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A337251 Positive integers k such that k^2 = A^2+B^2+C^2 and A^3+B^3+C^3 = m^3, where gcd(A,B,C) = 1 and A, B, C, m are positive integers.

75, 119, 551, 755, 4501, 4895, 16371, 56863, 61091, 74201, 201797, 336709, 534793, 596827, 879397, 1007541

Offset: 1

Mo Li, Aug 21 2020

From Chai Wah Wu, Sep 04 2020: (Start)
A. Martin and R. Davis showed that 91091088729334859 = sqrt(11868013975030087^2+16269106368215226^2+88837226814909894^2) is a term (see Links).
Table of values for k, A, B, C, m:
  k        A        B        C        m
  ---------------------------------------------
  75       14       23       70       71
  119      3        34       114      115
  551      18       349      426      493
  755      145      198      714      721
  4501     1016     2364     3693     4013
  4895     213      3450     3466     4357
  16371    3542     9286     13009    14497
  56863    6213     32194    46458    51157
  61091    29233    29574    44754    51985
  74201    32913    38444    54264    63185
  201797   106677   117252   124876   168373
  336709   110051   118044   295512   306467
  534793   116457   286752   436136   476393
  596827   202023   234550   510270   536023
  879397   43472    613560   628485   782597
  1007541  272267   417416   875656   914315
(End)

			56863 is in the sequence because 56863^2 = 6213^2 + 32194^2 + 46458^2, 6213^3 + 32194^3 + 46458^3 = 51157^3 and gcd(6213, 32194, 46458) = 1.

A. Martin and R. Davis, Solution of problem 143, Jahrbuch über die Fortschritte der Mathematik, Band 29, Jahrgang 1898, pub. 1900, p. 157.
Ed Pegg Jr.'s Math Puzzles, A^2 + B^2 + C^2 = Square, A^3 + B^3 + C^3 = Cube
Seiji Tomita, A simultaneous equation {x^2+y^2+z^2=u^2, x^3+y^3+z^3=v^3} has infinitely many integer solutions.

Cf. A096910.

A333391 Longest side of primitive integer triangles with nonzero rational distances between three vertices and first isogonic center, sorted.

Original entry on oeis.org

73, 95, 152, 205, 208, 280, 285, 287, 296, 343, 361, 387, 407, 437, 469, 473, 485, 497, 507, 608, 624, 633, 645, 713, 715, 728, 728, 817, 873, 931, 1016, 1273, 1288, 1311, 1313, 1343, 1368, 1387, 1443, 1457, 1463, 1469, 1477, 1488, 1519, 1519, 1560, 1584, 1591, 1591

Offset: 1

Mo Li, Mar 18 2020

nonn

			Case 1: When the isogonic center is inside the triangle, i.e., the three internal angles are all less than 120 degrees. Example: Length of three sides (a, b, c) = (57, 65, 73), rational distances with signs (x, y, z) = (325/7, 264/7, 195/7);
Case 2: When the isogonic center is outside the triangle, i.e., an internal angle is greater than 120 degrees. Example: Lengths of three sides (a, b, c) = (43, 248, 285), rational distances with signs (x, y, z) = (1800/7, 345/7, -136/7);
Thus 73 and 285 are in this sequence.
a(26) = a(27) = 728 is the smallest longest side that appears twice because: (a, b, c) = (57, 673, 728) is a triple with (x, y, z) = (9016/13, 840/13, -561/13), and (a, b, c) = (403, 725, 728) is a triple with (x, y, z) = (203000/349, 81928/349, 80475/349). - _Jinyuan Wang_, Feb 12 2025

Leisure Maths Entertainment Forum, The primitive integer triangles with nonzero rational distances between three vertices and 1st isogonic center, Chinese blog.
Project Euler, Problem 143. Investigating the Torricelli point of a triangle

Cf. A070082, A336332.

lista(nn) = my(d); for(c=4, nn, for(b=(c+2)\2, c-1, for(a=c-b+1, b-1, if(gcd([a, b, c])==1 && a^2+b^2+a*b!=c^2 && issquare(6*(a^2*b^2+b^2*c^2+c^2*a^2)-3*(a^4+b^4+c^4), &d) && issquare((a^2+b^2+c^2+d)/2), print1(c, ", "))))); \\ Jinyuan Wang, Feb 12 2025

More terms from Jinyuan Wang, Feb 12 2025

A328499 The number of primitive Pythagorean triangles with perimeter less than n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset: 1

Mo Li, Oct 17 2019

D. N. Lehmer has proved that the asymptotic density of a(n) is a(n)/n = log(2)/Pi^2 = 0.07023049... See A118858.

			For n=90, the triples are
   {3,  4,  5},  3 +  4 +  5 = 12 < 90
   {5, 12, 13},  5 + 12 + 13 = 30 < 90
   {7, 24, 25},  7 + 24 + 25 = 56 < 90
   {8, 15, 17},  8 + 15 + 17 = 40 < 90
   {9, 40, 41},  9 + 40 + 41 = 90
  {12, 35, 37}, 12 + 35 + 37 = 84 < 90
  {20, 21, 29}, 20 + 21 + 29 = 70 < 90
so a(90)=7.

Ron Knott, Pythagorean Triples and Online Calculators
D. N. Lehmer, Asymptotic evaluation of certain totient sums, Amer. J. Math. 22, 293-335, 1900.

Cf. A008846, A024364, A118858, A249750.

A307304 Number of inequivalent ways of placing 2 nonattacking rooks on n X n board up to rotations and reflections of the board.

Original entry on oeis.org

0, 1, 4, 13, 31, 66, 123, 214, 346, 535, 790, 1131, 1569, 2128, 2821, 3676, 4708, 5949, 7416, 9145, 11155, 13486, 16159, 19218, 22686, 26611, 31018, 35959, 41461, 47580, 54345, 61816, 70024, 79033, 88876, 99621, 111303, 123994, 137731, 152590, 168610, 185871

Offset: 1

Mo Li, Apr 19 2019

			For n = 4 the a(4) = 13 solutions are
{{1,0,0,0}}  {{1,0,0,0}}  {{1,0,0,0}}
{{0,1,0,0}}  {{0,0,1,0}}  {{0,0,0,1}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
—————————————————————————————————————
{{1,0,0,0}}  {{1,0,0,0}}  {{1,0,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
{{0,0,1,0}}  {{0,0,0,1}}  {{0,0,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,1}}
—————————————————————————————————————
{{0,1,0,0}}  {{0,1,0,0}}  {{0,1,0,0}}
{{1,0,0,0}}  {{0,0,1,0}}  {{0,0,0,1}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
—————————————————————————————————————
{{0,1,0,0}}  {{0,1,0,0}}  {{0,1,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,0,0}}
{{0,0,1,0}}  {{0,0,0,1}}  {{0,0,0,0}}
{{0,0,0,0}}  {{0,0,0,0}}  {{0,0,1,0}}
—————————————————————————————————————
{{0,0,0,0}}
{{0,1,0,0}}
{{0,0,1,0}}
{{0,0,0,0}}

Leisure Maths Entertainment Forum, 2 nonattacking rooks on n X n board, Chinese blog.

Cf. A000903, A163102, A035287, A179058, A144084.

Table[
Piecewise[{{(n (n^3 - 2 n^2 + 6 n - 4))/16, Mod[n, 2] == 0},
{((n - 1) (n^3 - n^2 + 5 n - 1))/16, Mod[n, 2] == 1}}],{n, 20}]

a(n) = (1/16)*n*(n^3-2n^2+6n-4) if n is even;
a(n) = (1/16)*(n-1)*(n^3-n^2+5n-1) if n is odd.
G.f.: -x^2*(x^2+1)*(x^2+x+1)/((x+1)^2*(x-1)^5). - Alois P. Heinz, Apr 26 2019

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A337251 Positive integers k such that k^2 = A^2+B^2+C^2 and A^3+B^3+C^3 = m^3, where gcd(A,B,C) = 1 and A, B, C, m are positive integers.

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A333391 Longest side of primitive integer triangles with nonzero rational distances between three vertices and first isogonic center, sorted.

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A328499 The number of primitive Pythagorean triangles with perimeter less than n.

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A307304 Number of inequivalent ways of placing 2 nonattacking rooks on n X n board up to rotations and reflections of the board.

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