A268509 Numbers x such that x^3 = y^2 + z for some y and some nonzero z with -x < z < x.

2, 3, 5, 13, 15, 17, 32, 35, 37, 40, 43, 46, 52, 56, 63, 65, 99, 101, 109, 136, 143, 145, 152, 158, 175, 190, 195, 197, 243, 255, 257, 312, 317, 323, 325, 331, 336, 351, 356, 366, 377, 399, 401, 422, 483, 485, 560, 568, 575, 577, 584, 592, 654, 675, 677, 717, 741, 783, 785, 799, 810, 891, 899, 901, 909, 937, 944, 978

Offset

1,1

Comments

List of x such as x^3 is a near square (see examples).

Note that z = 17 appears often (see A029728).

Links

Daniel Mondot, Table of n, a(n) for n = 1..10000

Example

2^3 = 3^2 - 1;
3^3 = 5^2 + 2;
5^3 = 11^2 + 4;
13^3 = 47^2 - 12;
15^3 = 58^2 + 11;
17^3 = 70^2 + 13;
32^3 = 181^2 + 7;
35^3 = 207^2 + 26;
37^3 = 225^2 + 28;
40^3 = 253^2 - 9;
43^3 = 282^2 - 17;
46^3 = 312^2 - 8;
52^3 = 375^2 - 17;
56^3 = 419^2 + 55;
63^3 = 500^2 + 47;
65^3 = 524^2 + 49;
99^3 = 985^2 + 74.

Prog

(C)
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MAX2 10000
/* list number x and y such that x^3 = y^2 ± delta (0 < delta < x) */
/* this generates A268509 and A268510 */
long long unsigned b, c, d;
long long signed ds;
unsigned long long list2[MAX2];
unsigned long long list3[MAX2];
long double b1, cd, dd;
void main(unsigned argc, char *argv[])
{
unsigned a, i;
  i=0;
  // I never actually calculate b^3 or c^2, but only b^(3/2) = c + ds
  // this allows me to indirectly check b^3 past 2^64
  for (b=0; b<100000000; ++b) // could go up to b<4294967295u; max
  {
    b1 = sqrtl(b);
    cd= b1 *(long double)b;
    c=(long long unsigned)(cd+(double)0.5);
    dd = 2 * c * (cd - c);
    if (dd<0) ds = (dd - 0.5);
    else ds = (dd + 0.5);
    d = llabs(ds);
    if (d<b) // d = abs(b^3 - c^2)
    {
      if (ds)
      {
        if (i<MAX2)
        {
          list2[i]= b;
          list3[i++]= c;
        }
      }
    }
  }
  // generate A268509 */
  for (a=0; a<i; ++a) printf("%u %llu\n", a+1, list2[a]);
  printf("\n\n");
  // generate A268510 */
  for (a=0; a<i; ++a) printf("%u %llu\n", a+1, list3[a]);
  printf("\n\n");
}
(PARI) is(n)=my(t=abs(n^3-round(n^1.5)^2)); 0<t && t<n \\ Charles R Greathouse IV, Feb 09 2016

Crossrefs

Cf. A029728, A253181, A268510.

Sequence in context: A251414 A145343 A058592 * A260594 A228940 A045311

Adjacent sequences: A268506 A268507 A268508 * A268510 A268511 A268512

Keyword

nonn

Author

Daniel Mondot, Feb 06 2016

Status

approved