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 1-7 of 7 results.

A336750 Triples for integer-sided triangles whose sides a < b < c are in arithmetic progression.

Original entry on oeis.org

2, 3, 4, 3, 4, 5, 3, 5, 7, 4, 5, 6, 4, 6, 8, 5, 6, 7, 4, 7, 10, 5, 7, 9, 6, 7, 8, 5, 8, 11, 6, 8, 10, 7, 8, 9, 5, 9, 13, 6, 9, 12, 7, 9, 11, 8, 9, 10, 6, 10, 14, 7, 10, 13, 8, 10, 12, 9, 10, 11, 6, 11, 16, 7, 11, 15, 8, 11, 14, 9, 11, 13, 10, 11, 12, 7, 12, 17, 8, 12, 16
Offset: 1

Views

Author

Bernard Schott, Aug 03 2020

Keywords

Comments

The triples are displayed in increasing order of perimeter, and if perimeters coincide then by increasing order of the smallest side, hence, each triple (a, b, c) is in increasing order.
Equivalently: triples of integer-sided triangles such that b = (a+c)/2 with a < c.
As the perimeter of these triangles = 3*b, the triples are also displayed in increasing order of middle side.
When a < b < c are in arithmetic progression with b - a = c - b = x, then 1 <= x <= floor((b-1)/2), hence, there exist for each side b >= 3, floor((b-1)/2) = A004526(b) triangles whose sides a < b < c are in arithmetic progression.
The only right integer-sided triangles such that a < b < c are in arithmetic progression correspond to the Pythagorean triples (3k, 4k, 5k) with k > 0.
There do not exist triangles whose sides a < b < c and angles A < B < C are both in arithmetic progression.
Three geometrical properties about these triangles, even if they are not integer-sided:
1) tan(A/2) * tan(C/2) = 1/3,
2) r = h_b/3, where r is the inradius and h_b the length of the altitude through B,
3) The line (IG) is parallel to side (AC), where I is the incenter and G is the centroid of the triangle.

Examples

			The smallest such triangle is (2, 3, 4).
The only triangle with perimeter = 12 corresponds to the Pythagorean triple: (3, 4, 5).
There exist two triangles with perimeter = 15 corresponding to triples (3, 5, 7) and (4, 5, 6).
There exist also two triangles with perimeter = 18 corresponding to triples (4, 6, 8) and (5, 6, 7).
The table begins:
  2, 3, 4;
  3, 4, 5;
  3, 5, 7;
  4, 5, 6;
  4, 6, 8;
  5, 6, 7;
  4, 7, 10;
  5, 7, 9;
  6, 7, 8;
		

References

  • V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-288 p. 120, André Desvigne.

Crossrefs

Cf. A336751 (smallest side), A307136 (middle side), A336753 (largest side), A336754 (perimeter), A024164 (number of triangles with perimeter = n), A336755 (primitive triples), A336756 (perimeter of primitive triangles), A336757 (number of primitive triangles with perimeter = n).
Cf. A004526 (number of triangles with middle side = b).
Cf. A103605 (similar, with Pythagorean triples).
Cf. A335893 (similar, with A, B, C in arithmetic progression).

Programs

  • Maple
    for b from 3 to 20 do
    for a from b-floor((b-1)/2) to b-1 do
    c := 2*b - a;
    print(a,b,c);
    end do;
    end do;
  • Mathematica
    Block[{nn = 12, a, b, c}, Reap[Do[Do[Sow@ {a, b, 2 b - a}, {a, b - Floor[(b - 1)/2], b - 1}], {b, 3, nn}]][[-1, 1]] ] // Flatten (* Michael De Vlieger, Oct 15 2020 *)
  • PARI
    tabf(nn) = {for (b = 3, nn, for (a = b-floor((b-1)/2), b-1, my(c = 2*b - a); print(a, " ", b, " ", c);););} \\ Michel Marcus, Sep 08 2020

Formula

T(n,1) = A336751(n); T(n,2) = A307136(n); T(n,3) = A336753(n).
A336754(n) = T(n,1) + T(n,2) + T(n,3).

A336754 Perimeters in increasing order of integer-sided triangles whose sides a < b < c are in arithmetic progression.

Original entry on oeis.org

9, 12, 15, 15, 18, 18, 21, 21, 21, 24, 24, 24, 27, 27, 27, 27, 30, 30, 30, 30, 33, 33, 33, 33, 33, 36, 36, 36, 36, 36, 39, 39, 39, 39, 39, 39, 42, 42, 42, 42, 42, 42, 45, 45, 45, 45, 45, 45, 45, 48, 48, 48, 48, 48, 48, 48, 51, 51, 51, 51, 51, 51, 51, 51
Offset: 1

Views

Author

Bernard Schott, Aug 31 2020

Keywords

Comments

Equivalently: perimeters of integer-sided triangles such that b = (a+c)/2 with a < c.
As perimeter = 3 * middle side, these perimeters p are all multiple of 3, and each term p appears floor((p-3)/6) = A004526((p-3)/3) consecutively.
For each perimeter = 12*k with k>0, there exists one right integer triangle whose triple is (3k, 4k, 5k).
For the corresponding primitive triples, miscellaneous properties and references, see A336750.

Examples

			Perimeter = 9 only for the smallest triangle (2, 3, 4).
Perimeter = 12 only for Pythagorean triple (3, 4, 5).
Perimeter = 15 for the two triples (3, 5, 7) and (4, 5, 6).
		

References

  • V. Lespinard and R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-290 p. 121, André Desvigne.

Crossrefs

Cf. A004526.
Cf. A336750 (triples), A336751 (smallest side), A307136 (middle side), A336753 (largest side), this sequence (perimeter), A024164 (number of such triangles whose perimeter = n), A336755 (primitive triples).
Cf. A335897 (perimeters when angles A, B and C are in arithmetic progression).

Programs

  • Maple
    for b from 3 to 30 do
    for a from b-floor((b-1)/2) to b-1 do
    c := 2*b - a;
    print(a+b+c);
    end do;
    end do;
  • Mathematica
    A336754[n_] := 3*Ceiling[2*Sqrt[n+Round[Sqrt[n]]]]; Array[A336754, 100] (* or *)
    Flatten[Array[ConstantArray[3*#, Floor[(#-1)/2]] &, 19, 3]] (* Paolo Xausa, Feb 29 2024 *)

Formula

a(n) = A336750(n, 1) + A336750(n, 2) + A336750(n, 3).
a(n) = 3 * A307136(n).

A024164 Number of integer-sided triangles with sides a,b,c, a

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 2, 0, 0, 3, 0, 0, 3, 0, 0, 4, 0, 0, 4, 0, 0, 5, 0, 0, 5, 0, 0, 6, 0, 0, 6, 0, 0, 7, 0, 0, 7, 0, 0, 8, 0, 0, 8, 0, 0, 9, 0, 0, 9, 0, 0, 10, 0, 0, 10, 0, 0, 11, 0, 0, 11, 0, 0, 12, 0, 0, 12, 0, 0, 13, 0, 0, 13, 0, 0, 14, 0, 0, 14, 0, 0, 15, 0, 0, 15, 0, 0, 16, 0, 0, 16
Offset: 1

Views

Author

Keywords

Comments

From Bernard Schott, Oct 10 2020: (Start)
Equivalently: number of integer-sided triangles whose sides a < b < c are in arithmetic progression with perimeter n.
Equivalently: number of integer-sided triangles such that b = (a+c)/2 with a < c and perimeter n.
All the perimeters are multiple of 3 because each perimeter = 3 * middle side b.
For each perimeter n = 12*k with k>0, there exists one and only one such right integer triangle whose triple is (3k, 4k, 5k).
For the corresponding primitive triples and miscellaneous properties and references, see A336750. (End)

Examples

			a(9) = 1 for the smallest such triangle (2, 3, 4).
a(12) = 1 for Pythagorean triple (3, 4, 5).
a(15) = 2 for the two triples (3, 5, 7) and (4, 5, 6).
		

Crossrefs

Cf. A336750 (triples), A336751 (smallest side), A307136 (middle side), A336753 (largest side), A336754 (perimeter), this sequence (number of triangles whose perimeter = n), A336755 (primitive triples), A336756 (primitive perimeters), A336757 (number of primitive triangles with perimeter = n).
Cf. A005044 (number of integer-sided triangles with perimeter = n).

Programs

  • Mathematica
    A024164[n_] := If[Mod[n, 3] == 0, Floor[(n - 3)/6], 0]; Array[A024164, 100] (* Wesley Ivan Hurt, Nov 01 2020 *)
    LinearRecurrence[{0,0,1,0,0,1,0,0,-1},{0,0,0,0,0,0,0,0,1},120] (* Harvey P. Dale, Jun 03 2021 *)

Formula

If n = 3*k, then a(n) = floor((n-3)/6) = A004526((n-3)/3), otherwise, a(3k+1) = a(3k+2) = 0. - Bernard Schott, Oct 10 2020
From Wesley Ivan Hurt, Nov 01 2020: (Start)
G.f.: x^9/((x^3 - 1)^2*(x^3 + 1)).
a(n) = a(n-3) + a(n-6) - a(n-9).
a(n) = (1 - ceiling(n/3) + floor(n/3)) * floor((n-3)/6). (End)
E.g.f.: (18 + (x - 6)*cosh(x) + (x - 3)*sinh(x) - exp(-x/2)*((9 + 3*exp(x) + x)*cos(sqrt(3)*x/2) + sqrt(3)*x*sin(sqrt(3)*x/2)))/18. - Stefano Spezia, Feb 29 2024

A336756 Perimeters in increasing order of primitive integer-sided triangles whose sides a < b < c are in arithmetic progression.

Original entry on oeis.org

9, 12, 15, 15, 18, 21, 21, 21, 24, 24, 27, 27, 27, 30, 30, 33, 33, 33, 33, 33, 36, 36, 39, 39, 39, 39, 39, 39, 42, 42, 42, 45, 45, 45, 45, 48, 48, 48, 48, 51, 51, 51, 51, 51, 51, 51, 51, 54, 54, 54, 57, 57, 57, 57, 57, 57, 57, 57, 57, 60, 60, 60, 60, 63, 63, 63, 63, 63, 63
Offset: 1

Views

Author

Bernard Schott, Sep 16 2020

Keywords

Comments

Equivalently: perimeters of primitive integer-sided triangles such that b = (a+c)/2 with a < c.
As perimeter = 3 * middle side, these perimeters p are all multiples of 3 and each term p appears consecutively A023022(p/3) = phi(p/3)/2 times for p >= 9.
Remark, when the middle side is prime, then the number of primitive triangles with a perimeter p = 3*b equals phi(p/3)/2 = (b-1)/2 = (p-3)/6 and in this case, all the triangles are primitive (see A336754).
For the corresponding primitive triples, miscellaneous properties, and references, see A336750.

Examples

			Perimeter = 9 only for the smallest triangle (2, 3, 4).
Perimeter = 12 only for the Pythagorean triple (3, 4, 5).
Perimeter = 15 for the two triples (3, 5, 7) and (4, 5, 6).
There only exists one primitive triangle with perimeter = 18 whose triple is (5, 6, 7), because (4, 6, 8) is not a primitive triple.
		

Crossrefs

Cf. A336754 (perimeters, primitive or not), A336755 (primitive triples), this sequence (perimeters of primitive triangles), A336757 (number of such primitive triangles whose perimeter = n).
Cf. A023022.

Programs

  • Maple
    for b from 3 to 21 do
    for a from b-floor((b-1)/2) to b -1 do
    c := 2*b - a;
    if gcd(a,b)=1 and gcd(b,c)=1 then print(a+b+c); end if;
    end do;
    end do;
  • Mathematica
    Flatten[Array[ConstantArray[3*#, EulerPhi[#]/2] &, 20, 3]] (* Paolo Xausa, Feb 29 2024 *)
  • PARI
    lista(nn) = {my(list=List()); for (b = 3, nn, for (a = b-floor((b-1)/2), b-1, my(c = 2*b - a); if (gcd([a, b, c]) == 1, listput(list, a+b+c);););); Vec(list);} \\ Michel Marcus, Sep 16 2020

A336757 Number of primitive integer-sided triangles whose sides a < b < c are in arithmetic progression with a perimeter = n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 3, 0, 0, 2, 0, 0, 3, 0, 0, 2, 0, 0, 5, 0, 0, 2, 0, 0, 6, 0, 0, 3, 0, 0, 4, 0, 0, 4, 0, 0, 8, 0, 0, 3, 0, 0, 4, 0, 0, 4, 0, 0, 6, 0, 0, 5, 0, 0, 11, 0, 0, 4
Offset: 1

Views

Author

Bernard Schott, Sep 20 2020

Keywords

Comments

Equivalently: number of primitive integer-sided triangles such that b = (a+c)/2 with a < c and perimeter = n.
As the perimeter of these triangles = 3*b where b is the middle side, a(n) >= 1 iff n = 3*b, with b >= 3.
When b is prime, all the triangles of perimeter n = 3*b are primitive, hence in this case: a(n) = A024164(n).
For the corresponding triples (primitive or not), miscellaneous properties and references, see A336750.

Examples

			a(9) = 1 for the smallest such triangle (2, 3, 4).
a(12) = 1 for the Pythagorean triple (3, 4, 5).
a(15) = 2 for the two triples (3, 5, 7) and (4, 5, 6).
a(18) = 1 for the triple (5, 6, 7); the other triple (4, 6, 8) corresponding to a perimeter = 18 is not a primitive triple.
		

Crossrefs

Cf. A336750 (triples, primitive or not), A336755 (primitive triples), A336756 (perimeters of primitive triangles).
Cf. A024164 (number of such triangles, primitive or not).
Similar sequences: A005044 (integer-sided triangles), A024155 (right triangles), A070201 (with integral inradius).

Formula

For n = 3*b, b >= 3, a(n) = A023022(b) = A000010(b)/2, otherwise a(n) = 0.

A339856 Primitive triples for integer-sided triangles whose sides a < b < c form a geometric progression.

Original entry on oeis.org

4, 6, 9, 9, 12, 16, 16, 20, 25, 25, 30, 36, 25, 35, 49, 25, 40, 64, 36, 42, 49, 49, 56, 64, 49, 63, 81, 49, 70, 100, 49, 77, 121, 64, 72, 81, 64, 88, 121, 81, 90, 100, 81, 99, 121, 81, 117, 169, 81, 126, 196, 100, 110, 121, 100, 130, 169, 121, 132, 144, 121, 143, 169
Offset: 1

Views

Author

Bernard Schott, Dec 19 2020

Keywords

Comments

These triangles are called "geometric triangles" in Project Euler problem 370 (see link).
The triples are displayed in increasing lexicographic order (a, b, c).
Equivalently: triples of integer-sided triangles such that b^2 = a*c with a < c and gcd(a, c) = 1.
When a < b < c are in geometric progression with b = a*q, c = b*q, q is the constant, then 1 < q < (1+sqrt(5))/2 = phi = A001622 = 1.6180... (this bound is used in Maple code).
For each triple (a, b, c), there exists (r, s), 0 < r < s such that a = r^2, b = r*s, c = s^2, q = s/r.
Angle C < 90 degrees if 1 < q < sqrt(phi) and angle C > 90 degrees if sqrt(phi) < q < phi with sqrt(phi) = A139339 = 1.2720...
For k >= 2, each triple (a, b, c) of the form (k^2, k*(k+1), (k+1)^2) is (A008133(3k+1), A008133(3k+2), A008133(3k+3)).
Three geometrical properties about these triangles:
1) The sinus satisfy sin^2(B) = sin(A) * sin(C) with sin(A) < sin(B) < sin(C) that form a geometric progression.
2) The heights satisfy h_b^2 = h_a * h_c with h_c < h_b < h_a that form a geometric progression.
3) b^2 = 2 * R * h_b, with R = circumradius of the triangle ABC.

Examples

			The smallest such triangle is (4, 6, 9) with 4*9 = 6^2.
There exist four triangles with small side = 49 corresponding to triples (49, 56, 64), (49, 63, 81), (49, 70, 100) and (49, 77, 121).
The table begins:
   4,  6,  9;
   9, 12, 16;
  16, 20, 25;
  25, 30, 36;
  25, 35, 49;
  25, 40, 64;
  36, 42, 49;
  ...
		

Crossrefs

Cf. A339857 (smallest side), A339858 (middle side), A339859 (largest side), A339860 (perimeter).
Cf. A336755 (similar for sides in arithmetic progression).
Cf. A335893 (similar for angles in arithmetic progression).
Cf. A001622 (phi), A139339 (sqrt(phi)), A008133.

Programs

  • Maple
    for a from 1 to 300 do
    for b from a+1 to floor((1+sqrt(5))/2 * a) do
    for c from b+1 to floor((1+sqrt(5))/2 * b) do
    k:=a*c;
    if k=b^2 and igcd(a,b,c)=1 then print(a,b,c); end if;
    end do;
    end do;
    end do;
  • PARI
    lista(nn) = {my(phi = (1+sqrt(5))/2); for (a=1, nn, for (b=a+1, floor(a*phi), for (c=b+1, floor(b*phi), if ((a*c == b^2) && (gcd([a,b,c])==1), print([a,b,c])););););} \\ Michel Marcus, Dec 25 2020
    
  • PARI
    upto(n) = my(res=List(), phi = (sqrt(5)+1) / 2); for(i = 2, sqrtint(n), for(j = i+1, (i*phi)\1, if(gcd(i, j)==1, listput(res, [i^2, i*j, j^2])))); concat(Vec(res)) \\ David A. Corneth, Dec 25 2020

Extensions

Data corrected by David A. Corneth, Dec 25 2020

A351178 Integral area of primitive integer-sided triangles whose sides a < b < c are in arithmetic progression.

Original entry on oeis.org

6, 84, 126, 156, 210, 456, 546, 570, 1116, 1170, 1176, 1554, 2046, 2220, 2394, 3096, 3216, 3294, 3354, 3924, 4740, 5124, 6006, 6180, 6510, 7326, 7446, 8760, 9030, 9264, 9906, 10374, 10920, 11466, 12684, 13104, 15210, 16170, 16296, 16716, 17556, 18060, 18090, 18354, 22134, 22860, 23550
Offset: 1

Views

Author

Bernard Schott, Feb 04 2022

Keywords

Comments

Middle side b is necessarily even, and the two other sides are odd, so all the areas are even numbers.
If b is the middle side with b even >= 4, if k odd = b-a = c-b with 1 <= k <= b/2 - 1, if gcd(b,k) = 1, then, we have area S = sqrt(3*b^2*(b^2-4*k^2))/4.

Examples

			a(1) = 6 corresponds to the Pythagorean triple (3, 4, 5), this is the unique right integer-sided triangle in this sequence.
a(2) = 84 for triple (13, 14, 15) (see MacNeill link).
a(3) = 126 for triple (15, 28, 41) (see Penguin reference, entry 126).
a(4) = 156 for triple (15, 26, 37) (see MacNeill link).
		

References

  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, Penguin Books, London, England, 1997, entry 126, page 122.

Crossrefs

Subsequence of A188158.

Programs

  • PARI
    lista(nn) = {my(list = List()); for (b = 3, nn, for (a = b-floor((b-1)/2), b-1, my(c = 2*b - a); if (gcd([a,b,c]) == 1, my(p = (a+b+c)/2); if (issquare(x=p*(p-a)*(p-b)*(p-c)), listput(list, sqrtint(x)));););); vecsort(Vec(list));} \\ Michel Marcus, Feb 05 2022

Extensions

Missing terms added by Michel Marcus, Feb 05 2022
Showing 1-7 of 7 results.