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A339856 Primitive triples for integer-sided triangles whose sides a < b < c form a geometric progression. 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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° if 1 < q < sqrt(phi) and angle C > 90° 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.

LINKS

David A. Corneth, Table of n, a(n) for n = 1..10002

Project Euler, Problem 370: Geometric Triangles.

EXAMPLE

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;

  ...

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;

PROG

(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

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.

Sequence in context: A201660 A341577 A094115 * A163297 A051679 A010378

Adjacent sequences:  A339853 A339854 A339855 * A339857 A339858 A339859

KEYWORD

nonn,easy,tabf

AUTHOR

Bernard Schott, Dec 19 2020

EXTENSIONS

Data corrected by David A. Corneth, Dec 25 2020

STATUS

approved

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Last modified July 27 11:04 EDT 2021. Contains 346304 sequences. (Running on oeis4.)