OFFSET
1,2
COMMENTS
Numbers n such that A070929(n) is a nonzero square.
The sequence of cubes a(n)^3 begins: 0, 8, 27, 216, 1728, 8000, 13824, 27000, 64000, 74088, 175616, 216000, 314432, ...
The sequence of m's begins: 1, 9, 36, 225, 1764, 8100, 13924, 27225, 64009, 74529, 176400, 216225, 314721, ...
The sequence of square roots of these m's begins: 1, 3, 6, 15, 42, 90, 118, 165, 253, 273, 420, 465, 561, 650, 689, 770, 812, ...
The sequence of squares m-n^3 begins: 1, 1, 9, 9, 36, 100, 100, 225, 9, 441, 784, 225, 289, 625, 169, 196, 841, ...
The sequence of their square roots begins: 1, 1, 3, 3, 6, 10, 10, 15, 3, 21, 28, 15, 17, 25, 13, 14, 29, 35, 43, 24, ... (note the first 12 terms are triangular numbers, A000217).
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
MATHEMATICA
fQ[n_]:=Module[{c=n^3, m}, m=(Floor[Sqrt[c]]+1)^2; IntegerQ[Sqrt[m-c]]]; Select[Range[0, 1650], fQ] (* Harvey P. Dale, Jan 03 2024 *)
PROG
(Python)
def isqrt(a):
sr = 1L << (long.bit_length(long(a)) >> 1)
while a < sr*sr: sr>>=1
b = sr>>1
while b:
s = sr+b
if a >= s*s: sr = s
b>>=1
return sr
def isSquare(a):
sr = isqrt(a)
return (a==sr*sr)
for n in range(77777):
n3 = n*n*n
a = isqrt(n3)+1
if isSquare(a*a-n3): print str(n)+', ',
(PARI) is(n)=issquare((sqrtint(n=n^3)+1)^2-n) \\ Charles R Greathouse IV, Dec 09 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Alex Ratushnyak, Dec 09 2013
STATUS
approved