, 14, 91, 92
Offset: 1

Views

Author

Charlie Marion, Dec 19 2011

Keywords

Comments

The definition can be generalized to define Pythagorean k-triples a<=bA198453 for more about Pythagorean k-triples.

Examples

			3*2 + 3*2 = 4*3
4*3 + 6*5 = 7*6
5*4 + 10*9 = 11*10
6*5 + 7*6 = 9*8
6*5 + 15*14 = 16*15
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

Crossrefs

A198468 Consider triples a<=b

Original entry on oeis.org

4, 7, 11, 9, 16, 12, 22, 29, 14, 37, 17, 24, 46, 29, 56, 19, 24, 67, 22, 28, 79, 47, 92, 21, 24, 37, 54, 106, 27, 42, 121, 137, 29, 53, 78, 154, 32, 59, 87, 172, 41, 50, 191, 34, 45, 55, 72, 211, 37, 42, 79, 117, 232, 128, 254, 39, 94, 277, 42, 63, 102, 301
Offset: 1

Views

Author

Charlie Marion, Dec 19 2011

Keywords

Comments

The definition can be generalized to define Pythagorean k-triples a<=bA198453 for more about Pythagorean k-triples.

Examples

			3*2 + 3*2 = 4*3
4*3 + 6*5 = 7*6
5*4 + 10*9 = 11*10
6*5 + 7*6 = 9*8
6*5 + 15*14 = 16*15
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

Crossrefs

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

Views

Author

Kival Ngaokrajang, Apr 09 2016

Keywords

Comments

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.

Examples

			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,
...
		

Crossrefs

Cf. A103606.

A377739 Array of positive integer triples (x,y,z) where the sum of their cubes equals another cubic number.

Original entry on oeis.org

3, 4, 5, 1, 6, 8, 6, 8, 10, 2, 12, 16, 9, 12, 15, 3, 10, 18, 7, 14, 17, 12, 16, 20, 4, 17, 22, 3, 18, 24, 18, 19, 21, 11, 15, 27, 15, 20, 25, 4, 24, 32, 18, 24, 30, 6, 20, 36, 14, 28, 34, 2, 17, 40, 6, 32, 33, 21, 28, 35, 16, 23, 41, 5, 30, 40, 3, 36, 37, 27, 30, 37, 24, 32, 40, 8, 34, 44, 29, 34, 44, 6, 36, 48, 12, 19, 53, 27, 36, 45, 36, 38, 42
Offset: 1

Views

Author

Luke Voyles, Nov 05 2024

Keywords

Comments

The Shiraishi theorem demonstrated that there were an infinite number of cubic number triples whose sum equaled a cubic number (A226903). Through an analysis of the cubic number triples found by Russell and Gwyther, another way to prove that there are an infinite number of cubic number triples who sum equals a cubic number appeared. For any triple, one can add a zero to the end of the three numbers. The new three numbers will also equal a cubic number. For example, 3^3+4^3+5^3=6^3 can be transformed into 30^3+40^3+50^3=60^3. The number of zeros that are consistently applied to each of the numbers who cubic numbers will always create new cubic numbers. For example, 30^3+40^3+50^3=60^3 can become 300^3+400^3+500^3=600^3 and 3000^3+4000^3+5000^3=6000^3, and so on. Through experiments, the formula holds true for Pythagorean triples and Pythagorean quadruples as well. To apply the method to Pythagorean triples, 3^2+4^2=5^2 can be transformed into 30^2+40^2=50^2, 300^2+400^2=500^2, and so on. For Pythagorean quadruples, 3^2+4^2+12^2=13^2 can be transformed into 30^2+40^2+120^2=130^2 and then to 300^2+400^2+1200^2=1300^2. The property holds even beyond the second and third powers. For example, 3530^4=300^4+1200^4+2720^4+3150^4 just as 353^4=30^4+120^4+272^4+315^4. Additionally, 1440^5=270^5+840^5+1100^5+1330^5 just as 144^5=27^5+84^5+110^5+133^5. Once one set is found, it appears there can be an infinite number of similar sets for any power through this method.
The list of Russell and Gwyther also reveals that the cube of 38 can be represented as the sum of the cubes of nine unique positive integers. This is because 38^3=3^3+4^3+5^3+7^3+14^3+17^3+18^3+24^3+30^3.

Examples

			3^3+4^3+5^3=6^3
1^3+6^3+8^3=9^3
6^3+8^3+10^3=12^3
2^3+12^3+16^3=18^3
9^3+12^3+15^3=18^3
		

Crossrefs

The sum of each cubic number triple produce the sequence A023042. The comments produce another method to produce an infinite number of cubic number triples whose sum equals a cube that the method shown by Shiraishi according to A226903. The comments discuss qualities of Pythagorean triples A103606 and Pythagorean quadruples A096907. The title's structure drew inspiration from A291694.

Formula

If a^3+b^3+c^3=d^3, then any specific number k that has a zero as the last digit will make k(d^3) another cubic number through the formula k(a^3)+k(b^3)+k(c^3)=k(d^3)

A198466 Consider triples a<=b

Original entry on oeis.org

3, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 10, 10, 11, 11, 12, 12, 12, 13, 13, 13, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 17, 18, 18, 18, 18, 19, 19, 19, 19, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 22, 22, 23, 23, 24, 24, 24, 25, 25, 25, 25, 26, 26, 26, 26, 27, 27, 27, 27, 28, 28, 28, 28, 28
Offset: 1

Views

Author

Charlie Marion, Dec 19 2011

Keywords

Comments

The definition can be generalized to define Pythagorean k-triples a<=bA198453 for more about Pythagorean k-triples.

Examples

			3*2 + 3*2 = 4*3
4*3 + 6*5 = 7*6
5*4 + 10*9 = 11*10
6*5 + 7*6 = 9*8
6*5 + 15*14 = 16*15
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

Crossrefs

A198467 Consider triples a<=b

Original entry on oeis.org

3, 6, 10, 7, 15, 10, 21, 28, 11, 36, 14, 22, 45, 27, 55, 15, 21, 66, 18, 25, 78, 45, 91, 15, 19, 34, 52, 105, 22, 39, 120, 136, 23, 50, 76, 153, 26, 56, 85, 171, 36, 46, 190, 27, 40, 51, 69, 210, 30, 36, 76, 115, 231, 126, 253, 31, 91, 276, 34, 58, 99, 300
Offset: 1

Views

Author

Charlie Marion, Dec 19 2011

Keywords

Comments

The definition can be generalized to define Pythagorean k-triples a<=bA198453 for more about Pythagorean k-triples.

Examples

			3*2 + 3*2 = 4*3
4*3 + 6*5 = 7*6
5*4 + 10*9 = 11*10
6*5 + 7*6 = 9*8
6*5 + 15*14 = 16*15
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

Crossrefs

A330138 List of pairs [j,k] ordered by increasing j and k according to the procedure described in the Comments.

Original entry on oeis.org

3, 4, 5, 15, 8, 17, 5, 12, 13, 35, 12, 37, 21, 20, 29, 7, 24, 25, 63, 16, 65, 45, 28, 53, 27, 36, 45, 9, 40, 41, 99, 20, 101, 77, 36, 85, 55, 48, 73, 33, 56, 65, 11, 60, 61, 143, 24, 145, 117, 44, 125, 91, 60, 109, 65, 72, 97, 39, 80, 89, 13, 84, 85
Offset: 1

Views

Author

Joseph Damico, Dec 02 2019

Keywords

Comments

Procedure: The sequence lists two indices, j and k. Let (a, b, and c) represent the two legs and the hypotenuse of a right-angled triangle. The required computations are as follows:
(1) compute 2*j^2 (call this p, it is c-a)
(2) compute (2*k-1)^2 (call this q, it is c-b)
(3) compute 2j*(2k-1) (call this r, it is a+b-c)
So c = p+q+r; a = c-p; b = c-q.
Most, but not all, of the Pythagorean triples generated here are primitive (i.e., they have sides that are relatively prime). In the first 105 Pythagorean triples, there are 14 that are not primitive. The first exception is the 9th in the list: 27,36,45, where j=3 and k=2.

Examples

			As the values of j and k increase, so do the lengths of the sides of the triangles. So it makes sense to order the triangles based on the sum (j+k).  When j=k=1, the sum (j+k)=2, and there is only one triangle (3,4,5).  When the sum of j and k is 3, there are two possibilities: j=1, k=2 (15,8,17) or j=2, k=1 (5,12,13).  When the sum is 4, there are three possibilities. Each time the sum (j+k) increases by one, the number of triangles also increases by one. And the total number of triangles grows according to sequence A000217 (the triangular numbers).
Value of j+k:    2, 3, 4,  5,  6,  7,  8,  9, 10, 11, 12, ...
Total triangles: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, ...
So it is clear that triangle number 61 would be in the range of triangles ranked 56 through 66, where (j+k)=12.  In fact, exactly 6 beyond the triangle associated with the values j=10, k=1.  So j=6, k=6 is associated with the sixty-first triangle. And the computations give us this Pythagorean triple: a=253, b=204, c=325.
		

Crossrefs

Cf. A103606.

Programs

  • Mathematica
    Table[Map[{#2 + #3, #1 + #3, #1 + #2 + #3} & @@ {2 #1^2, (2 #2 - 1)^2, 2 #1 (2 #2 - 1)} & @@ {#, n - #} &, Range[n - 1]], {n, 7}] // Flatten (* Michael De Vlieger, Dec 14 2019 *)
Previous Showing 21-29 of 29 results.