A351477 -id:A351477 - cp's OEIS frontend

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).

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

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.

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).

A336329 When F is the Fermat point of a triangle ABC, this sequence lists the integer total distances FA + FB + FC corresponding to primitive triangles in A336328.

Original entry on oeis.org

112, 147, 185, 283, 273, 331, 331, 403, 559, 485, 645, 520, 691, 592, 637, 965, 1047, 1560, 1415, 1688, 1649, 2093, 1895, 2045, 1687, 1843, 2073, 1839, 1768, 1805, 1729, 1729, 2593, 2337, 2792, 2408, 2709, 2696, 2813, 2704, 2960, 3192, 3007, 3681, 3217, 3752, 2855

Offset: 1

Bernard Schott, Jul 18 2020

nonn

Inspired by Project Euler, Problem 143 (see link).
The triples of sides (a,b,c) with a < b < c 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.
For the terms of the data, every FA, FB, FC is a fraction but FA + FB + FC is an integer (see example).
This sequence is not increasing. For example, a(5) = 283 for triangle with largest side = 205 while a(6) = 273 for triangle with largest side = 208. Also, a(6) = a(7) = 331 show that two distinct triangles can have the same minimum possible integer distance FA + FB + FC.

			For first triple (57, 65, 73), d = 112 is solution of 3*(57^4 + 65^4 + 73^4 + d^4) = (57^2 + 65^2 + 73^2 + d^2)^2, hence, 112 is a term because d = FA + FB + FC = 264/7 + 195/7 + 325/7 = 112.

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), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), A351477.

Cf. A061281 (supersequence with non-primitive terms).

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(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, &d), print1(d, ", "))))); \\ Jinyuan Wang, Jul 20 2020

For triangle (a, b, c) whose area is S, and d = FA+FB+FC, then
d = sqrt((1/2)*(a^2+b^2+c^2) + 2*S*sqrt(3)), also,
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
3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.

More terms from Jinyuan Wang, Jul 20 2020

A352360 Three-column array giving list of primitive triples for integer-sided triangles with A < B < C < 2*Pi/3 and such that FA, FB, FC are also integers where F is the Fermat point of the triangle.

Original entry on oeis.org

399, 455, 511, 511, 616, 665, 1591, 5439, 5624, 35941, 47544, 58015, 8827, 16835, 18928, 36741, 73151, 92680, 16219, 94335, 97976, 1235, 4056, 4459, 12728, 13545, 15523, 14744, 33271, 37539, 13889, 16856, 17501, 1911, 4901, 5681, 196935, 320624, 324079, 9435, 12691, 17501, 22477, 37128, 44135

Offset: 1

Bernard Schott, Mar 17 2022

Inspired by Project Euler, Problem 143 (see link) where such a triangle is called a Torricelli triangle.
Differs from A336328 where FA + FB + FC is an integer, but FA, FB and FC are fractions. Jinyuan Wang has found that the 37th and 58th triples are the first triples for which the common denominator of these fractions is 1 (A351477).
Each triple (a, b, c) is in increasing order, and the triples are displayed in the same increasing order of the corresponding triples in A336328 (see formulas).
+-------+-------+-------+---------+--------+-------+-------+--------+--------+
|    a  |   b   |   c   |gcd(a,b,c)|   FA  |   FB  |   FC  |    d   | a+b+c  |
+-------+-------+-------+----------+-------+-------+-------+--------+--------+
|   399 |   455 |   511 |     7    |   325 |   264 |   195 |    784 |   1365 |
|   511 |   616 |   665 |     7    |   440 |   325 |   264 |   1029 |   1792 |
|  1591 |  5439 |  5624 |    37    |  5016 |  1064 |   765 |   6845 |  12654 |
| 35941 | 47544 | 58015 |   283    | 39360 | 27265 | 13464 |  80089 | 141500 |
|  8827 | 16835 | 18928 |    91    | 14800 |  6528 |  3515 |  24843 |  44590 |
| 36741 | 73151 | 92680 |   331    | 70720 | 34200 |  4641 | 109561 | 202572 |
| 16219 | 94335 | 97976 |   331    | 91200 | 12376 |  5985 | 109561 | 208530 |
|  1235 |  4056 |  4459 |    13    |  3864 |  1015 |   360 |   5239 |   9750 |
| 12728 | 13545 | 15523 |    43    |  9405 |  8512 |  6120 |  24037 |  41796 |
| 14744 | 33271 | 37539 |    97    | 30429 | 11520 |  5096 |  47045 |  87554 |
..............................................................................
The sequences with terms of this table are listed in Crossrefs section; here, d = FA + FB + FC. The perimeter corresponding to n-th triple a+b+c = A336333(n) * A351477(n).

			The array begins:
    399,   455,   511;
    511,   616,   665;
   1591,  5439,  5624;
  35941, 47544, 58015;
   8827, 16835, 18928;
  36741, 73151, 92680;
  .....................
For 1st triple (399, 455, 511) with gcd(399, 455, 511) = 7, we get FA = 325, FB = 264 and FC = 195. This smallest triangle such that a, b, c, FA, FB, FC are all integers is the example proposed in Project Euler's link.

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

Cf. A336328.

Cf. A351477 (gcd(a,b,c)), A351801 (FA), A351802 (FB), A351803 (FC), A351476 (FA+FB+FC).

a(3n-2) = A336328(3n-2) * A351477(n), a(3n-1) = A336328(3n-1) * A351477(n), a(3n) = A336328(3n) * A351477(n), for n >= 1.

cp's OEIS Frontend

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).

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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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PARI

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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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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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A336329 When F is the Fermat point of a triangle ABC, this sequence lists the integer total distances FA + FB + FC corresponding to primitive triangles in A336328.

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