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.

Showing 1-6 of 6 results.

A072052 Largest of 3 integer distances referred to in A061281.

Original entry on oeis.org

73, 95, 152, 146, 208, 205, 190, 280, 296, 219, 304, 343, 285, 292, 387, 437, 416, 456, 361, 365, 410, 380, 473, 485, 407, 560, 592, 438, 469, 475, 608, 511, 686, 624, 615, 570, 584, 760, 624, 774, 840, 888, 657, 665, 874, 728, 832, 912, 722, 730, 820, 760, 946
Offset: 1

Views

Author

David Wasserman, Jun 10 2002

Keywords

Examples

			a(1) = 73 because the equilateral triangle with side length A061281(1) = 112 has an interior point whose distances from the vertices are 73, 65 and 57.

Crossrefs

Cf. A061281, A072053, A072054.

Extensions

More terms from Jinyuan Wang, Jul 20 2020

A072053 Median of 3 integer distances referred to in A061281.

Original entry on oeis.org

65, 88, 147, 130, 185, 168, 176, 221, 285, 195, 294, 312, 264, 260, 343, 377, 370, 441, 315, 325, 336, 352, 343, 408, 392, 442, 570, 390, 464, 440, 588, 455, 624, 555, 504, 528, 520, 735, 589, 686, 663, 855, 585, 616, 754, 725, 740, 882, 630, 650, 672, 704, 686
Offset: 1

Views

Author

David Wasserman, Jun 10 2002

Keywords

Examples

			a(2) = 88 because the equilateral triangle with side length A061281(2) = 147 has an interior point whose distances from the vertices are 95, 88 and 73.

Crossrefs

Cf. A061281, A072052, A072054.

Extensions

More terms from Jinyuan Wang, Jul 20 2020

A072054 Smallest of 3 integer distances referred to in A061281.

Original entry on oeis.org

57, 73, 43, 114, 97, 127, 146, 111, 49, 171, 86, 95, 219, 228, 152, 147, 194, 129, 296, 285, 254, 292, 255, 247, 323, 222, 98, 342, 285, 365, 172, 399, 190, 291, 381, 438, 456, 215, 469, 304, 333, 147, 513, 511, 294, 403, 388, 258, 592, 570, 508, 584, 510, 285
Offset: 1

Views

Author

David Wasserman, Jun 10 2002

Keywords

Examples

			a(3) = 43 because the equilateral triangle with side length A061281(3) = 185 has an interior point whose distances from the vertices are 152, 147 and 43.

Crossrefs

Cf. A061281, A072052, A072053.

Extensions

More terms from Jinyuan Wang, Jul 20 2020

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

Views

Author

Bernard Schott, Jul 17 2020

Keywords

Comments

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).

Examples

			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.

References

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

Links

Crossrefs

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.

Formula

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.

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

Views

Author

Bernard Schott, Jul 18 2020

Keywords

Comments

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.

Examples

			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.

References

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

Links

Crossrefs

Cf. A336328 (triples), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), A351477.
Cf. A061281 (supersequence with non-primitive terms).

Programs

  • PARI
    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

Formula

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.

Extensions

More terms from Jinyuan Wang, Jul 20 2020

A088977 Side of primitive equilateral triangle with prime cevian p=A002476(n) cutting an edge into two integral parts.

Original entry on oeis.org

8, 15, 21, 35, 40, 48, 65, 77, 80, 91, 112, 117, 119, 133, 160, 168, 171, 187, 207, 209, 221, 224, 253, 255, 264, 280, 312, 323, 325, 341, 352, 377, 391, 403, 408, 425, 435, 440, 455, 465, 483, 504, 525, 527, 560, 576, 595, 609, 624, 645, 651, 665, 667, 703
Offset: 1

Views

Author

Lekraj Beedassy, Oct 31 2003

Keywords

Comments

The edge a(n) is partitioned into q=s^2 - t^2=A088243(n)*A088296(n) and r=t(2s+t)=A088242(n)*A088299(n) by a cevian of length p. [Alternatively, (p,q,r) form a triangle with angle 2pi/3 opposite side p.] The quadruple {p,q,r,a(n)=q+r} satisfies the triangle relation: see A061281, or the simpler relation a(n)^2 = p^2 + q*r.

Links

Crossrefs

Cf. A002476, A088241, A088242, A088243, A088296, A088298, A088299.

Programs

  • Mathematica
    sol[p_] := Solve[0 < t < s && s^2 + s t + t^2 == p, {s, t}, Integers];
Union[Reap[For[n = 1, n <= 10000, n++, If[PrimeQ[p = 6n + 1], an = s(s + 2t) /. sol[p][[1]]]; Sow[an]]][[2, 1]]] (* Jean-François Alcover, Mar 06 2020 *)

Formula

a(n) = A088241(n)*A088298(n) = s(s+2t), where s^2 + st + t^2, with s>t, form the primes p = 1 (mod 6) = A002476(n).

Extensions

More terms from Ray Chandler, Nov 01 2003
Showing 1-6 of 6 results.

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