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.

A335896 Largest side of integer-sided primitive triangles whose angles A < B < C are in arithmetic order.

Original entry on oeis.org

8, 8, 15, 15, 21, 21, 35, 35, 40, 40, 48, 48, 55, 55, 65, 65, 77, 77, 80, 80, 91, 91, 96, 96, 99, 99, 112, 112, 117, 117, 119, 119, 133, 133, 143, 143, 153, 153, 160, 160, 171, 171, 168, 168, 187, 187, 176, 176, 209, 209, 207, 207, 221, 221, 224, 224, 225, 225
Offset: 1

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Author

Bernard Schott, Jul 10 2020

Keywords

Comments

The triples of sides (a,b,c) with a < b < c are in nondecreasing order of middle side, and if middle sides coincide then by increasing order of the largest side. This sequence lists the c's.
Equivalently, lengths of the largest side c of primitive non-equilateral triangles that have an angle of Pi/3; indeed, this side is opposite to the largest angle C.
Also, solutions c of the Diophantine equation b^2 = a^2 - a*c + c^2 with gcd(a,b) = 1 and a < b.
For the corresponding primitive triples and miscellaneous properties and references, see A335893.
When (a, b, c) is a triple, then (c-a, b, c) is another triple, so every c in the data is twice consecutively present according to the corresponding pair (b, c) (see examples).
As B = Pi/3 and C runs from Pi/3 to 2*Pi/3, sin(C) gets a maximum when C = Pi/2 with sin(C) = 1, hence, from law of sinus, b/sin(B) = c/sin(C), c < b/sin(Pi/3) = b * 2/sqrt(3) < 6*b/5. This bound is used in PARI and Maple programs.
This sequence is not increasing. For example, a(41) = a(42) = 171 for triangle with middle side = 151 while a(43) = a(44) = 168 for triangle with middle side = 157.

Examples

			c = 8 appears twice because:
  7^2 = 3^2 - 3*8 + 8^2, with triple (3, 7, 8),
  7^2 = 5^2 - 5*8 + 8^2, with triple (5, 7, 8).
c = 96 and c = 99 each appear twice associated with b = 91 because:
  91^2 = 11^2 - 11*96 + 96^2, with triple (11, 91, 96),
  91^2 = 85^2 - 85*96 + 96^2, with triple (85, 91, 96),
  91^2 = 19^2 - 19*99 + 99^2, with triple (19, 91, 99),
  91^2 = 80^2 - 80*99 + 99^2, with triple (80, 91, 99).
		

References

  • V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-298 p. 124, André Desvigne.

Crossrefs

Cf. A089025 (terms in increasing order without repetition).
Cf. A335893 (triples), A335894 (smallest side), A335895 (middle side), this sequence (largest side), A335897 (perimeter).

Programs

  • Maple
    for b from 3 to 250 by 2 do
    for c from b+1 to 6*b/5 do
    a := (c - sqrt(4*b^2-3*c^2))/2;
    if gcd(a,b)=1 and issqr(4*b^2-3*c^2) then print(c,c); end if;
    end do;
    end do;
  • PARI
    lista(nn) = {forstep(b=1, nn, 2, for(c=b+1, 6*b\5, if (issquare(d=4*b^2 - 3*c^2), my(a = (c - sqrtint(d))/2); if ((denominator(a)==1) && (gcd(a, b) == 1), print(c, ", ", c, ", ");););););} \\ Michel Marcus, Jul 15 2020

Formula

a(n) = A335893(n, 3).
c satisfies c^2 - a*c + a^2 - b^2 = 0 with gcd(a,b) = 1 and a < b.