A233400 Number n such that a2 - n^3 is a triangular number (A000217), where a2 is the least square above n^3.

0, 1, 2, 9, 12, 107, 109, 120, 244, 337, 381, 407, 565, 592, 937, 1209, 1224, 1341, 1717, 2032, 2402, 3280, 4957, 5149, 5265, 5644, 7065, 7240, 8181, 8820, 9712, 10732, 11901, 15059, 18300, 19120, 20436, 22672, 24516, 25139, 28044, 28550, 36145, 38221, 66201, 72335, 77100

Offset: 1

Author

Alex Ratushnyak, Dec 09 2013

Keywords

• nonn

Comments

The sequence of cubes begins: 0, 1, 8, 729, 1728, 1225043, 1295029, 1728000, 14526784, 38272753, 55306341, ...

The sequence of squares begins: 1, 4, 9, 784, 1764, 1225449, 1295044, 1729225, 14531344, 38278969, 55308969, ...

The sequence of roots of these squares begins: 1, 2, 3, 28, 42, 1107, 1138, 1315, 3812, 6187, 7437, 8211, 13430, 14404, 28682, ...

Links

• Robert Israel, Table of n, a(n) for n = 1..500

Crossrefs

Cf. A000217, A000290, A000578, A233401.

Programs

Maple

istri:= proc(n) issqr(1+8*n) end proc:
filter:= proc(n) local a2, t;
  a2:= (floor(sqrt(n^3))+1)^2;
  istri(a2-n^3)
end proc:
select(filter, [$0..10^5]); # Robert Israel, Sep 10 2024

Python

def isqrt(a):
    sr = 1 << (int.bit_length(int(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 isTriangular(a):
    a+=a
    sr = isqrt(a)
    return (a==sr*(sr+1))
for n in range(77777):
    n3 = n*n*n
    a = isqrt(n3)+1
    if isTriangular(a*a-n3):  print(str(n), end=', ')