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A360427
Values of the argument at successive record minima of the function R defined as follows. For any integer x >= 1, let y > x be the smallest integer such that there exist integers x < c < d < y such that x^3 + y^3 = c^3 + d^3. Then R(x) = y/x.
1
1, 2, 8, 9, 10, 17, 30, 42, 51, 135, 156, 285, 792, 1634, 3751, 4026, 6192, 14934, 15768, 16147, 45121, 58230, 61389, 79876, 167757, 177560, 213652, 525537, 917324, 1050787, 2237052, 3954983, 4157802
OFFSET
1,2
COMMENTS
For a given integer x, the identity x^3 + (12x)^3 = (9x)^3 + (10x)^3 holds, so R(x) <= 12.
A quadruple x = 2*N^4 - 4*N^3 + 9*N^2 - 8*N +10, y = 2*N^4 + 6*N^2 + N + 9, c = 2*N^4 - 3*N^3 + 12*N^2 - 5*N + 12, d = 2*N^4 - N^3 + 6*N^2 + N + 1 (for integer N) shows that the sequence is infinite.
EXAMPLE
For x = 1, y = 12, 1^3 + 12^3 = 9^3 + 10^3, R(1) = 12. So, a(1) = 1.
For x = 2, y = 16, 2^3 + 16^3 = 9^3 + 15^3, R(2) = 8. So, a(2) = 2.
For x = 3, y = 36, 3^3 + 36^3 = 27^3 + 30^3, R(3) = 12. So, this does not provide a record minimum. The same negative outcome happens for x = 4, x = 5, x = 6, x = 7.
For x = 8, y = 53, 8^3 + 53^3 = 29^3 + 50^3, R(8) = 6.625. So, a(4) = 8.
For n = 8, a(8) = 42, since 42^3 + 69^3 = 56^3 + 61^3, and the ratio R(42) = 69/42 = 1.6428571... is an absolute minimum (eighth successive) for the function R(x) for 1 <= x <= 42.
PROG
(Python)
xm, ym, x, n = 0, 1, 0, 1
while True:
x, y = x+1, x+4
while y*xm < ym*x:
c, d, s = x+1, y-1, x**3+y**3
while c<d:
t=c**3+d**3
if t<s:
c+=1
elif t>s:
d-=1
else:
break
if t==s:
print("a({})={} x={} c={} d={} y={}".format(n, x, x, c, d, y))
xm, ym, n = x, y, n+1
break
y+=1
# Bert Dobbelaere, Mar 18 2023
CROSSREFS
Sequence in context: A283774 A037456 A277857 * A369204 A237280 A282636
KEYWORD
nonn,more
AUTHOR
Giedrius Alkauskas, Feb 07 2023
EXTENSIONS
a(25)-a(33) from Bert Dobbelaere, Mar 18 2023
STATUS
approved