A224515 a(n) = least k such that sqrt(k^2 XOR (k+1)^2) = 2*n+1, a(n) = -1 if there is no such k.

0, 4, 3, 24, 23, 44, 43, 112, 111, 180, 76, 264, 248, 348, 164, 480, 479, 411, 611, 327, 183, 115, 139, 943, 1103, 747, 787, 1111, 1447, 323, 699, 1984, 1983, 1851, 2243, 2008, 1576, 1388, 1684, 1072, 976, 1268, 499, 3383, 3271, 4124, 4068, 3679, 4511, 4315, 3804, 4999

Offset: 0

Author

Alex Ratushnyak, Apr 08 2013

Keywords

• nonn
• base
• less

Comments

Conjectures:

1. a(n) >= 0.

2. Least k is also the only such k.

If both conjectures are true, then the sequence is a permutation of A221643.

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A221643.

Programs

Mathematica

a[n_] := For[k=0, k <= 3*n^2+1, k++, If[ Sqrt[ BitXor[k^2, (k+1)^2]] == 2*n+1, Return[k]]] /. Null -> -1; a /@ Range[0, 51] (* Jean-François Alcover, Jun 05 2013 *)

PARI

a(n)=my(k=sqrtint(2*n^2),t);while(!issquare(bitxor(k^2,(k+1)^2),&t)||t!=2*n+1,k++);k \\ Charles R Greathouse IV, Jun 05 2013

Python

import math
needTerms = n = 1024
i = 0
terms = [-1] * n
while n:
  s = (i*i) ^ ((i+1)*(i+1))
  r = int(math.sqrt(s))
  if s == r*r:
    if (r&1)==0:  break
    r = (r-1)//2
    if r < needTerms:
        if terms[r] >= 0:  break
        terms[r] = i
        n -= 1
  i += 1
if n:  print('Error')
else:
  for i in range(needTerms):
    t = terms[i]
    print(t, end=', ') # math.sqrt((t*t) ^ ((t+1)*(t+1)))