A336329 -id:A336329 - cp's OEIS frontend

A351477 a(n) is the common denominator of FA, FB and FC, where F is the Fermat point of the integer-sided triangle ABC with A < B < C < 2*Pi/3 such that FA + FB + FC = A336329(n).

7, 7, 37, 283, 91, 331, 331, 13, 43, 97, 43, 13, 691, 37, 91, 193, 349, 13, 283, 211, 97, 91, 379, 409, 7, 97, 691, 613, 13, 19, 13, 91, 2593, 19, 349, 43, 1, 337, 97, 169, 37, 19, 31, 409, 3217, 67, 571, 169, 241, 43, 67, 157, 4171, 3601, 889, 1591, 811, 1, 139

Offset: 1

Bernard Schott, Feb 12 2022

nonn

Inspired by Project Euler, Problem 143 (see link).

For the corresponding primitive triples, miscellaneous properties and references, see A336328.

			For 1st triple (57, 65, 73) in A336328, we get A336329(1) = FA + FB + FC = 325/7 + 264/7 + 195/7 = 112, hence a(1) = 7.
For 3rd triple (43, 147, 152) in A336328, we get A336329(3) = FA + FB + FC = 5016/37 + 1064/37 + 765/37 = 185, hence a(3) = 37.

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.
Eric Weisstein's World of Mathematics, Fermat points.

Cf. A336328 (primitive triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter).

Cf. A351476.

a(n) = A351476(n)/A336329(n).

a(n) is the common denominator of fractions FA, FB, FC when FA = sqrt(((2*b*c)^2 - (b^2+c^2-d^2)^2)/3) / d, FB = sqrt(((2*a*c)^2 - (a^2+c^2-d^2)^2)/3) / d, FC = sqrt(((2*a*b)^2 - (a^2+b^2-d^2)^2)/3) / d, with a = (A336328(n,1), b = (A336328(n,2), c = (A336328(n,3)) and d = A336329(n) (formulas FA, FB, FC from Jinyuan Wang, Feb 17 2022).

More terms from Jinyuan Wang, Feb 17 2022

A351801 a(n) = A351477(n) * FA where F is the Fermat point of a primitive integer-sided triangle ABC with A < B < C < 2*Pi/3 and FA + FB + FC = A336329(n).

Original entry on oeis.org

325, 440, 5016, 39360, 14800, 70720, 91200, 3864, 9405, 30429, 11704, 4669, 250096, 11704, 32640, 81840, 203000, 7208, 218120, 199325, 99360, 76760, 359352, 342912, 8184, 122200, 595595, 621387, 12600, 26040, 19320, 137344, 3108105, 24955, 409640, 58400, 1520

Offset: 1

Bernard Schott, Feb 19 2022

nonn

Inspired by Project Euler, Problem 143 (see link) where such a triangle is called a "Torricelli triangle".
For the corresponding primitive triples, miscellaneous properties and references, see A336328.
Equivalently, a(n) is the numerator of the fraction FA = a(n) / A351477(n).
Also, if F is the Fermat point of a triangle ABC with A < B < C < 2*Pi/3, where AB, BC, CA, FA, FB and FC are all positive integers, then, when FA + FB + FC = d = A351476(n), we have FA = a(n).
FA is the largest length with FC < FB < FA (remember a < b < c).

			For the 1st triple in A336328, i.e., (57, 65, 73), we get A336329(1) = FA + FB + FC = 325/7 + 264/7 + 195/7 = 112, hence A351477(1) = 7 and a(1) = 325.

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.
Wikipedia, Fermat point.

Cf. A336328 (primitive triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), this sequence (FA numerator), A351802 (FB numerator), A351803 (FC numerator), A351477 (common denominator of FA, FB, FC), A351476 (other 'FA + FB + FC').

lista(nn) = {my(d); for(c=4, nn, for(b=ceil(c/sqrt(3)), c-1, for(a=1+(sqrt(4*c^2-3*b^2)-b)\2, b-1, if(gcd([a, b, c])==1 && issquare(d=6*(a^2*b^2+b^2*c^2+c^2*a^2)-3*(a^4+b^4+c^4)) && issquare(d=(a^2+b^2+c^2+sqrtint(d))/2), d = sqrtint(d); print1(numerator(sqrtint(((2*b*c)^2 - (b^2 + c^2 - d^2)^2)/3)/d), ", ");););););} \\ Michel Marcus, Mar 02 2022

a(n) = A351476(n) - A351802(n) - A351803(n).

FA = sqrt(((2*b*c)^2 - (b^2 + c^2 - d^2)^2)/3) / d. - Jinyuan Wang, Feb 19 2022

More terms from Jinyuan Wang, Feb 19 2022

A351802 a(n) = A351477(n) * FB where F is the Fermat point of a primitive integer-sided triangle ABC with A < B < C < 2*Pi/3 and FA + FB + FC = A336329(n).

Original entry on oeis.org

264, 325, 1064, 27265, 6528, 34200, 12376, 1015, 8512, 11520, 8415, 1656, 116025, 8415, 17575, 56448, 81928, 6765, 107712, 106128, 43953, 60903, 235008, 311885, 3105, 32571, 571648, 411320, 9499, 4991, 1800, 13875, 1894144, 16320, 402375, 42735, 805, 218925

Offset: 1

Bernard Schott, Feb 19 2022

nonn

Inspired by Project Euler, Problem 143 (see link) where such a triangle is called a "Torricelli triangle".
For the corresponding primitive triples, miscellaneous properties and references, see A336328.
Equivalently, a(n) is the numerator of the fraction FB = a(n) / A351477(n).
Also, if F is the Fermat point of a triangle ABC with A < B < C < 2*Pi/3, where AB, BC, CA, FA, FB and FC are all positive integers, then, when FA + FB + FC = d = A351476(n), we have FB = a(n).
FB is the middle length with FC < FB < FA (remember a < b < c).

			For the 2nd triple in A336328, i.e., (73, 88, 95), we get A336329(2) = FA + FB + FC = 440/7 + 325/7 + 264/7 = 147, hence A351477(2) = 7 and a(2) = 325.

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.

Cf. A336328 (primitive triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), A351801 (FA numerator), this sequence (FB numerator), A351803 (FC numerator), A351477 (common denominator of FA, FB, FC), A351476 (other 'FA + FB + FC').

lista(nn) = {my(d); for(c=4, nn, for(b=ceil(c/sqrt(3)), c-1, for(a=1+(sqrt(4*c^2-3*b^2)-b)\2, b-1, if(gcd([a, b, c])==1 && issquare(d=6*(a^2*b^2+b^2*c^2+c^2*a^2)-3*(a^4+b^4+c^4)) && issquare(d=(a^2+b^2+c^2+sqrtint(d))/2), d = sqrtint(d); print1(numerator(sqrtint(((2*a*c)^2 - (a^2+c^2-d^2)^2)/3)/d), ", ");););););} \\ Michel Marcus, Mar 01 2022

a(n) = A351476(n) - A351801(n) - A351803(n).

FB = sqrt(((2*a*c)^2 - (a^2+c^2-d^2)^2)/3) / d. - Jinyuan Wang, Feb 19 2022

More terms from Jinyuan Wang, Feb 19 2022

A351803 a(n) = A351477(n) * FC where F is the Fermat point of a primitive integer-sided triangle ABC with A < B < C < 2*Pi/3 and FA + FB + FC = A336329(n).

Original entry on oeis.org

195, 264, 765, 13464, 3515, 4641, 5985, 360, 6120, 5096, 7616, 435, 111360, 1785, 7752, 47957, 80475, 6307, 74613, 50715, 16640, 52800, 123845, 181608, 520, 24000, 265200, 94600, 885, 3264, 1357, 6120, 1721400, 3128, 162393, 2409, 384, 122507, 27720, 22575, 12383

Offset: 1

Bernard Schott, Feb 19 2022

nonn

Inspired by Project Euler, Problem 143 (see link) where such a triangle is called a "Torricelli triangle".
For the corresponding primitive triples, miscellaneous properties and references, see A336328.
Equivalently, a(n) is the numerator of the fraction FC = a(n) / A351477(n).
Also, if F is the Fermat point of a triangle ABC with A < B < C < 2*Pi/3, where AB, BC, CA, FA, FB and FC are all positive integers, then, when FA + FB + FC = d = A351476(n), we have FC = a(n).
FC is the smallest length with FC < FB < FA (remember a < b < c).

			For the 3rd triple in A336328, i.e., (43, 147, 152), we get A336329(3) = FA + FB + FC = 5016/37 + 1064/37 + 765/37 = 185, hence A351477(3) = 37 and a(3) = 765.

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.

Cf. A336328 (primitive triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), A351801 (FA numerator), A351802 (FB numerator), this sequence (FC numerator), A351477 (common denominator of FA, FB, FC), A351476 (FA + FB + FC other).

lista(nn) = {my(d); for(c=4, nn, for(b=ceil(c/sqrt(3)), c-1, for(a=1+(sqrt(4*c^2-3*b^2)-b)\2, b-1, if(gcd([a, b, c])==1 && issquare(d=6*(a^2*b^2+b^2*c^2+c^2*a^2)-3*(a^4+b^4+c^4)) && issquare(d=(a^2+b^2+c^2+sqrtint(d))/2), d = sqrtint(d); print1(numerator(sqrtint(((2*a*b)^2 - (a^2 + b^2 - d^2)^2)/3)/d), ", ");););););} \\ Michel Marcus, Mar 02 2022

a(n) = A351476(n) - A351801(n) - A351802(n).

FC = sqrt(((2*a*b)^2 - (a^2 + b^2 - d^2)^2)/3) / d. - Jinyuan Wang, Feb 19 2022

More terms from Jinyuan Wang, Feb 19 2022

A336328 Primitive triples for integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

Original entry on oeis.org

57, 65, 73, 73, 88, 95, 43, 147, 152, 127, 168, 205, 97, 185, 208, 111, 221, 280, 49, 285, 296, 95, 312, 343, 296, 315, 361, 152, 343, 387, 323, 392, 407, 147, 377, 437, 285, 464, 469, 255, 343, 473, 247, 408, 485, 469, 589, 624, 403, 725, 728, 871, 901, 931

Offset: 1

Bernard Schott, Jul 17 2020

Inspired by Project Euler, Problem 143 (see link).
The triples are displayed in increasing order of largest side c, and if largest sides coincide then by increasing order of the middle side b; so, each triple (a, b, c) is in increasing order.
If one angle of the triangle, for example C, is >= 2*Pi/3 then the Fermat point F is this vertex C, so, FA + FB + FC becomes CA + CB, while when all angles are < 2*Pi/3, then the Fermat point is inside the triangle (see link Fermat points), this last condition means that c^2 < a^2 + a*b + b^2.
As a < b < c, then FA > FB > FC.
If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d: 3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2 (see Martin Gardner).
Equivalently: if a point M is inside an equilateral triangle A'B'C' and integer distances to vertices are MA' = a = A072054(n), MB' = b = A072053(n), MC' = c = A072052(n), then the side of this equilateral triangle A'B'C' is equal to d = FA + FB + FC = A061281(n) where F is the Fermat point of the triangle ABC with sides (a,b,c) (see Martin Gardner).
+-----+-----+-----+-----------+-----------+-----------+-----+-------+
|  a  |  b  |  c  |    FA     |    FB     |    FC     |  d  | a+b+c |
+-----------+-----+-----------+-----------+-----------+-----+-------+
|  57 |  65 |  73 |   325/7   |   264/7   |   195/7   | 112 |  195  |
|  73 |  88 |  95 |   440/7   |   325/7   |   264/7   | 147 |  256  |
|  43 | 147 | 152 |  5016/37  |  1064/37  |   765/37  | 185 |  342  |
| 127 | 168 | 205 | 39360/283 | 27265/283 | 13464/283 | 283 |  500  |
|  97 | 185 | 208 | 14800/91  |  6528/91  |  3515/91  | 273 |  490  |
| 111 | 221 | 280 | 70720/331 | 34200/331 |  4641/331 | 331 |  612  |
|  49 | 285 | 296 | 91200/331 | 12376/331 |  5985/331 | 331 |  630  |
|  95 | 312 | 343 |  3864/13  |  1015/13  |   360/13  | 403 |  750  |
| 296 | 315 | 361 |  9405/43  |  8512/43  |  6120/43  | 559 |  972  |
| 152 | 343 | 387 | 30429/97  | 11520/97  |  5096/97  | 485 |  882  |
.....................................................................
From the previous table, we observe that every FA, FB, FC is a fraction while FA + FB + FC = d is an integer (A336329). Jinyuan Wang has found that the 37th triple is the first for which the common denominator of these fractions is 1 (A351477).

			The table begins:
   57,  65,  73;
   73,  88,  95;
   43, 147, 152;
  127, 168, 205;
   97, 185, 208;
  111, 221, 280;
   49, 285, 296;
  .............
For first triple (57, 65, 73) and corresponding d = FA + FB + FC = 325/7 + 264/7 + 195/7 = 112, relation gives: 3*(57^4 + 65^4 + 73^4 + 112^4) = (57^2 + 65^2 + 73^2 + 112^2)^2 = 642470409.

Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.
Eric Weisstein's World of Mathematics, Fermat points.

Cf. A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), A351801 (FA numerator), A351802 (FB numerator), A351803 (FC numerator), A351477 (common denominator of FA, FB, FC), A351476 (other FA + FB + FC).
Cf. A333391 (with isogonic center).
Cf. A061281, A072052, A072053, A072054.

If FA + FB + FC = d, then
d = sqrt(((a^2 + b^2 + c^2)/2) + (1/2) * sqrt(6*(a^2*b^2 + b^2*c^2 + c^2*a^2) - 3*(a^4 + b^4 + c^4))), or,
d^2 = (1/2) * (a^2 + b^2 + c^2) + 2 * S * sqrt(3) where S = area of triangle ABC.

A336332 Largest side, in increasing order, of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

Original entry on oeis.org

73, 95, 152, 205, 208, 280, 296, 343, 361, 387, 407, 437, 469, 473, 485, 624, 728, 931, 1016, 1273, 1311, 1313, 1368, 1387, 1443, 1457, 1463, 1469, 1477, 1519, 1560, 1591, 1687, 1895, 2015, 2045, 2045, 2085, 2197, 2231, 2289, 2347, 2363, 2416, 2465, 2553, 2728, 2821, 2923

Offset: 1

Bernard Schott, Jul 20 2020

nonn

Inspired by Project Euler, Problem 143 (see link).
This sequence is increasing because triples are in increasing order of largest side.
For the corresponding primitive triples and miscellaneous properties and references, see A336328.
If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner): 3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.

			a(36) = a(37) = 2045 is the smallest largest side that appears twice because:
   (1023, 1387, 2045) is a triple with FA+FB+FC = 2408, and
   (1051, 1744, 2045) is a triple with FA+FB+FC = 2709.

Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.

Cf. A336328 (triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), this sequence (largest side), A336333 (perimeter).

Cf. A072052 (largest sides: primitives and multiples), A333391.

a(n) = A336328(n, 3).

A336333 Perimeter of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

Original entry on oeis.org

195, 256, 342, 500, 490, 612, 630, 750, 972, 882, 1122, 961, 1218, 1071, 1140, 1682, 1856, 2703, 2508, 3015, 2990, 3636, 3348, 3572, 3136, 3364, 3640, 3328, 3249, 3362, 3312, 3330, 4530, 4250, 4921, 4455, 4840, 4565, 5054, 4945, 5307, 5655, 5440, 6440, 5746, 6561, 5588

Offset: 1

Bernard Schott, Jul 21 2020

nonn

Inspired by Project Euler, Problem 143 (see link).
For the corresponding primitive triples and miscellaneous properties and references, see A336328.
If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):
   3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
This sequence is not increasing. For example, a(4) = 500 for triangle with largest side = 205 while a(5) = 490 for triangle with largest side = 208.

			a(1) = 195 because the first triple is (57, 65, 73) with corresponding d = FA + FB + FC = 264/7 + 195/7 + 325/7 = 112 and 57 + 65 + 73 = 195.

Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.

Cf. A336328 (triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A351476, A351477.

a(n) = A336328(n, 1) + A336328(n, 2) + A336328(n, 3).

a(n) = A336330(n) + A336331(n) + A336332(n).

A336330 Smallest side of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

Original entry on oeis.org

57, 73, 43, 127, 97, 111, 49, 95, 296, 152, 323, 147, 285, 255, 247, 469, 403, 871, 561, 657, 559, 1083, 833, 1057, 485, 507, 1072, 760, 767, 379, 211, 195, 1208, 952, 1443, 1023, 1051, 889, 1240, 1209, 1249, 1423, 1005, 1679, 1568, 1843, 193, 485, 1512

Offset: 1

Bernard Schott, Jul 21 2020

nonn

Inspired by Project Euler, Problem 143 (see link).
For the corresponding primitive triples and miscellaneous properties and references, see A336328.
If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):
   3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
This sequence is not increasing. For example, a(2) = 73 for triangle with largest side = 95 while a(3) = 43 for triangle with largest side = 152.

			a(1) = 57 because the first triple is (57, 65, 73) with corresponding d = FA + FB + FC = 264/7 + 195/7 + 325/7 = 112 and the symmetric relation satisfies: 3*(57^4 + 65^4 + 73^4 + 112^4) = (57^2 + 65^2 + 73^2 + 112^2)^2 = 642470409.

Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.

Cf. A336328 (triples), A336329 (FA + FB + FC), this sequence (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter).

Cf. A072054 (smallest sides: primitives and multiples).

a(n) = A336328(n, 1).

A336331 Middle side of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

Original entry on oeis.org

65, 88, 147, 168, 185, 221, 285, 312, 315, 343, 392, 377, 464, 343, 408, 589, 725, 901, 931, 1085, 1120, 1240, 1147, 1128, 1208, 1400, 1105, 1099, 1005, 1464, 1541, 1544, 1635, 1403, 1463, 1387, 1744, 1891, 1617, 1505, 1769, 1885, 2072, 2345, 1713, 2165, 2667, 2784, 2855

Offset: 1

Bernard Schott, Jul 20 2020

nonn

Inspired by Project Euler, Problem 143 (see link).
For the corresponding primitive triples and miscellaneous properties and references, see A336328.
If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):
   3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
This sequence is not increasing. For example, a(11) = 392 for triangle with largest side = 407 while a(12) = 377 for triangle with largest side = 437.

			a(1) = 65 because the first triple is (57, 65, 73) with corresponding d = FA + FB + FC = 264/7 + 195/7 + 325/7 = 112 and the symmetric relation satisfies: 3*(57^4 + 65^4 + 73^4 + 112^4) = (57^2 + 65^2 + 73^2 + 112^2)^2 = 642470409.

Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.

Cf. A336328 (triples), A336329 (FA + FB + FC), A336330 (smallest side), this sequence (middle side), A336332 (largest side), A336333 (perimeter).

Cf. A072053 (middle sides: primitives and multiples).

a(n) = A336328(n, 2).

A351476 If F is the Fermat point of a triangle ABC with A < B < C < 2*Pi/3, where AB, BC, CA, FA, FB and FC are all positive integers, then, this sequence gives the sum FA + FB + FC when gcd(a, b, c) = A351477(n).

Original entry on oeis.org

784, 1029, 6845, 80089, 24843, 109561, 109561, 5239, 24037, 47045, 27735, 6760, 477481, 21904, 57967, 186245, 365403, 20280, 400445, 356168, 159953, 190463, 718205, 836405, 11809, 178771, 1432443, 1127307, 22984, 34295, 22477, 157339, 6723649, 44403, 974408

Offset: 1

Bernard Schott, Feb 12 2022

nonn

Inspired by Project Euler, Problem 143 (see link) where such a triangle is called a "Torricelli triangle".
Differs from A336329 where ABC is a primitive integer-sided triangle with A < B < C < 2*Pi/3 and only FA+FB+FC is an integer; in fact, terms of A336329 are the sum of 3 fractions of the form FA = p/t, FB = q/t, FC = r/t but (p+q+r)/t is an integer. Here, FA, FB and FC are all integers and the sums FA+FB+FC are displayed according to same order as in A336329. The corresponding common denominators t of the fractions (p/t, q/t, r/t) are in A351477.
If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):
3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
=> d = sqrt(((a^2 + b^2 + c^2)/2) + (1/2) * sqrt(6*(a^2*b^2 + b^2*c^2 + c^2*a^2) - 3*(a^4 + b^4 + c^4))).

			a(1) = FA + FB + FC = 325 + 264 + 195 = 784, corresponding to first triple (399, 455, 511) whose gcd = 7.
a(6) = FA + FB + FC = 70720 + 34200 + 4641 = 109561, corresponding to triple (36741, 73151, 92680) whose gcd = 331.
a(7) = FA + FB + FC = 91200 + 12376 + 8985 = 109561, corresponding to triple (16219, 94335, 97976) whose gcd = 331.

Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.
Eric Weisstein's World of Mathematics, Fermat points.

Cf. A336328, A336329, A336330, A336331, A336332, A336333, A351477, A351801, A351802, A351803.

lista(nn) = {my(d); for(c=4, nn, for(b=ceil(c/sqrt(3)), c-1, for(a=1+(sqrt(4*c^2-3*b^2)-b)\2, b-1, if(gcd([a, b, c])==1 && issquare(d=6*(a^2*b^2+b^2*c^2+c^2*a^2)-3*(a^4+b^4+c^4)) && issquare(d=(a^2+b^2+c^2+sqrtint(d))/2), d = sqrtint(d); my(s = numerator(sqrtint(((2*b*c)^2 - (b^2+c^2-d^2)^2)/3)/d) + numerator(sqrtint(((2*a*b)^2 - (a^2+b^2-d^2)^2)/3)/d) + numerator(sqrtint(((2*a*c)^2 - (a^2+c^2-d^2)^2)/3)/d)); print1(s, ", ");););););} \\ Michel Marcus, Mar 02 2022

a(n) = A336329(n) * A351477(n).
If FA + FB + FC = d, then
d^2 = (1/2) * (a^2 + b^2 + c^2) + 2 * S * sqrt(3) where S = area of triangle ABC.

More terms from Jinyuan Wang, Feb 17 2022

cp's OEIS Frontend

A351477 a(n) is the common denominator of FA, FB and FC, where F is the Fermat point of the integer-sided triangle ABC with A < B < C < 2*Pi/3 such that FA + FB + FC = A336329(n).

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A351801 a(n) = A351477(n) * FA where F is the Fermat point of a primitive integer-sided triangle ABC with A < B < C < 2*Pi/3 and FA + FB + FC = A336329(n).

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A351802 a(n) = A351477(n) * FB where F is the Fermat point of a primitive integer-sided triangle ABC with A < B < C < 2*Pi/3 and FA + FB + FC = A336329(n).

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A351803 a(n) = A351477(n) * FC where F is the Fermat point of a primitive integer-sided triangle ABC with A < B < C < 2*Pi/3 and FA + FB + FC = A336329(n).

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A336328 Primitive triples for integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

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A336332 Largest side, in increasing order, of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

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A336333 Perimeter of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

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A336330 Smallest side of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

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