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A383109
Consider the isosceles triangle whose vertices are the Gaussian integers z1 = 0, z2 = x+i*y, z3 = x-i*y. The sequence lists the pairs of positive integer foci (f_i, f_j), f_i < f_j of Steiner inellipse for some z2, z3.
0
3, 5, 6, 10, 9, 15, 13, 15, 12, 20, 15, 25, 18, 30, 15, 37, 15, 41, 17, 39, 21, 35, 26, 30, 24, 40, 27, 45, 25, 51, 30, 50, 39, 45, 33, 55, 36, 60, 29, 75, 30, 74, 39, 65, 51, 53, 30, 82, 34, 78, 42, 70, 52, 60, 45, 75, 39, 85, 48, 80, 51, 85, 65, 75, 54, 90, 61
OFFSET
1,1
COMMENTS
With z1 = 0, z2 = x+i*y, z3 = x-i*y and P(z) = (z-z1)*(z-z2)*(z-z3) we obtain:
(1) P(z) = z^3 - 2*x*z^2 + (x^2+y^2)*z.
(2) P'(z) = 3*z^2 - 4*x*z + x^2+y^2.
The two zeros of P'(z) are: (2*x-sqrt(x^2-3*y^2))/3 and (2*x+sqrt(x^2-3*y^2))/3.
REFERENCES
Beniamin Bogosel, A Geometric Proof of the Siebeck-Marden Theorem, Amer. Math. Monthly, vol. 125, no 4, 2017, p. 459-463.
A. Eydelzon, On a New Property of the Steiner Inellipse, Amer. Math. Monthly, vol. 127, no 10, 2020, p. 933-935.
EXAMPLE
(a(1),a(2)) = (3,5) because from (2) with (x,y) = (6,3), the two zeros of P'(z) are: (2*6-sqrt(6^2-3*3^2))/3 = (12 - sqrt(9))/3 = 3 and (2*6+sqrt(6^2-3*3^2))/3 = (12 + sqrt(9))/3 = 5. The two foci are integers.
(a(7),a(8)) = (13,15) because from (2) with (x,y) = (21,12), the two zeros of P'(z) are (2*21-sqrt(21^2-3*12^2))/3 = (42 - sqrt(9))/3 = 13 and (2*21+sqrt(21^2-3*12^2))/3 = (12 + sqrt(9))/3 = 15. The two foci are integers.
MAPLE
nn:=200:
for x from 1 to nn do:
for y from 1 to nn while(x^2>3*y^2) do:
u:=sqrt(x^2+y^2):v:=2*b:s:=sqrt(x^2+y^2)+y:
z1:=(2*x-sqrt(x^2-3*y^2))/3:z2:=(2*x+sqrt(x^2-3*y^2))/3:
if z1=floor(z1) and z2=floor(z2) then printf(`%d, `, z1):
printf(`%d, `, z2):
else fi:
od:
od:
CROSSREFS
Sequence in context: A248644 A242197 A283051 * A376377 A244671 A391349
KEYWORD
nonn
AUTHOR
Michel Lagneau, Apr 16 2025
STATUS
approved