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.

A088514 Smaller section of edge A089025(n) by integral cevian A088513(n).

Original entry on oeis.org

3, 7, 5, 11, 7, 13, 16, 9, 32, 17, 40, 11, 19, 55, 40, 24, 13, 23, 65, 69, 25, 56, 15, 75, 32, 104, 56, 29, 17, 87, 85, 31, 119, 72, 93, 64, 144, 95, 40, 35, 133, 21, 136, 105, 37, 105, 111, 88, 23, 185, 152, 176, 80, 41, 115
Offset: 1

Views

Author

Lekraj Beedassy, Nov 14 2003

Keywords

Comments

Obviously, larger section is A089025(n) - a(n).

Programs

  • Mathematica
    findPrimIntEquiSmallSection[maxC_] :=
Reap[Do[Do[
     With[{cevian = Abs[c E^((2 \[Pi] I)/6) - a]},
      If[FractionalPart[cevian] == 0 && GCD[a, c] == 1,
       Sow[a]; Break[]]], {a, Floor[c/2],
      1, -1}], {c, maxC}]][[2, 1]]
(* Andrew Turner, Aug 04 2017 *)

Formula

Larger section is A089025(n) - a(n) = A088586(n). a(n) = (A089025(n) - A088587(n))/2. - Lekraj Beedassy, Nov 25 2003

Extensions

a(7) changed from 18 to 16 by W. D. Smith, Mar 04 2012 (N. J. A. Sloane, Dec 05 2012)
a(42) changed from 105 to 21, and a(52)=176 inserted by Andrew Turner, Aug 04 2017

A088586 Larger section of edge A089025(n) by integral cevian A088513(n).

Original entry on oeis.org

5, 8, 16, 24, 33, 35, 39, 56, 45, 63, 51, 85, 80, 57, 77, 95, 120, 120, 88, 91, 143, 115, 161, 112, 175, 105, 165, 195, 208, 160, 168, 224, 145, 203, 187, 221, 155, 217, 279, 288, 192, 320, 209, 247, 323, 272, 280, 315, 385, 231, 273, 259, 357, 399, 333
Offset: 1

Views

Author

Lekraj Beedassy, Nov 20 2003

Keywords

Programs

  • Mathematica
    findPrimIntEquiLargeSection[maxC_] :=
Reap[Do[Do[
     With[{cevian = Abs[c E^((2 \[Pi] I)/6) - a]},
      If[FractionalPart[cevian] == 0 && GCD[a, c] == 1,
       Sow[c - a]; Break[]]], {a, Floor[c/2],
      1, -1}], {c, maxC}]][[2, 1]]
(* Andrew Turner, Aug 04 2017 *)

Formula

a(n) = (A089025(n) + A088587(n))/2. - Lekraj Beedassy, Nov 25 2003

Extensions

Repeated 112 at a(24) and a(25) removed, and order of a(35)=187 and a(36)=221 reversed by Andrew Turner, Aug 04 2017

A088587 Difference between the two segments of edge d = A089025(n) partitioned by cevian c = A088513(n).

Original entry on oeis.org

2, 1, 11, 13, 26, 22, 23, 47, 13, 46, 11, 74, 61, 2, 37, 71, 107, 97, 23, 22, 118, 59, 146, 37, 143, 1, 109, 166, 191, 73, 83, 193, 26, 131, 94, 157, 11, 122, 239, 253, 59, 299, 73, 142, 286, 167, 169, 227, 362, 46, 121, 83, 277, 358, 218
Offset: 1

Views

Author

Lekraj Beedassy, Nov 20 2003

Keywords

Comments

a(n) is minimal, i.e., equals 1 for values c = A001570(n), d = A028230(n).

Crossrefs

Cf. A088514.

Programs

  • Mathematica
    findPrimIntEquiSectionDiff[maxC_] :=
Reap[Do[Do[
     With[{cevian = Abs[c E^((2 \[Pi] I)/6) - a]},
      If[FractionalPart[cevian] == 0 && GCD[a, c] == 1,
       Sow[c - 2 a]; Break[]]], {a, Floor[c/2],
      1, -1}], {c, maxC}]][[2, 1]]
(* Andrew Turner, Aug 04 2017 *)

Formula

a(n) = sqrt(4c^2 - 3d^2).

Extensions

a(35)=94 and a(36)=157 order reversed by Andrew Turner, Aug 04 2017

A089025 Side of primitive equilateral triangle bearing at least one integral cevian that partitions an edge into two integral sections.

Original entry on oeis.org

8, 15, 21, 35, 40, 48, 55, 65, 77, 80, 91, 96, 99, 112, 117, 119, 133, 143, 153, 160, 168, 171, 176, 187, 207, 209, 221, 224, 225, 247, 253, 255, 264, 275, 280, 285, 299, 312, 319, 323, 325, 341, 345, 352, 360, 377, 391, 403, 408, 416, 425, 435, 437, 440, 448
Offset: 1

Views

Author

Lekraj Beedassy, Nov 12 2003

Keywords

Comments

The cevians are numbers divisible only by primes of form 6n+1:A002476 (i.e., correspond to entries of A004611).
Composite cevians c belong to more than one equilateral triangle, actually to 2^(omega(c)-1) of them, where omega(n)=A001221(n). For instance, cevian 1813=7^2*37, with omega(1813)=2, belongs to 2^(2-1)=2 equilateral triangles, their sides being 1927=255+1627 and 1960=343+1617, while cevian 1729=7*13*19, with omega(1729)=3, belongs to 2^(3-1)=4 equilateral triangles whose sides are 1775=96+1679, 1824=209+1615, 1840=249+1591, 1859=299+1560.
Given a triangle with integer side lengths a, b, c relatively prime with a < b, c < b, and angle opposite c of 60 degrees then a*a - a*b + b*b = c*c from law of cosines and called a primitive Eisenstein triple by Gordon. This sequence is the possible side lengths of b. - Michael Somos, Apr 11 2012

Examples

			The equilateral triangle with side 280, for instance, has cevian 247 partitioning an edge into 93+187, as well as cevian 271 that sections the edge into 19+261.

Links

Crossrefs

Cf. A088513, A088514, A088977.

Programs

  • Mathematica
    findPrimIntEquiSide[maxC_] :=
Reap[Do[Do[
     With[{cevian = Abs[c E^((2 \[Pi] I)/6) - a]},
      If[FractionalPart[cevian] == 0 && GCD[a, c] == 1,
       Sow[c]; Break[]]], {a, Floor[c/2],
      1, -1}], {c, maxC}]][[2, 1]]
(* Andrew Turner, Aug 04 2017 *)

A357277 Largest side c of primitive triples, in nondecreasing order, for integer-sided triangles with angles A < B < C = 2*Pi/3 = 120 degrees.

Original entry on oeis.org

7, 13, 19, 31, 37, 43, 49, 61, 67, 73, 79, 91, 91, 97, 103, 109, 127, 133, 133, 139, 151, 157, 163, 169, 181, 193, 199, 211, 217, 217, 223, 229, 241, 247, 247, 259, 259, 271, 277, 283, 301, 301, 307, 313, 331, 337, 343, 349, 361, 367, 373, 379, 397, 403, 403, 409, 421, 427, 427, 433, 439, 457
Offset: 1

Views

Author

Bernard Schott, Oct 01 2022

Keywords

Comments

For the corresponding primitive triples and miscellaneous properties and references, see A357274.
Solutions c of the Diophantine equation c^2 = a^2 + a*b + b^2 with gcd(a,b) = 1 and a < b.
Also, side c can be generated with integers u, v such that gcd(u,v) = 1 and 0 < v < u, c = u^2 + u*v + v^2.
Some properties:
-> Terms are primes of the form 6k+1, or products of primes of the form 6k+1.
-> The lengths c are in A004611 \ {1} without repetition, in increasing order.
-> Every term appears 2^(k-1) (k>=1) times consecutively.
-> The smallest term that appears 2^(k-1) times is precisely A121940(k): see examples.
-> The terms that appear only once in this sequence are in A133290.
-> The terms are the same as in A335895 but frequency is not the same: when a term appears m times consecutively here, it appears 2m times consecutively in A335895. This is because if (a,b,c) is a primitive 120-triple, then both (a,a+b,c) and (a+b,b,c) are 60-triples in A335893 (see Emrys Read link, lemma 2 p. 302).
Differs from A088513, the first 20 terms are the same then a(21) = 151 while A088513(21) = 157.
A050931 gives all the possible values of the largest side c, in increasing order without repetition, for all triangles with an angle of 120 degrees, but not necessarily primitive.

Examples

			c = 7 appears once because A121940(1) = 7 with triple (3,5,7) and 7^2 = 3^2 + 3*5 + 5^2.
c = 91 is the smallest term to appear twice because A121940(2) = 91 with primitive 120-triples (11, 85, 91) and (19, 80, 91).
c = 1729 is the smallest term to appear four times because A121940(3) = 1729 with triples (96, 1679, 1729), (249, 1591, 1729), (656, 1305, 1729), (799, 1185, 1729).

Links

Crossrefs

Cf. A357274 (triples), A357275(smallest side), A357276 (middle side), A357278 (perimeter).
Cf. Also A004611, A050931, A088513, A121940, A133290, A335895.

Programs

  • Maple
    for c from 5 to 500 by 2 do
for a from 3 to c-2 do
b := (-a + sqrt(4*c^2-3*a^2))/2;
if b=floor(b) and gcd(a, b)=1 and a

Formula

a(n) = A357274(n, 3).

A291420 Numbers n such that there exist exactly four distinct Pythagorean triangles, at least one of them primitive, with area n.

Original entry on oeis.org

341880, 8168160, 14636160, 17957940, 52492440, 116396280, 1071572040, 1187525640, 1728483120, 5988702720, 6609482880, 22539095040, 29239970760, 136496680320, 258670630680, 398648544840, 494892478080, 592003418160, 1329673884000, 1343798407560, 2190884461920
Offset: 1

Views

Author

Sture Sjöstedt, Aug 23 2017

Keywords

Comments

Numbers n such that there exist positive integers x, y with x > y and n = x*y*(x-y)*(x+y).
Many of them consist of a Pythagorean triangle plus a triple which is a solution to Carroll's problem: Find three Pythagorean triangles with the same area.

Examples

			p^2 - p*q + q^2 = r^2;
p = 208, q = 418, r = 362, q - p = 210;
n = p*r*q*(q-p) = 208*418*362*210 = 6609482880.
x = 640, y = 627 gives the same area:
n = x*y*(x-y)*(x+y) = 640*627*13*1267 = 6609482880.

Crossrefs

Cf. A009127, A024407, A055193, A088513, A088977, A089025, A177021, A291591.

Extensions

a(12)-a(21) from Giovanni Resta, Aug 28 2017
Showing 1-6 of 6 results.

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